标签归档:directory

Python中的目录树列表

问题:Python中的目录树列表

如何获取Python给定目录中所有文件(和目录)的列表?

How do I get a list of all files (and directories) in a given directory in Python?


回答 0

这是遍历目录树中每个文件和目录的一种方式:

import os

for dirname, dirnames, filenames in os.walk('.'):
    # print path to all subdirectories first.
    for subdirname in dirnames:
        print(os.path.join(dirname, subdirname))

    # print path to all filenames.
    for filename in filenames:
        print(os.path.join(dirname, filename))

    # Advanced usage:
    # editing the 'dirnames' list will stop os.walk() from recursing into there.
    if '.git' in dirnames:
        # don't go into any .git directories.
        dirnames.remove('.git')

This is a way to traverse every file and directory in a directory tree:

import os

for dirname, dirnames, filenames in os.walk('.'):
    # print path to all subdirectories first.
    for subdirname in dirnames:
        print(os.path.join(dirname, subdirname))

    # print path to all filenames.
    for filename in filenames:
        print(os.path.join(dirname, filename))

    # Advanced usage:
    # editing the 'dirnames' list will stop os.walk() from recursing into there.
    if '.git' in dirnames:
        # don't go into any .git directories.
        dirnames.remove('.git')

回答 1

您可以使用

os.listdir(path)

作为参考和更多的os函数,请看这里:

You can use

os.listdir(path)

For reference and more os functions look here:


回答 2

这是我经常使用的辅助函数:

import os

def listdir_fullpath(d):
    return [os.path.join(d, f) for f in os.listdir(d)]

Here’s a helper function I use quite often:

import os

def listdir_fullpath(d):
    return [os.path.join(d, f) for f in os.listdir(d)]

回答 3

import os

for filename in os.listdir("C:\\temp"):
    print  filename
import os

for filename in os.listdir("C:\\temp"):
    print  filename

回答 4

如果您需要遍历功能,那么还可以使用一个模块。例如:

import glob
glob.glob('./[0-9].*')

将返回类似:

['./1.gif', './2.txt']

请参阅此处的文档。

If you need globbing abilities, there’s a module for that as well. For example:

import glob
glob.glob('./[0-9].*')

will return something like:

['./1.gif', './2.txt']

See the documentation here.


回答 5

尝试这个:

import os
for top, dirs, files in os.walk('./'):
    for nm in files:       
        print os.path.join(top, nm)

Try this:

import os
for top, dirs, files in os.walk('./'):
    for nm in files:       
        print os.path.join(top, nm)

回答 6

对于当前工作目录中的文件,但未指定路径

Python 2.7:

import os
os.listdir(os.getcwd())

Python 3.x:

import os
os.listdir()

感谢Stam Kaly对python 3.x的评论

For files in current working directory without specifying a path

Python 2.7:

import os
os.listdir(os.getcwd())

Python 3.x:

import os
os.listdir()

Thanks to Stam Kaly for comment on python 3.x


回答 7

递归实现

import os

def scan_dir(dir):
    for name in os.listdir(dir):
        path = os.path.join(dir, name)
        if os.path.isfile(path):
            print path
        else:
            scan_dir(path)

A recursive implementation

import os

def scan_dir(dir):
    for name in os.listdir(dir):
        path = os.path.join(dir, name)
        if os.path.isfile(path):
            print path
        else:
            scan_dir(path)

回答 8

我写了一个很长的版本,其中包含了我可能需要的所有选项:http : //sam.nipl.net/code/python/find.py

我想它也适合这里:

#!/usr/bin/env python

import os
import sys

def ls(dir, hidden=False, relative=True):
    nodes = []
    for nm in os.listdir(dir):
        if not hidden and nm.startswith('.'):
            continue
        if not relative:
            nm = os.path.join(dir, nm)
        nodes.append(nm)
    nodes.sort()
    return nodes

def find(root, files=True, dirs=False, hidden=False, relative=True, topdown=True):
    root = os.path.join(root, '')  # add slash if not there
    for parent, ldirs, lfiles in os.walk(root, topdown=topdown):
        if relative:
            parent = parent[len(root):]
        if dirs and parent:
            yield os.path.join(parent, '')
        if not hidden:
            lfiles   = [nm for nm in lfiles if not nm.startswith('.')]
            ldirs[:] = [nm for nm in ldirs  if not nm.startswith('.')]  # in place
        if files:
            lfiles.sort()
            for nm in lfiles:
                nm = os.path.join(parent, nm)
                yield nm

def test(root):
    print "* directory listing, with hidden files:"
    print ls(root, hidden=True)
    print
    print "* recursive listing, with dirs, but no hidden files:"
    for f in find(root, dirs=True):
        print f
    print

if __name__ == "__main__":
    test(*sys.argv[1:])

I wrote a long version, with all the options I might need: http://sam.nipl.net/code/python/find.py

I guess it will fit here too:

#!/usr/bin/env python

import os
import sys

def ls(dir, hidden=False, relative=True):
    nodes = []
    for nm in os.listdir(dir):
        if not hidden and nm.startswith('.'):
            continue
        if not relative:
            nm = os.path.join(dir, nm)
        nodes.append(nm)
    nodes.sort()
    return nodes

def find(root, files=True, dirs=False, hidden=False, relative=True, topdown=True):
    root = os.path.join(root, '')  # add slash if not there
    for parent, ldirs, lfiles in os.walk(root, topdown=topdown):
        if relative:
            parent = parent[len(root):]
        if dirs and parent:
            yield os.path.join(parent, '')
        if not hidden:
            lfiles   = [nm for nm in lfiles if not nm.startswith('.')]
            ldirs[:] = [nm for nm in ldirs  if not nm.startswith('.')]  # in place
        if files:
            lfiles.sort()
            for nm in lfiles:
                nm = os.path.join(parent, nm)
                yield nm

def test(root):
    print "* directory listing, with hidden files:"
    print ls(root, hidden=True)
    print
    print "* recursive listing, with dirs, but no hidden files:"
    for f in find(root, dirs=True):
        print f
    print

if __name__ == "__main__":
    test(*sys.argv[1:])

回答 9

这是另一种选择。

os.scandir(path='.')

它返回os.DirEntry对象的迭代器,该对象与path所给目录中的条目(以及文件属性信息)相对应。

例:

with os.scandir(path) as it:
    for entry in it:
        if not entry.name.startswith('.'):
            print(entry.name)

使用scandir()而不是listdir()可以显着提高还需要文件类型或文件属性信息的代码的性能,因为如果操作系统在扫描目录时提供了os.DirEntry对象,则该信息会公开。所有的os.DirEntry方法都可以执行系统调用,但是is_dir()和is_file()通常只需要系统调用即可进行符号链接。os.DirEntry.stat()在Unix上始终需要系统调用,而在Windows上只需要一个系统调用即可。

Python文档

Here is another option.

os.scandir(path='.')

It returns an iterator of os.DirEntry objects corresponding to the entries (along with file attribute information) in the directory given by path.

Example:

with os.scandir(path) as it:
    for entry in it:
        if not entry.name.startswith('.'):
            print(entry.name)

Using scandir() instead of listdir() can significantly increase the performance of code that also needs file type or file attribute information, because os.DirEntry objects expose this information if the operating system provides it when scanning a directory. All os.DirEntry methods may perform a system call, but is_dir() and is_file() usually only require a system call for symbolic links; os.DirEntry.stat() always requires a system call on Unix but only requires one for symbolic links on Windows.

Python Docs


回答 10

虽然os.listdir()可以很好地生成文件名和目录名列表,但是一旦拥有了这些文件名,就经常想做更多的事情-在Python3中,pathlib使其他琐事变得简单。让我们看一下,看看您是否像我一样喜欢它。

要列出目录内容,请构造一个Path对象并获取迭代器:

In [16]: Path('/etc').iterdir()
Out[16]: <generator object Path.iterdir at 0x110853fc0>

如果我们只想要事物名称列表:

In [17]: [x.name for x in Path('/etc').iterdir()]
Out[17]:
['emond.d',
 'ntp-restrict.conf',
 'periodic',

如果您只想要Dirs:

In [18]: [x.name for x in Path('/etc').iterdir() if x.is_dir()]
Out[18]:
['emond.d',
 'periodic',
 'mach_init.d',

如果您想要该树中所有conf文件的名称:

In [20]: [x.name for x in Path('/etc').glob('**/*.conf')]
Out[20]:
['ntp-restrict.conf',
 'dnsextd.conf',
 'syslog.conf',

如果要在树中的conf文件列表> = 1K中:

In [23]: [x.name for x in Path('/etc').glob('**/*.conf') if x.stat().st_size > 1024]
Out[23]:
['dnsextd.conf',
 'pf.conf',
 'autofs.conf',

解决相对路径变得容易:

In [32]: Path('../Operational Metrics.md').resolve()
Out[32]: PosixPath('/Users/starver/code/xxxx/Operational Metrics.md')

使用路径导航非常清晰(尽管出乎意料):

In [10]: p = Path('.')

In [11]: core = p / 'web' / 'core'

In [13]: [x for x in core.iterdir() if x.is_file()]
Out[13]:
[PosixPath('web/core/metrics.py'),
 PosixPath('web/core/services.py'),
 PosixPath('web/core/querysets.py'),

While os.listdir() is fine for generating a list of file and dir names, frequently you want to do more once you have those names – and in Python3, pathlib makes those other chores simple. Let’s take a look and see if you like it as much as I do.

To list dir contents, construct a Path object and grab the iterator:

In [16]: Path('/etc').iterdir()
Out[16]: <generator object Path.iterdir at 0x110853fc0>

If we want just a list of names of things:

In [17]: [x.name for x in Path('/etc').iterdir()]
Out[17]:
['emond.d',
 'ntp-restrict.conf',
 'periodic',

If you want just the dirs:

In [18]: [x.name for x in Path('/etc').iterdir() if x.is_dir()]
Out[18]:
['emond.d',
 'periodic',
 'mach_init.d',

If you want the names of all conf files in that tree:

In [20]: [x.name for x in Path('/etc').glob('**/*.conf')]
Out[20]:
['ntp-restrict.conf',
 'dnsextd.conf',
 'syslog.conf',

If you want a list of conf files in the tree >= 1K:

In [23]: [x.name for x in Path('/etc').glob('**/*.conf') if x.stat().st_size > 1024]
Out[23]:
['dnsextd.conf',
 'pf.conf',
 'autofs.conf',

Resolving relative paths become easy:

In [32]: Path('../Operational Metrics.md').resolve()
Out[32]: PosixPath('/Users/starver/code/xxxx/Operational Metrics.md')

Navigating with a Path is pretty clear (although unexpected):

In [10]: p = Path('.')

In [11]: core = p / 'web' / 'core'

In [13]: [x for x in core.iterdir() if x.is_file()]
Out[13]:
[PosixPath('web/core/metrics.py'),
 PosixPath('web/core/services.py'),
 PosixPath('web/core/querysets.py'),

回答 11

一个很好的衬垫,可以递归地仅列出文件。我在setup.py package_data指令中使用了此命令:

import os

[os.path.join(x[0],y) for x in os.walk('<some_directory>') for y in x[2]]

我知道这不是问题的答案,但可能会派上用场

A nice one liner to list only the files recursively. I used this in my setup.py package_data directive:

import os

[os.path.join(x[0],y) for x in os.walk('<some_directory>') for y in x[2]]

I know it’s not the answer to the question, but may come in handy


回答 12

对于Python 2

#!/bin/python2

import os

def scan_dir(path):
    print map(os.path.abspath, os.listdir(pwd))

对于Python 3

对于过滤器和地图,您需要使用list()包装它们

#!/bin/python3

import os

def scan_dir(path):
    print(list(map(os.path.abspath, os.listdir(pwd))))

现在的建议是,用生成器表达式或列表推导替换map和filter的用法:

#!/bin/python

import os

def scan_dir(path):
    print([os.path.abspath(f) for f in os.listdir(path)])

For Python 2

#!/bin/python2

import os

def scan_dir(path):
    print map(os.path.abspath, os.listdir(pwd))

For Python 3

For filter and map, you need wrap them with list()

#!/bin/python3

import os

def scan_dir(path):
    print(list(map(os.path.abspath, os.listdir(pwd))))

The recommendation now is that you replace your usage of map and filter with generators expressions or list comprehensions:

#!/bin/python

import os

def scan_dir(path):
    print([os.path.abspath(f) for f in os.listdir(path)])

回答 13

这是一行Pythonic版本:

import os
dir = 'given_directory_name'
filenames = [os.path.join(os.path.dirname(os.path.abspath(__file__)),dir,i) for i in os.listdir(dir)]

此代码列出了给定目录名称中所有文件和目录的完整路径。

Here is a one line Pythonic version:

import os
dir = 'given_directory_name'
filenames = [os.path.join(os.path.dirname(os.path.abspath(__file__)),dir,i) for i in os.listdir(dir)]

This code lists the full path of all files and directories in the given directory name.


回答 14

我知道这是一个老问题。如果您使用的是liunx机器,这是我遇到的一种巧妙方法。

import subprocess
print(subprocess.check_output(["ls", "/"]).decode("utf8"))

I know this is an old question. This is a neat way I came across if you are on a liunx machine.

import subprocess
print(subprocess.check_output(["ls", "/"]).decode("utf8"))

回答 15

#import modules
import os

_CURRENT_DIR = '.'


def rec_tree_traverse(curr_dir, indent):
    "recurcive function to traverse the directory"
    #print "[traverse_tree]"

    try :
        dfList = [os.path.join(curr_dir, f_or_d) for f_or_d in os.listdir(curr_dir)]
    except:
        print "wrong path name/directory name"
        return

    for file_or_dir in dfList:

        if os.path.isdir(file_or_dir):
            #print "dir  : ",
            print indent, file_or_dir,"\\"
            rec_tree_traverse(file_or_dir, indent*2)

        if os.path.isfile(file_or_dir):
            #print "file : ",
            print indent, file_or_dir

    #end if for loop
#end of traverse_tree()

def main():

    base_dir = _CURRENT_DIR

    rec_tree_traverse(base_dir," ")

    raw_input("enter any key to exit....")
#end of main()


if __name__ == '__main__':
    main()
#import modules
import os

_CURRENT_DIR = '.'


def rec_tree_traverse(curr_dir, indent):
    "recurcive function to traverse the directory"
    #print "[traverse_tree]"

    try :
        dfList = [os.path.join(curr_dir, f_or_d) for f_or_d in os.listdir(curr_dir)]
    except:
        print "wrong path name/directory name"
        return

    for file_or_dir in dfList:

        if os.path.isdir(file_or_dir):
            #print "dir  : ",
            print indent, file_or_dir,"\\"
            rec_tree_traverse(file_or_dir, indent*2)

        if os.path.isfile(file_or_dir):
            #print "file : ",
            print indent, file_or_dir

    #end if for loop
#end of traverse_tree()

def main():

    base_dir = _CURRENT_DIR

    rec_tree_traverse(base_dir," ")

    raw_input("enter any key to exit....")
#end of main()


if __name__ == '__main__':
    main()

回答 16

仅供参考,添加扩展名或扩展名文件过滤器os

path = '.'
for dirname, dirnames, filenames in os.walk(path):
    # print path to all filenames with extension py.
    for filename in filenames:
        fname_path = os.path.join(dirname, filename)
        fext = os.path.splitext(fname_path)[1]
        if fext == '.py':
            print fname_path
        else:
            continue

FYI Add a filter of extension or ext file import os

path = '.'
for dirname, dirnames, filenames in os.walk(path):
    # print path to all filenames with extension py.
    for filename in filenames:
        fname_path = os.path.join(dirname, filename)
        fext = os.path.splitext(fname_path)[1]
        if fext == '.py':
            print fname_path
        else:
            continue

回答 17

如果知道的话,我会把它扔进去。通配符搜索的简单而肮脏的方法。

import re
import os

[a for a in os.listdir(".") if re.search("^.*\.py$",a)]

If figured I’d throw this in. Simple and dirty way to do wildcard searches.

import re
import os

[a for a in os.listdir(".") if re.search("^.*\.py$",a)]

回答 18

下面的代码将列出目录和目录中的文件

def print_directory_contents(sPath):
        import os                                       
        for sChild in os.listdir(sPath):                
            sChildPath = os.path.join(sPath,sChild)
            if os.path.isdir(sChildPath):
                print_directory_contents(sChildPath)
            else:
                print(sChildPath)

Below code will list directories and the files within the dir

def print_directory_contents(sPath):
        import os                                       
        for sChild in os.listdir(sPath):                
            sChildPath = os.path.join(sPath,sChild)
            if os.path.isdir(sChildPath):
                print_directory_contents(sChildPath)
            else:
                print(sChildPath)

回答 19

和我一起工作的人是上述萨利赫回答的一种修改版本。

代码如下:

“ dir =’given_directory_name’文件名= [os.listdir(dir)中用于i的os.path.abspath(os.path.join(dir,i))]”

The one worked with me is kind of a modified version from Saleh answer above.

The code is as follows:

“dir = ‘given_directory_name’ filenames = [os.path.abspath(os.path.join(dir,i)) for i in os.listdir(dir)]”


获取当前目录中所有子目录的列表

问题:获取当前目录中所有子目录的列表

有没有办法在Python中返回当前目录中所有子目录的列表?

我知道您可以使用文件来执行此操作,但是我需要获取目录列表。

Is there a way to return a list of all the subdirectories in the current directory in Python?

I know you can do this with files, but I need to get the list of directories instead.


回答 0

您是指直接子目录,还是树下的每个目录?

无论哪种方式,您都可以使用os.walk以下方法:

os.walk(directory)

将为每个子目录生成一个元组。该三元组中的第一个条目是目录名称,因此

[x[0] for x in os.walk(directory)]

应该递归地给你所有的子目录。

请注意,该元组中的第二个条目是该条目的子目录列表的第一个位置,因此您可以改用它,但是不太可能为您节省很多。

但是,您可以使用它来为您提供直接的子目录:

next(os.walk('.'))[1]

或者使用os.listdir和查看已经发布的其他解决方案os.path.isdir,包括“ 如何在Python中获取所有直接子目录 ”中的解决方案。

Do you mean immediate subdirectories, or every directory right down the tree?

Either way, you could use os.walk to do this:

os.walk(directory)

will yield a tuple for each subdirectory. Ths first entry in the 3-tuple is a directory name, so

[x[0] for x in os.walk(directory)]

should give you all of the subdirectories, recursively.

Note that the second entry in the tuple is the list of child directories of the entry in the first position, so you could use this instead, but it’s not likely to save you much.

However, you could use it just to give you the immediate child directories:

next(os.walk('.'))[1]

Or see the other solutions already posted, using os.listdir and os.path.isdir, including those at “How to get all of the immediate subdirectories in Python“.


回答 1

import os

d = '.'
[os.path.join(d, o) for o in os.listdir(d) 
                    if os.path.isdir(os.path.join(d,o))]
import os

d = '.'
[os.path.join(d, o) for o in os.listdir(d) 
                    if os.path.isdir(os.path.join(d,o))]

回答 2

你可以用 glob.glob

from glob import glob
glob("/path/to/directory/*/")

别忘了/之后的尾随*

You could just use glob.glob

from glob import glob
glob("/path/to/directory/*/")

Don’t forget the trailing / after the *.


回答 3

比上面的要好得多,因为您不需要几个os.path.join(),并且可以直接获取完整路径(如果需要),因此可以在Python 3.5及更高版本中执行此操作。

subfolders = [ f.path for f in os.scandir(folder) if f.is_dir() ]

这将提供到子目录的完整路径。如果只希望使用子目录的名称,f.name而不是f.path

https://docs.python.org/3/library/os.html#os.scandir


OT:如果您需要递归所有子文件夹和/或所有文件,请看一下此功能,它比os.walk&快,glob并且将返回所有子文件夹以及这些(子)子文件夹中的所有文件的列表:https://stackoverflow.com/a/59803793/2441026

如果您只需要递归所有子文件夹

def fast_scandir(dirname):
    subfolders= [f.path for f in os.scandir(dirname) if f.is_dir()]
    for dirname in list(subfolders):
        subfolders.extend(fast_scandir(dirname))
    return subfolders

返回所有子文件夹及其完整路径的列表。这又比os.walk和快得多glob


所有功能分析

tl; dr:
-如果要获取文件夹使用的所有直接子目录os.scandir
-如果要获取所有子目录,甚至嵌套的子目录,请使用os.walk或-稍微快一点- fast_scandir上面的函数。
-从来不使用os.walk只顶级子目录,因为它可以是数百倍慢于(!) os.scandir

  • 如果您运行下面的代码,请确保运行一次,以便您的操作系统可以访问该文件夹,丢弃结果并运行测试,否则结果将被弄乱。
  • 您可能想混淆函数调用,但是我对其进行了测试,但这并不重要。
  • 所有示例都将提供文件夹的完整路径。pathlib示例作为(Windows)Path对象。
  • 的第一个元素os.walk将是基本文件夹。因此,您将不会仅获得子目录。您可以使用fu.pop(0)将其删除。
  • 所有结果都不会使用自然排序。这意味着将对结果进行如下排序:1、10、2。要进行自然排序(1、2、10),请查看https://stackoverflow.com/a/48030307/2441026


结果

os.scandir      took   1 ms. Found dirs: 439
os.walk         took 463 ms. Found dirs: 441 -> it found the nested one + base folder.
glob.glob       took  20 ms. Found dirs: 439
pathlib.iterdir took  18 ms. Found dirs: 439
os.listdir      took  18 ms. Found dirs: 439

已在W7x64,Python 3.8.1中测试。

# -*- coding: utf-8 -*-
# Python 3


import time
import os
from glob import glob
from pathlib import Path


directory = r"<insert_folder>"
RUNS = 1


def run_os_walk():
    a = time.time_ns()
    for i in range(RUNS):
        fu = [x[0] for x in os.walk(directory)]
    print(f"os.walk\t\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_glob():
    a = time.time_ns()
    for i in range(RUNS):
        fu = glob(directory + "/*/")
    print(f"glob.glob\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_pathlib_iterdir():
    a = time.time_ns()
    for i in range(RUNS):
        dirname = Path(directory)
        fu = [f for f in dirname.iterdir() if f.is_dir()]
    print(f"pathlib.iterdir\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_os_listdir():
    a = time.time_ns()
    for i in range(RUNS):
        dirname = Path(directory)
        fu = [os.path.join(directory, o) for o in os.listdir(directory) if os.path.isdir(os.path.join(directory, o))]
    print(f"os.listdir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_os_scandir():
    a = time.time_ns()
    for i in range(RUNS):
        fu = [f.path for f in os.scandir(directory) if f.is_dir()]
    print(f"os.scandir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms.\tFound dirs: {len(fu)}")


if __name__ == '__main__':
    run_os_scandir()
    run_os_walk()
    run_glob()
    run_pathlib_iterdir()
    run_os_listdir()

Much nicer than the above, because you don’t need several os.path.join() and you will get the full path directly (if you wish), you can do this in Python 3.5 and above.

subfolders = [ f.path for f in os.scandir(folder) if f.is_dir() ]

This will give the complete path to the subdirectory. If you only want the name of the subdirectory use f.name instead of f.path

https://docs.python.org/3/library/os.html#os.scandir


Slightly OT: In case you need all subfolder recursively and/or all files recursively, have a look at this function, that is faster than os.walk & glob and will return a list of all subfolders as well as all files inside those (sub-)subfolders: https://stackoverflow.com/a/59803793/2441026

In case you want only all subfolders recursively:

def fast_scandir(dirname):
    subfolders= [f.path for f in os.scandir(dirname) if f.is_dir()]
    for dirname in list(subfolders):
        subfolders.extend(fast_scandir(dirname))
    return subfolders

Returns a list of all subfolders with their full paths. This again is faster than os.walk and a lot faster than glob.


An analysis of all functions

tl;dr:
– If you want to get all immediate subdirectories for a folder use os.scandir.
– If you want to get all subdirectories, even nested ones, use os.walk or – slightly faster – the fast_scandir function above.
– Never use os.walk for only top-level subdirectories, as it can be hundreds(!) of times slower than os.scandir.

  • If you run the code below, make sure to run it once so that your OS will have accessed the folder, discard the results and run the test, otherwise results will be screwed.
  • You might want to mix up the function calls, but I tested it, and it did not really matter.
  • All examples will give the full path to the folder. The pathlib example as a (Windows)Path object.
  • The first element of os.walk will be the base folder. So you will not get only subdirectories. You can use fu.pop(0) to remove it.
  • None of the results will use natural sorting. This means results will be sorted like this: 1, 10, 2. To get natural sorting (1, 2, 10), please have a look at https://stackoverflow.com/a/48030307/2441026


Results:

os.scandir      took   1 ms. Found dirs: 439
os.walk         took 463 ms. Found dirs: 441 -> it found the nested one + base folder.
glob.glob       took  20 ms. Found dirs: 439
pathlib.iterdir took  18 ms. Found dirs: 439
os.listdir      took  18 ms. Found dirs: 439

Tested with W7x64, Python 3.8.1.

# -*- coding: utf-8 -*-
# Python 3


import time
import os
from glob import glob
from pathlib import Path


directory = r"<insert_folder>"
RUNS = 1


def run_os_walk():
    a = time.time_ns()
    for i in range(RUNS):
        fu = [x[0] for x in os.walk(directory)]
    print(f"os.walk\t\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_glob():
    a = time.time_ns()
    for i in range(RUNS):
        fu = glob(directory + "/*/")
    print(f"glob.glob\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_pathlib_iterdir():
    a = time.time_ns()
    for i in range(RUNS):
        dirname = Path(directory)
        fu = [f for f in dirname.iterdir() if f.is_dir()]
    print(f"pathlib.iterdir\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_os_listdir():
    a = time.time_ns()
    for i in range(RUNS):
        dirname = Path(directory)
        fu = [os.path.join(directory, o) for o in os.listdir(directory) if os.path.isdir(os.path.join(directory, o))]
    print(f"os.listdir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_os_scandir():
    a = time.time_ns()
    for i in range(RUNS):
        fu = [f.path for f in os.scandir(directory) if f.is_dir()]
    print(f"os.scandir\t\ttook {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms.\tFound dirs: {len(fu)}")


if __name__ == '__main__':
    run_os_scandir()
    run_os_walk()
    run_glob()
    run_pathlib_iterdir()
    run_os_listdir()

回答 4

如果您需要一个可在子目录中找到所有子目录的递归解决方案,请按照之前的建议使用walk。

如果仅需要当前目录的子目录,请os.listdiros.path.isdir

If you need a recursive solution that will find all the subdirectories in the subdirectories, use walk as proposed before.

If you only need the current directory’s child directories, combine os.listdir with os.path.isdir


回答 5

我更喜欢使用过滤器(https://docs.python.org/2/library/functions.html#filter),但这只是一个问题。

d='.'
filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d))

I prefer using filter (https://docs.python.org/2/library/functions.html#filter), but this is just a matter of taste.

d='.'
filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d))

回答 6

使用python-os-walk实现了这一点。(http://www.pythonforbeginners.com/code-snippets-source-code/python-os-walk/

import os

print("root prints out directories only from what you specified")
print("dirs prints out sub-directories from root")
print("files prints out all files from root and directories")
print("*" * 20)

for root, dirs, files in os.walk("/var/log"):
    print(root)
    print(dirs)
    print(files)

Implemented this using python-os-walk. (http://www.pythonforbeginners.com/code-snippets-source-code/python-os-walk/)

import os

print("root prints out directories only from what you specified")
print("dirs prints out sub-directories from root")
print("files prints out all files from root and directories")
print("*" * 20)

for root, dirs, files in os.walk("/var/log"):
    print(root)
    print(dirs)
    print(files)

回答 7

您可以使用os.listdir(path)获取Python 2.7中的子目录(和文件)列表

import os
os.listdir(path)  # list of subdirectories and files

You can get the list of subdirectories (and files) in Python 2.7 using os.listdir(path)

import os
os.listdir(path)  # list of subdirectories and files

回答 8

仅列出目录

print("\nWe are listing out only the directories in current directory -")
directories_in_curdir = filter(os.path.isdir, os.listdir(os.curdir))
print(directories_in_curdir)

仅列出当前目录中的文件

files = filter(os.path.isfile, os.listdir(os.curdir))
print("\nThe following are the list of all files in the current directory -")
print(files)

Listing Out only directories

print("\nWe are listing out only the directories in current directory -")
directories_in_curdir = filter(os.path.isdir, os.listdir(os.curdir))
print(directories_in_curdir)

Listing Out only files in current directory

files = filter(os.path.isfile, os.listdir(os.curdir))
print("\nThe following are the list of all files in the current directory -")
print(files)

回答 9

蟒3.4引入pathlib模块到标准库,它提供了一个面向对象的方法来处理的文件系统的路径:

from pathlib import Path

p = Path('./')

# List comprehension
[f for f in p.iterdir() if f.is_dir()]

# The trailing slash to glob indicated directories
# This will also include the current directory '.'
list(p.glob('**/'))

也可以通过PyPi上的pathlib2模块在 Python 2.7 上使用Pathlib。

Python 3.4 introduced the pathlib module into the standard library, which provides an object oriented approach to handle filesystem paths:

from pathlib import Path

p = Path('./')

# List comprehension
[f for f in p.iterdir() if f.is_dir()]

# The trailing slash to glob indicated directories
# This will also include the current directory '.'
list(p.glob('**/'))

Pathlib is also available on Python 2.7 via the pathlib2 module on PyPi.


回答 10

由于我使用Python 3.4和Windows UNC路径偶然发现了此问题,因此以下是此环境的变体:

from pathlib import WindowsPath

def SubDirPath (d):
    return [f for f in d.iterdir() if f.is_dir()]

subdirs = SubDirPath(WindowsPath(r'\\file01.acme.local\home$'))
print(subdirs)

Pathlib是Python 3.4中的新增功能,它使在不同操作系统下使用路径变得更加容易:https ://docs.python.org/3.4/library/pathlib.html

Since I stumbled upon this problem using Python 3.4 and Windows UNC paths, here’s a variant for this environment:

from pathlib import WindowsPath

def SubDirPath (d):
    return [f for f in d.iterdir() if f.is_dir()]

subdirs = SubDirPath(WindowsPath(r'\\file01.acme.local\home$'))
print(subdirs)

Pathlib is new in Python 3.4 and makes working with paths under different OSes much easier: https://docs.python.org/3.4/library/pathlib.html


回答 11

尽管很久以前就回答了这个问题。我想建议使用该pathlib模块,因为这是在Windows和Unix OS上工作的可靠方法。

因此,要获取特定目录(包括子目录)中的所有路径:

from pathlib import Path
paths = list(Path('myhomefolder', 'folder').glob('**/*.txt'))

# all sorts of operations
file = paths[0]
file.name
file.stem
file.parent
file.suffix

等等

Although this question is answered a long time ago. I want to recommend to use the pathlib module since this is a robust way to work on Windows and Unix OS.

So to get all paths in a specific directory including subdirectories:

from pathlib import Path
paths = list(Path('myhomefolder', 'folder').glob('**/*.txt'))

# all sorts of operations
file = paths[0]
file.name
file.stem
file.parent
file.suffix

etc.


回答 12

谢谢提醒伙计。我遇到了一个以dirs返回的软链接(无限递归)的问题。软链接?我们不希望没有臭味的软链接!所以…

这仅显示目录,而不显示软链接:

>>> import os
>>> inf = os.walk('.')
>>> [x[0] for x in inf]
['.', './iamadir']

Thanks for the tips, guys. I ran into an issue with softlinks (infinite recursion) being returned as dirs. Softlinks? We don’t want no stinkin’ soft links! So…

This rendered just the dirs, not softlinks:

>>> import os
>>> inf = os.walk('.')
>>> [x[0] for x in inf]
['.', './iamadir']

回答 13

复制粘贴友好ipython

import os
d='.'
folders = list(filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d)))

来自的输出print(folders)

['folderA', 'folderB']

Copy paste friendly in ipython:

import os
d='.'
folders = list(filter(lambda x: os.path.isdir(os.path.join(d, x)), os.listdir(d)))

Output from print(folders):

['folderA', 'folderB']

回答 14

这就是我的方法。

    import os
    for x in os.listdir(os.getcwd()):
        if os.path.isdir(x):
            print(x)

This is how I do it.

    import os
    for x in os.listdir(os.getcwd()):
        if os.path.isdir(x):
            print(x)

回答 15

这是基于@Blair Conrad的示例的几个简单函数-

import os

def get_subdirs(dir):
    "Get a list of immediate subdirectories"
    return next(os.walk(dir))[1]

def get_subfiles(dir):
    "Get a list of immediate subfiles"
    return next(os.walk(dir))[2]

Here are a couple of simple functions based on @Blair Conrad’s example –

import os

def get_subdirs(dir):
    "Get a list of immediate subdirectories"
    return next(os.walk(dir))[1]

def get_subfiles(dir):
    "Get a list of immediate subfiles"
    return next(os.walk(dir))[2]

回答 16

在Eli Bendersky解决方案的基础上,使用以下示例:

import os
test_directory = <your_directory>
for child in os.listdir(test_directory):
    test_path = os.path.join(test_directory, child)
    if os.path.isdir(test_path):
        print test_path
        # Do stuff to the directory "test_path"

<your_directory>您要遍历的目录的路径在哪里。

Building upon Eli Bendersky’s solution, use the following example:

import os
test_directory = <your_directory>
for child in os.listdir(test_directory):
    test_path = os.path.join(test_directory, child)
    if os.path.isdir(test_path):
        print test_path
        # Do stuff to the directory "test_path"

where <your_directory> is the path to the directory you want to traverse.


回答 17

有了完整路径和占路感...\\..\\..\\subfolder,等:

import os, pprint
pprint.pprint([os.path.join(os.path.abspath(path), x[0]) \
    for x in os.walk(os.path.abspath(path))])

With full path and accounting for path being ., .., \\, ..\\..\\subfolder, etc:

import os, pprint
pprint.pprint([os.path.join(os.path.abspath(path), x[0]) \
    for x in os.walk(os.path.abspath(path))])

回答 18

这个答案似乎还不存在。

directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]

This answer didn’t seem to exist already.

directories = [ x for x in os.listdir('.') if os.path.isdir(x) ]

回答 19

我最近有一个类似的问题,我发现python 3.6(作为用户havlock添加)的最佳答案是使用os.scandir。由于似乎没有使用它的解决方案,因此我将添加自己的解决方案。首先,一种非递归解决方案,仅列出直接位于根目录下的子目录。

def get_dirlist(rootdir):

    dirlist = []

    with os.scandir(rootdir) as rit:
        for entry in rit:
            if not entry.name.startswith('.') and entry.is_dir():
                dirlist.append(entry.path)

    dirlist.sort() # Optional, in case you want sorted directory names
    return dirlist

递归版本如下所示:

def get_dirlist(rootdir):

    dirlist = []

    with os.scandir(rootdir) as rit:
        for entry in rit:
            if not entry.name.startswith('.') and entry.is_dir():
                dirlist.append(entry.path)
                dirlist += get_dirlist(entry.path)

    dirlist.sort() # Optional, in case you want sorted directory names
    return dirlist

请记住,这entry.path将使用子目录的绝对路径。如果只需要文件夹名称,则可以entry.name改用。有关该对象的更多详细信息,请参考os.DirEntryentry

I’ve had a similar question recently, and I found out that the best answer for python 3.6 (as user havlock added) is to use os.scandir. Since it seems there is no solution using it, I’ll add my own. First, a non-recursive solution that lists only the subdirectories directly under the root directory.

def get_dirlist(rootdir):

    dirlist = []

    with os.scandir(rootdir) as rit:
        for entry in rit:
            if not entry.name.startswith('.') and entry.is_dir():
                dirlist.append(entry.path)

    dirlist.sort() # Optional, in case you want sorted directory names
    return dirlist

The recursive version would look like this:

def get_dirlist(rootdir):

    dirlist = []

    with os.scandir(rootdir) as rit:
        for entry in rit:
            if not entry.name.startswith('.') and entry.is_dir():
                dirlist.append(entry.path)
                dirlist += get_dirlist(entry.path)

    dirlist.sort() # Optional, in case you want sorted directory names
    return dirlist

keep in mind that entry.path wields the absolute path to the subdirectory. In case you only need the folder name, you can use entry.name instead. Refer to os.DirEntry for additional details about the entry object.


回答 20

os.path.isdiros.listdir() 这样的东西使用过滤器功能filter(os.path.isdir,[os.path.join(os.path.abspath('PATH'),p) for p in os.listdir('PATH/')])

use a filter function os.path.isdir over os.listdir() something like this filter(os.path.isdir,[os.path.join(os.path.abspath('PATH'),p) for p in os.listdir('PATH/')])


回答 21

这将在文件树的右边列出所有子目录。

import pathlib


def list_dir(dir):
    path = pathlib.Path(dir)
    dir = []
    try:
        for item in path.iterdir():
            if item.is_dir():
                dir.append(item)
                dir = dir + list_dir(item)
        return dir
    except FileNotFoundError:
        print('Invalid directory')

pathlib 是3.4版的新功能

This will list all subdirectories right down the file tree.

import pathlib


def list_dir(dir):
    path = pathlib.Path(dir)
    dir = []
    try:
        for item in path.iterdir():
            if item.is_dir():
                dir.append(item)
                dir = dir + list_dir(item)
        return dir
    except FileNotFoundError:
        print('Invalid directory')

pathlib is new in version 3.4


回答 22

返回给定文件路径中所有子目录的列表的函数。将搜索整个文件树。

import os

def get_sub_directory_paths(start_directory, sub_directories):
    """
    This method iterates through all subdirectory paths of a given 
    directory to collect all directory paths.

    :param start_directory: The starting directory path.
    :param sub_directories: A List that all subdirectory paths will be 
        stored to.
    :return: A List of all sub-directory paths.
    """

    for item in os.listdir(start_directory):
        full_path = os.path.join(start_directory, item)

        if os.path.isdir(full_path):
            sub_directories.append(full_path)

            # Recursive call to search through all subdirectories.
            get_sub_directory_paths(full_path, sub_directories)

return sub_directories

Function to return a List of all subdirectories within a given file path. Will search through the entire file tree.

import os

def get_sub_directory_paths(start_directory, sub_directories):
    """
    This method iterates through all subdirectory paths of a given 
    directory to collect all directory paths.

    :param start_directory: The starting directory path.
    :param sub_directories: A List that all subdirectory paths will be 
        stored to.
    :return: A List of all sub-directory paths.
    """

    for item in os.listdir(start_directory):
        full_path = os.path.join(start_directory, item)

        if os.path.isdir(full_path):
            sub_directories.append(full_path)

            # Recursive call to search through all subdirectories.
            get_sub_directory_paths(full_path, sub_directories)

return sub_directories

回答 23

我们可以使用os.walk()获取所有文件夹的列表

import os

path = os.getcwd()

pathObject = os.walk(path)

这个pathObject是一个对象,我们可以通过

arr = [x for x in pathObject]

arr is of type [('current directory', [array of folder in current directory], [files in current directory]),('subdirectory', [array of folder in subdirectory], [files in subdirectory]) ....]

我们可以通过遍历arr并打印中间数组来获取所有子目录的列表

for i in arr:
   for j in i[1]:
      print(j)

这将打印所有子目录。

要获取所有文件:

for i in arr:
   for j in i[2]:
      print(i[0] + "/" + j)

we can get list of all the folders by using os.walk()

import os

path = os.getcwd()

pathObject = os.walk(path)

this pathObject is a object and we can get an array by

arr = [x for x in pathObject]

arr is of type [('current directory', [array of folder in current directory], [files in current directory]),('subdirectory', [array of folder in subdirectory], [files in subdirectory]) ....]

We can get list of all the subdirectory by iterating through the arr and printing the middle array

for i in arr:
   for j in i[1]:
      print(j)

This will print all the subdirectory.

To get all the files:

for i in arr:
   for j in i[2]:
      print(i[0] + "/" + j)

回答 24

具有给定父级的该函数以递归方式directory遍历其所有函数directories以及printsfilenames其中找到的所有函数。太有用了。

import os

def printDirectoryFiles(directory):
   for filename in os.listdir(directory):  
        full_path=os.path.join(directory, filename)
        if not os.path.isdir(full_path): 
            print( full_path + "\n")


def checkFolders(directory):

    dir_list = next(os.walk(directory))[1]

    #print(dir_list)

    for dir in dir_list:           
        print(dir)
        checkFolders(directory +"/"+ dir) 

    printDirectoryFiles(directory)       

main_dir="C:/Users/S0082448/Desktop/carpeta1"

checkFolders(main_dir)


input("Press enter to exit ;")

This function, with a given parent directory iterates over all its directories recursively and prints all the filenames which it founds inside. Too useful.

import os

def printDirectoryFiles(directory):
   for filename in os.listdir(directory):  
        full_path=os.path.join(directory, filename)
        if not os.path.isdir(full_path): 
            print( full_path + "\n")


def checkFolders(directory):

    dir_list = next(os.walk(directory))[1]

    #print(dir_list)

    for dir in dir_list:           
        print(dir)
        checkFolders(directory +"/"+ dir) 

    printDirectoryFiles(directory)       

main_dir="C:/Users/S0082448/Desktop/carpeta1"

checkFolders(main_dir)


input("Press enter to exit ;")


回答 25

通过从此处加入多个解决方案,这就是我最终使用的方法:

import os
import glob

def list_dirs(path):
    return [os.path.basename(x) for x in filter(
        os.path.isdir, glob.glob(os.path.join(path, '*')))]

By joining multiple solutions from here, this is what I ended up using:

import os
import glob

def list_dirs(path):
    return [os.path.basename(x) for x in filter(
        os.path.isdir, glob.glob(os.path.join(path, '*')))]

如何遍历给定目录中的文件?

问题:如何遍历给定目录中的文件?

我需要遍历.asm给定目录内的所有文件并对它们执行一些操作。

如何有效地做到这一点?

I need to iterate through all .asm files inside a given directory and do some actions on them.

How can this be done in a efficient way?


回答 0

原始答案:

import os

for filename in os.listdir(directory):
    if filename.endswith(".asm") or filename.endswith(".py"): 
         # print(os.path.join(directory, filename))
        continue
    else:
        continue

上面答案的Python 3.6版本,使用os-假设您将目录路径作为str对象包含在名为的变量中directory_in_str

import os

directory = os.fsencode(directory_in_str)

for file in os.listdir(directory):
     filename = os.fsdecode(file)
     if filename.endswith(".asm") or filename.endswith(".py"): 
         # print(os.path.join(directory, filename))
         continue
     else:
         continue

或递归使用pathlib

from pathlib import Path

pathlist = Path(directory_in_str).glob('**/*.asm')
for path in pathlist:
     # because path is object not string
     path_in_str = str(path)
     # print(path_in_str)

Original answer:

import os

for filename in os.listdir(directory):
    if filename.endswith(".asm") or filename.endswith(".py"): 
         # print(os.path.join(directory, filename))
        continue
    else:
        continue

Python 3.6 version of the above answer, using os – assuming that you have the directory path as a str object in a variable called directory_in_str:

import os

directory = os.fsencode(directory_in_str)

for file in os.listdir(directory):
     filename = os.fsdecode(file)
     if filename.endswith(".asm") or filename.endswith(".py"): 
         # print(os.path.join(directory, filename))
         continue
     else:
         continue

Or recursively, using pathlib:

from pathlib import Path

pathlist = Path(directory_in_str).glob('**/*.asm')
for path in pathlist:
     # because path is object not string
     path_in_str = str(path)
     # print(path_in_str)

回答 1

这将遍历所有后代文件,而不仅仅是目录的直接子级:

import os

for subdir, dirs, files in os.walk(rootdir):
    for file in files:
        #print os.path.join(subdir, file)
        filepath = subdir + os.sep + file

        if filepath.endswith(".asm"):
            print (filepath)

This will iterate over all descendant files, not just the immediate children of the directory:

import os

for subdir, dirs, files in os.walk(rootdir):
    for file in files:
        #print os.path.join(subdir, file)
        filepath = subdir + os.sep + file

        if filepath.endswith(".asm"):
            print (filepath)

回答 2

您可以尝试使用glob模块:

import glob

for filepath in glob.iglob('my_dir/*.asm'):
    print(filepath)

从Python 3.5开始,您还可以搜索子目录:

glob.glob('**/*.txt', recursive=True) # => ['2.txt', 'sub/3.txt']

从文档:

glob模块根据Unix shell使用的规则查找与指定模式匹配的所有路径名,尽管结果以任意顺序返回。没有波浪符号扩展,但是* 、?和用[]表示的字符范围将正确匹配。

You can try using glob module:

import glob

for filepath in glob.iglob('my_dir/*.asm'):
    print(filepath)

and since Python 3.5 you can search subdirectories as well:

glob.glob('**/*.txt', recursive=True) # => ['2.txt', 'sub/3.txt']

From the docs:

The glob module finds all the pathnames matching a specified pattern according to the rules used by the Unix shell, although results are returned in arbitrary order. No tilde expansion is done, but *, ?, and character ranges expressed with [] will be correctly matched.


回答 3

从Python 3.5开始,使用os.scandir()可以轻松得多

with os.scandir(path) as it:
    for entry in it:
        if entry.name.endswith(".asm") and entry.is_file():
            print(entry.name, entry.path)

使用scandir()而不是listdir()可以显着提高还需要文件类型或文件属性信息的代码的性能,因为如果操作系统在扫描目录时提供了os.DirEntry对象,则os.DirEntry对象将公开此信息。所有的os.DirEntry方法都可以执行系统调用,但是is_dir()和is_file()通常只需要系统调用即可进行符号链接。os.DirEntry.stat()在Unix上始终需要系统调用,而在Windows上只需要一个系统调用即可。

Since Python 3.5, things are much easier with os.scandir()

with os.scandir(path) as it:
    for entry in it:
        if entry.name.endswith(".asm") and entry.is_file():
            print(entry.name, entry.path)

Using scandir() instead of listdir() can significantly increase the performance of code that also needs file type or file attribute information, because os.DirEntry objects expose this information if the operating system provides it when scanning a directory. All os.DirEntry methods may perform a system call, but is_dir() and is_file() usually only require a system call for symbolic links; os.DirEntry.stat() always requires a system call on Unix but only requires one for symbolic links on Windows.


回答 4

Python 3.4和更高版本在标准库中提供pathlib。您可以这样做:

from pathlib import Path

asm_pths = [pth for pth in Path.cwd().iterdir()
            if pth.suffix == '.asm']

或者,如果您不喜欢列表推导:

asm_paths = []
for pth in Path.cwd().iterdir():
    if pth.suffix == '.asm':
        asm_pths.append(pth)

Path 对象可以轻松转换为字符串。

Python 3.4 and later offer pathlib in the standard library. You could do:

from pathlib import Path

asm_pths = [pth for pth in Path.cwd().iterdir()
            if pth.suffix == '.asm']

Or if you don’t like list comprehensions:

asm_paths = []
for pth in Path.cwd().iterdir():
    if pth.suffix == '.asm':
        asm_pths.append(pth)

Path objects can easily be converted to strings.


回答 5

这是我遍历Python中文件的方式:

import os

path = 'the/name/of/your/path'

folder = os.fsencode(path)

filenames = []

for file in os.listdir(folder):
    filename = os.fsdecode(file)
    if filename.endswith( ('.jpeg', '.png', '.gif') ): # whatever file types you're using...
        filenames.append(filename)

filenames.sort() # now you have the filenames and can do something with them

这些技术均无法保证任何迭代顺序

是的,超级变幻莫测。请注意,我对文件名进行了排序,这在文件顺序很重要的情况下很重要,例如,对于视频帧或与时间有关的数据收集。不过,请务必在文件名中添加索引!

Here’s how I iterate through files in Python:

import os

path = 'the/name/of/your/path'

folder = os.fsencode(path)

filenames = []

for file in os.listdir(folder):
    filename = os.fsdecode(file)
    if filename.endswith( ('.jpeg', '.png', '.gif') ): # whatever file types you're using...
        filenames.append(filename)

filenames.sort() # now you have the filenames and can do something with them

NONE OF THESE TECHNIQUES GUARANTEE ANY ITERATION ORDERING

Yup, super unpredictable. Notice that I sort the filenames, which is important if the order of the files matters, i.e. for video frames or time dependent data collection. Be sure to put indices in your filenames though!


回答 6

您可以使用glob来引用目录和列表:

import glob
import os

#to get the current working directory name
cwd = os.getcwd()
#Load the images from images folder.
for f in glob.glob('images\*.jpg'):   
    dir_name = get_dir_name(f)
    image_file_name = dir_name + '.jpg'
    #To print the file name with path (path will be in string)
    print (image_file_name)

要获取数组中所有目录的列表,可以使用os

os.listdir(directory)

You can use glob for referring the directory and the list :

import glob
import os

#to get the current working directory name
cwd = os.getcwd()
#Load the images from images folder.
for f in glob.glob('images\*.jpg'):   
    dir_name = get_dir_name(f)
    image_file_name = dir_name + '.jpg'
    #To print the file name with path (path will be in string)
    print (image_file_name)

To get the list of all directory in array you can use os :

os.listdir(directory)

回答 7

我对该实现还不太满意,我想拥有一个自定义构造函数,DirectoryIndex._make(next(os.walk(input_path)))该构造函数可以使您只传递要为其列出文件的路径。欢迎编辑!

import collections
import os

DirectoryIndex = collections.namedtuple('DirectoryIndex', ['root', 'dirs', 'files'])

for file_name in DirectoryIndex(*next(os.walk('.'))).files:
    file_path = os.path.join(path, file_name)

I’m not quite happy with this implementation yet, I wanted to have a custom constructor that does DirectoryIndex._make(next(os.walk(input_path))) such that you can just pass the path you want a file listing for. Edits welcome!

import collections
import os

DirectoryIndex = collections.namedtuple('DirectoryIndex', ['root', 'dirs', 'files'])

for file_name in DirectoryIndex(*next(os.walk('.'))).files:
    file_path = os.path.join(path, file_name)

回答 8

我真的很喜欢使用库中scandir内置的指令os。这是一个工作示例:

import os

i = 0
with os.scandir('/usr/local/bin') as root_dir:
    for path in root_dir:
        if path.is_file():
            i += 1
            print(f"Full path is: {path} and just the name is: {path.name}")
print(f"{i} files scanned successfully.")

I really like using the scandir directive that is built into the os library. Here is a working example:

import os

i = 0
with os.scandir('/usr/local/bin') as root_dir:
    for path in root_dir:
        if path.is_file():
            i += 1
            print(f"Full path is: {path} and just the name is: {path.name}")
print(f"{i} files scanned successfully.")

如何获取当前文件目录的完整路径?

问题:如何获取当前文件目录的完整路径?

我想获取当前文件的目录路径。我试过了:

>>> os.path.abspath(__file__)
'C:\\python27\\test.py'

但是如何检索目录的路径?

例如:

'C:\\python27\\'

I want to get the current file’s directory path. I tried:

>>> os.path.abspath(__file__)
'C:\\python27\\test.py'

But how can I retrieve the directory’s path?

For example:

'C:\\python27\\'

回答 0

Python 3

对于正在运行的脚本的目录:

import pathlib
pathlib.Path(__file__).parent.absolute()

对于当前工作目录:

import pathlib
pathlib.Path().absolute()

Python 2和3

对于正在运行的脚本的目录:

import os
os.path.dirname(os.path.abspath(__file__))

如果您的意思是当前工作目录:

import os
os.path.abspath(os.getcwd())

请注意,前后分别file是两个下划线,而不仅仅是一个。

另请注意,如果您正在交互运行或已从文件以外的内容(例如数据库或在线资源)中加载了代码,则__file__可能不会设置,因为没有“当前文件”的概念。上面的答案假设运行文件中的python脚本的最常见情况。

参考文献

  1. python文档中的pathlib
  2. os.path 2.7os.path 3.8
  3. os.getcwd 2.7os.getcwd 3.8
  4. __file__变量的含义/作用是什么?

Python 3

For the directory of the script being run:

import pathlib
pathlib.Path(__file__).parent.absolute()

For the current working directory:

import pathlib
pathlib.Path().absolute()

Python 2 and 3

For the directory of the script being run:

import os
os.path.dirname(os.path.abspath(__file__))

If you mean the current working directory:

import os
os.path.abspath(os.getcwd())

Note that before and after file is two underscores, not just one.

Also note that if you are running interactively or have loaded code from something other than a file (eg: a database or online resource), __file__ may not be set since there is no notion of “current file”. The above answer assumes the most common scenario of running a python script that is in a file.

References

  1. pathlib in the python documentation.
  2. os.path 2.7, os.path 3.8
  3. os.getcwd 2.7, os.getcwd 3.8
  4. what does the __file__ variable mean/do?

回答 1

使用Path是因为Python 3的推荐方式:

from pathlib import Path
print("File      Path:", Path(__file__).absolute())
print("Directory Path:", Path().absolute())  

文档:pathlib

注意:如果使用Jupyter Notebook,__file__则不会返回期望值,因此Path().absolute()必须使用。

Using Path is the recommended way since Python 3:

from pathlib import Path
print("File      Path:", Path(__file__).absolute())
print("Directory Path:", Path().absolute())  

Documentation: pathlib

Note: If using Jupyter Notebook, __file__ doesn’t return expected value, so Path().absolute() has to be used.


回答 2

在Python 3.x中,我这样做:

from pathlib import Path

path = Path(__file__).parent.absolute()

说明:

  • Path(__file__) 是当前文件的路径。
  • .parent为您提供文件所在的目录
  • .absolute()给您完整的绝对路径。

使用pathlib是使用路径的现代方法。如果以后由于某种原因需要它作为字符串,只需执行str(path)

In Python 3.x I do:

from pathlib import Path

path = Path(__file__).parent.absolute()

Explanation:

  • Path(__file__) is the path to the current file.
  • .parent gives you the directory the file is in.
  • .absolute() gives you the full absolute path to it.

Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).


回答 3

import os
print os.path.dirname(__file__)
import os
print os.path.dirname(__file__)

回答 4

您可以轻松地使用os和存储os.path库,如下所示

import os
os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname从当前目录返回上一级目录。它使我们可以在不传递任何文件参数且不知道绝对路径的情况下切换到更高级别。

You can use os and os.path library easily as follows

import os
os.chdir(os.path.dirname(os.getcwd()))

os.path.dirname returns upper directory from current one. It lets us change to an upper level without passing any file argument and without knowing absolute path.


回答 5

尝试这个:

import os
dir_path = os.path.dirname(os.path.realpath(__file__))

Try this:

import os
dir_path = os.path.dirname(os.path.realpath(__file__))

回答 6

我发现以下命令将全部返回Python 3.6脚本的父目录的完整路径。

Python 3.6脚本:

#!/usr/bin/env python3.6
# -*- coding: utf-8 -*-

from pathlib import Path

#Get the absolute path of a Python3.6 script
dir1 = Path().resolve()  #Make the path absolute, resolving any symlinks.
dir2 = Path().absolute() #See @RonKalian answer 
dir3 = Path(__file__).parent.absolute() #See @Arminius answer 

print(f'dir1={dir1}\ndir2={dir2}\ndir3={dir3}')

说明链接:.resolve() .absolute() 路径(文件).parent()绝对的()。

I found the following commands will all return the full path of the parent directory of a Python 3.6 script.

Python 3.6 Script:

#!/usr/bin/env python3.6
# -*- coding: utf-8 -*-

from pathlib import Path

#Get the absolute path of a Python3.6 script
dir1 = Path().resolve()  #Make the path absolute, resolving any symlinks.
dir2 = Path().absolute() #See @RonKalian answer 
dir3 = Path(__file__).parent.absolute() #See @Arminius answer 

print(f'dir1={dir1}\ndir2={dir2}\ndir3={dir3}')

Explanation links: .resolve(), .absolute(), Path(file).parent().absolute()


回答 7

系统:MacOS

版本:Python 3.6 w / Anaconda

import os

rootpath = os.getcwd()

os.chdir(rootpath)

System: MacOS

Version: Python 3.6 w/ Anaconda

import os

rootpath = os.getcwd()

os.chdir(rootpath)

回答 8

PYTHON中的有用路径属性:

 from pathlib import Path

    #Returns the path of the directory, where your script file is placed
    mypath = Path().absolute()
    print('Absolute path : {}'.format(mypath))

    #if you want to go to any other file inside the subdirectories of the directory path got from above method
    filePath = mypath/'data'/'fuel_econ.csv'
    print('File path : {}'.format(filePath))

    #To check if file present in that directory or Not
    isfileExist = filePath.exists()
    print('isfileExist : {}'.format(isfileExist))

    #To check if the path is a directory or a File
    isadirectory = filePath.is_dir()
    print('isadirectory : {}'.format(isadirectory))

    #To get the extension of the file
    fileExtension = mypath/'data'/'fuel_econ.csv'
    print('File extension : {}'.format(filePath.suffix))

输出: 绝对路径是放置Python文件的路径

绝对路径:D:\ Study \ Machine Learning \ Jupitor Notebook \ JupytorNotebookTest2 \ Udacity_Scripts \ Matplotlib和seaborn Part2

文件路径:D:\ Study \ Machine Learning \ Jupitor Notebook \ JupytorNotebookTest2 \ Udacity_Scripts \ Matplotlib和seaborn Part2 \ data \ fuel_econ.csv

isfileExist:真

isadirectory:错误

文件扩展名:.csv

USEFUL PATH PROPERTIES IN PYTHON:

 from pathlib import Path

    #Returns the path of the directory, where your script file is placed
    mypath = Path().absolute()
    print('Absolute path : {}'.format(mypath))

    #if you want to go to any other file inside the subdirectories of the directory path got from above method
    filePath = mypath/'data'/'fuel_econ.csv'
    print('File path : {}'.format(filePath))

    #To check if file present in that directory or Not
    isfileExist = filePath.exists()
    print('isfileExist : {}'.format(isfileExist))

    #To check if the path is a directory or a File
    isadirectory = filePath.is_dir()
    print('isadirectory : {}'.format(isadirectory))

    #To get the extension of the file
    fileExtension = mypath/'data'/'fuel_econ.csv'
    print('File extension : {}'.format(filePath.suffix))

OUTPUT: ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED

Absolute path : D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2

File path : D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2\data\fuel_econ.csv

isfileExist : True

isadirectory : False

File extension : .csv


回答 9

如果您只想查看当前的工作目录

import os
print(os.getcwd)

如果要更改当前工作目录

os.chdir(path)

path是一个字符串,其中包含要移动的所需路径。例如

path = "C:\\Users\\xyz\\Desktop\\move here"

If you just want to see the current working directory

import os
print(os.getcwd)

If you want to change the current working directory

os.chdir(path)

path is a string containing the required path to be moved. e.g.

path = "C:\\Users\\xyz\\Desktop\\move here"

回答 10

IPython有一个神奇的命令%pwd来获取当前的工作目录。它可以按以下方式使用:

from IPython.terminal.embed import InteractiveShellEmbed

ip_shell = InteractiveShellEmbed()

present_working_directory = ip_shell.magic("%pwd")

在IPython Jupyter Notebook上%pwd可以直接使用,如下所示:

present_working_directory = %pwd

IPython has a magic command %pwd to get the present working directory. It can be used in following way:

from IPython.terminal.embed import InteractiveShellEmbed

ip_shell = InteractiveShellEmbed()

present_working_directory = ip_shell.magic("%pwd")

On IPython Jupyter Notebook %pwd can be used directly as following:

present_working_directory = %pwd

回答 11

要保持跨平台(macOS / Windows / Linux)的迁移一致性,请尝试:

path = r'%s' % os.getcwd().replace('\\','/')

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\\','/')

回答 12

为了获取当前文件夹,我已经在CGI的IIS下运行python时使用了一个函数:

import os 
   def getLocalFolder():
        path=str(os.path.dirname(os.path.abspath(__file__))).split('\\')
        return path[len(path)-1]

I have made a function to use when running python under IIS in CGI in order to get the current folder:

import os 
   def getLocalFolder():
        path=str(os.path.dirname(os.path.abspath(__file__))).split('\\')
        return path[len(path)-1]

回答 13

假设您具有以下目录结构:-

主/折1折2折3 …

folders = glob.glob("main/fold*")

for fold in folders:
    abspath = os.path.dirname(os.path.abspath(fold))
    fullpath = os.path.join(abspath, sch)
    print(fullpath)

Let’s assume you have the following directory structure: –

main/ fold1 fold2 fold3…

folders = glob.glob("main/fold*")

for fold in folders:
    abspath = os.path.dirname(os.path.abspath(fold))
    fullpath = os.path.join(abspath, sch)
    print(fullpath)

回答 14

## IMPORT MODULES
import os

## CALCULATE FILEPATH VARIABLE
filepath = os.path.abspath('') ## ~ os.getcwd()
## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..
## ..OR DIFFERENT IN SOME CIRCUMSTANCES
## IMPORT MODULES
import os

## CALCULATE FILEPATH VARIABLE
filepath = os.path.abspath('') ## ~ os.getcwd()
## TEST TO MAKE SURE os.getcwd() is EQUIVALENT ALWAYS..
## ..OR DIFFERENT IN SOME CIRCUMSTANCES

如何删除文件或文件夹?

问题:如何删除文件或文件夹?

如何在Python中删除文件或文件夹?

How to delete a file or folder in Python?


回答 0


PathPython 3.4+ pathlib模块中的对象还公开了这些实例方法:


Path objects from the Python 3.4+ pathlib module also expose these instance methods:


回答 1

Python语法删除文件

import os
os.remove("/tmp/<file_name>.txt")

要么

import os
os.unlink("/tmp/<file_name>.txt")

要么

适用于Python版本> 3.5的pathlib

file_to_rem = pathlib.Path("/tmp/<file_name>.txt")
file_to_rem.unlink()

Path.unlink(missing_ok = False)

Unlink方法用于删除文件或符号链接。

如果missing_ok为false(默认值),则在路径不存在时引发FileNotFoundError。
如果missing_ok为true,则将忽略FileNotFoundError异常(与POSIX rm -f命令相同的行为)。
在版本3.8中更改:添加了missing_ok参数。

最佳实践

  1. 首先,检查文件或文件夹是否存在,然后仅删除该文件。这可以通过两种方式实现:
    一。os.path.isfile("/path/to/file")
    b。采用exception handling.

实例os.path.isfile

#!/usr/bin/python
import os
myfile="/tmp/foo.txt"

## If file exists, delete it ##
if os.path.isfile(myfile):
    os.remove(myfile)
else:    ## Show an error ##
    print("Error: %s file not found" % myfile)

异常处理

#!/usr/bin/python
import os

## Get input ##
myfile= raw_input("Enter file name to delete: ")

## Try to delete the file ##
try:
    os.remove(myfile)
except OSError as e:  ## if failed, report it back to the user ##
    print ("Error: %s - %s." % (e.filename, e.strerror))

预期输出

输入要删除的文件名:demo.txt
错误:demo.txt-没有这样的文件或目录。

输入要删除的文件名:rrr.txt
错误:rrr.txt-不允许操作。

输入要删除的文件名:foo.txt

删除文件夹的Python语法

shutil.rmtree()

范例 shutil.rmtree()

#!/usr/bin/python
import os
import sys
import shutil

# Get directory name
mydir= raw_input("Enter directory name: ")

## Try to remove tree; if failed show an error using try...except on screen
try:
    shutil.rmtree(mydir)
except OSError as e:
    print ("Error: %s - %s." % (e.filename, e.strerror))

Python syntax to delete a file

import os
os.remove("/tmp/<file_name>.txt")

Or

import os
os.unlink("/tmp/<file_name>.txt")

Or

pathlib Library for Python version > 3.5

file_to_rem = pathlib.Path("/tmp/<file_name>.txt")
file_to_rem.unlink()

Path.unlink(missing_ok=False)

Unlink method used to remove the file or the symbolik link.

If missing_ok is false (the default), FileNotFoundError is raised if the path does not exist.
If missing_ok is true, FileNotFoundError exceptions will be ignored (same behavior as the POSIX rm -f command).
Changed in version 3.8: The missing_ok parameter was added.

Best practice

  1. First, check whether the file or folder exists or not then only delete that file. This can be achieved in two ways :
    a. os.path.isfile("/path/to/file")
    b. Use exception handling.

EXAMPLE for os.path.isfile

#!/usr/bin/python
import os
myfile="/tmp/foo.txt"

## If file exists, delete it ##
if os.path.isfile(myfile):
    os.remove(myfile)
else:    ## Show an error ##
    print("Error: %s file not found" % myfile)

Exception Handling

#!/usr/bin/python
import os

## Get input ##
myfile= raw_input("Enter file name to delete: ")

## Try to delete the file ##
try:
    os.remove(myfile)
except OSError as e:  ## if failed, report it back to the user ##
    print ("Error: %s - %s." % (e.filename, e.strerror))

RESPECTIVE OUTPUT

Enter file name to delete : demo.txt
Error: demo.txt - No such file or directory.

Enter file name to delete : rrr.txt
Error: rrr.txt - Operation not permitted.

Enter file name to delete : foo.txt

Python syntax to delete a folder

shutil.rmtree()

Example for shutil.rmtree()

#!/usr/bin/python
import os
import sys
import shutil

# Get directory name
mydir= raw_input("Enter directory name: ")

## Try to remove tree; if failed show an error using try...except on screen
try:
    shutil.rmtree(mydir)
except OSError as e:
    print ("Error: %s - %s." % (e.filename, e.strerror))

回答 2

采用

shutil.rmtree(path[, ignore_errors[, onerror]])

(请参阅关于shutil的完整文档)和/或

os.remove

os.rmdir

(关于os的完整文档。)

Use

shutil.rmtree(path[, ignore_errors[, onerror]])

(See complete documentation on shutil) and/or

os.remove

and

os.rmdir

(Complete documentation on os.)


回答 3

这是同时使用os.remove和的强大功能shutil.rmtree

def remove(path):
    """ param <path> could either be relative or absolute. """
    if os.path.isfile(path) or os.path.islink(path):
        os.remove(path)  # remove the file
    elif os.path.isdir(path):
        shutil.rmtree(path)  # remove dir and all contains
    else:
        raise ValueError("file {} is not a file or dir.".format(path))

Here is a robust function that uses both os.remove and shutil.rmtree:

def remove(path):
    """ param <path> could either be relative or absolute. """
    if os.path.isfile(path) or os.path.islink(path):
        os.remove(path)  # remove the file
    elif os.path.isdir(path):
        shutil.rmtree(path)  # remove dir and all contains
    else:
        raise ValueError("file {} is not a file or dir.".format(path))

回答 4

您可以使用内置的pathlib模块(需要Python 3.4+,但也有旧版本PyPI上的反向移植:pathlibpathlib2)。

要删除文件,可以使用以下unlink方法:

import pathlib
path = pathlib.Path(name_of_file)
path.unlink()

rmdir删除文件夹的方法:

import pathlib
path = pathlib.Path(name_of_folder)
path.rmdir()

You can use the built-in pathlib module (requires Python 3.4+, but there are backports for older versions on PyPI: pathlib, pathlib2).

To remove a file there is the unlink method:

import pathlib
path = pathlib.Path(name_of_file)
path.unlink()

Or the rmdir method to remove an empty folder:

import pathlib
path = pathlib.Path(name_of_folder)
path.rmdir()

回答 5

如何在Python中删除文件或文件夹?

对于Python 3,要分别删除文件和目录,请分别使用unlink和对象方法:rmdir Path

from pathlib import Path
dir_path = Path.home() / 'directory' 
file_path = dir_path / 'file'

file_path.unlink() # remove file

dir_path.rmdir()   # remove directory

请注意,您还可以将相对路径与Path对象一起使用,并且可以使用来检查当前的工作目录Path.cwd

要在Python 2中删除单个文件和目录,请参见下面标记的部分。

要删除包含目录的目录,请使用shutil.rmtree,请注意,该目录在Python 2和3中可用:

from shutil import rmtree

rmtree(dir_path)

示范

Path对象是Python 3.4中的新增功能。

让我们用一个目录和文件来演示用法。请注意,我们使用/来连接路径的各个部分,这解决了操作系统之间的问题以及Windows上使用反斜杠(在其中您需要将反斜杠加倍,\\或者使用原始字符串,如r"foo\bar")引起的问题:

from pathlib import Path

# .home() is new in 3.5, otherwise use os.path.expanduser('~')
directory_path = Path.home() / 'directory'
directory_path.mkdir()

file_path = directory_path / 'file'
file_path.touch()

现在:

>>> file_path.is_file()
True

现在让我们删除它们。首先文件:

>>> file_path.unlink()     # remove file
>>> file_path.is_file()
False
>>> file_path.exists()
False

我们可以使用通配符删除多个文件-首先,我们为此创建一些文件:

>>> (directory_path / 'foo.my').touch()
>>> (directory_path / 'bar.my').touch()

然后只需遍历全局模式:

>>> for each_file_path in directory_path.glob('*.my'):
...     print(f'removing {each_file_path}')
...     each_file_path.unlink()
... 
removing ~/directory/foo.my
removing ~/directory/bar.my

现在,演示删除目录:

>>> directory_path.rmdir() # remove directory
>>> directory_path.is_dir()
False
>>> directory_path.exists()
False

如果我们要删除目录及其中的所有内容怎么办?对于此用例,请使用shutil.rmtree

让我们重新创建目录和文件:

file_path.parent.mkdir()
file_path.touch()

并注意rmdir除非它为空,否则它将失败,这就是rmtree如此方便的原因:

>>> directory_path.rmdir()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "~/anaconda3/lib/python3.6/pathlib.py", line 1270, in rmdir
    self._accessor.rmdir(self)
  File "~/anaconda3/lib/python3.6/pathlib.py", line 387, in wrapped
    return strfunc(str(pathobj), *args)
OSError: [Errno 39] Directory not empty: '/home/username/directory'

现在,导入rmtree并将目录传递给该功能:

from shutil import rmtree
rmtree(directory_path)      # remove everything 

我们可以看到整个内容已被删除:

>>> directory_path.exists()
False

Python 2

如果您使用的是Python 2,则有一个名为pathlib2的pathlib模块的反向端口,可以使用pip进行安装:

$ pip install pathlib2

然后您可以将库别名为 pathlib

import pathlib2 as pathlib

或者直接导入Path对象(如此处所示):

from pathlib2 import Path

如果太多,您可以使用删除文件os.removeos.unlink

from os import unlink, remove
from os.path import join, expanduser

remove(join(expanduser('~'), 'directory/file'))

要么

unlink(join(expanduser('~'), 'directory/file'))

您可以使用以下命令删除目录os.rmdir

from os import rmdir

rmdir(join(expanduser('~'), 'directory'))

请注意,还有一个os.removedirs-它仅以递归方式删除空目录,但它可能适合您的用例。

How do I delete a file or folder in Python?

For Python 3, to remove the file and directory individually, use the unlink and rmdir Path object methods respectively:

from pathlib import Path
dir_path = Path.home() / 'directory' 
file_path = dir_path / 'file'

file_path.unlink() # remove file

dir_path.rmdir()   # remove directory

Note that you can also use relative paths with Path objects, and you can check your current working directory with Path.cwd.

For removing individual files and directories in Python 2, see the section so labeled below.

To remove a directory with contents, use shutil.rmtree, and note that this is available in Python 2 and 3:

from shutil import rmtree

rmtree(dir_path)

Demonstration

New in Python 3.4 is the Path object.

Let’s use one to create a directory and file to demonstrate usage. Note that we use the / to join the parts of the path, this works around issues between operating systems and issues from using backslashes on Windows (where you’d need to either double up your backslashes like \\ or use raw strings, like r"foo\bar"):

from pathlib import Path

# .home() is new in 3.5, otherwise use os.path.expanduser('~')
directory_path = Path.home() / 'directory'
directory_path.mkdir()

file_path = directory_path / 'file'
file_path.touch()

and now:

>>> file_path.is_file()
True

Now let’s delete them. First the file:

>>> file_path.unlink()     # remove file
>>> file_path.is_file()
False
>>> file_path.exists()
False

We can use globbing to remove multiple files – first let’s create a few files for this:

>>> (directory_path / 'foo.my').touch()
>>> (directory_path / 'bar.my').touch()

Then just iterate over the glob pattern:

>>> for each_file_path in directory_path.glob('*.my'):
...     print(f'removing {each_file_path}')
...     each_file_path.unlink()
... 
removing ~/directory/foo.my
removing ~/directory/bar.my

Now, demonstrating removing the directory:

>>> directory_path.rmdir() # remove directory
>>> directory_path.is_dir()
False
>>> directory_path.exists()
False

What if we want to remove a directory and everything in it? For this use-case, use shutil.rmtree

Let’s recreate our directory and file:

file_path.parent.mkdir()
file_path.touch()

and note that rmdir fails unless it’s empty, which is why rmtree is so convenient:

>>> directory_path.rmdir()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "~/anaconda3/lib/python3.6/pathlib.py", line 1270, in rmdir
    self._accessor.rmdir(self)
  File "~/anaconda3/lib/python3.6/pathlib.py", line 387, in wrapped
    return strfunc(str(pathobj), *args)
OSError: [Errno 39] Directory not empty: '/home/username/directory'

Now, import rmtree and pass the directory to the funtion:

from shutil import rmtree
rmtree(directory_path)      # remove everything 

and we can see the whole thing has been removed:

>>> directory_path.exists()
False

Python 2

If you’re on Python 2, there’s a backport of the pathlib module called pathlib2, which can be installed with pip:

$ pip install pathlib2

And then you can alias the library to pathlib

import pathlib2 as pathlib

Or just directly import the Path object (as demonstrated here):

from pathlib2 import Path

If that’s too much, you can remove files with os.remove or os.unlink

from os import unlink, remove
from os.path import join, expanduser

remove(join(expanduser('~'), 'directory/file'))

or

unlink(join(expanduser('~'), 'directory/file'))

and you can remove directories with os.rmdir:

from os import rmdir

rmdir(join(expanduser('~'), 'directory'))

Note that there is also a os.removedirs – it only removes empty directories recursively, but it may suit your use-case.


回答 6

import os

folder = '/Path/to/yourDir/'
fileList = os.listdir(folder)

for f in fileList:
    filePath = folder + '/'+f

    if os.path.isfile(filePath):
        os.remove(filePath)

    elif os.path.isdir(filePath):
        newFileList = os.listdir(filePath)
        for f1 in newFileList:
            insideFilePath = filePath + '/' + f1

            if os.path.isfile(insideFilePath):
                os.remove(insideFilePath)
import os

folder = '/Path/to/yourDir/'
fileList = os.listdir(folder)

for f in fileList:
    filePath = folder + '/'+f

    if os.path.isfile(filePath):
        os.remove(filePath)

    elif os.path.isdir(filePath):
        newFileList = os.listdir(filePath)
        for f1 in newFileList:
            insideFilePath = filePath + '/' + f1

            if os.path.isfile(insideFilePath):
                os.remove(insideFilePath)

回答 7

shutil.rmtree是异步函数,因此,如果要检查它是否完成,可以使用while … loop

import os
import shutil

shutil.rmtree(path)

while os.path.exists(path):
  pass

print('done')

shutil.rmtree is the asynchronous function, so if you want to check when it complete, you can use while…loop

import os
import shutil

shutil.rmtree(path)

while os.path.exists(path):
  pass

print('done')

回答 8

删除文件:

os.unlink(path, *, dir_fd=None)

要么

os.remove(path, *, dir_fd=None)

这两个功能在语义上是相同的。此功能删除(删除)文件路径。如果path不是文件,而是目录,则会引发异常。

删除文件夹:

shutil.rmtree(path, ignore_errors=False, onerror=None)

要么

os.rmdir(path, *, dir_fd=None)

为了删除整个目录树,shutil.rmtree()可以使用。os.rmdir仅在目录为空且存在时才起作用。

要递归删除父文件夹:

os.removedirs(name)

它用self删除每个空的父目录,直到有一些内容的父目录为止

例如 os.removedirs(’abc / xyz / pqr’)如果目录为空,则会按顺序abc / xyz / pqr,abc / xyz和abc删除目录。

欲了解更多信息检查官方文档:os.unlinkos.removeos.rmdirshutil.rmtreeos.removedirs

For deleting files:

os.unlink(path, *, dir_fd=None)

or

os.remove(path, *, dir_fd=None)

Both functions are semantically same. This functions removes (deletes) the file path. If path is not a file and it is directory, then exception is raised.

For deleting folders:

shutil.rmtree(path, ignore_errors=False, onerror=None)

or

os.rmdir(path, *, dir_fd=None)

In order to remove whole directory trees, shutil.rmtree() can be used. os.rmdir only works when the directory is empty and exists.

For deleting folders recursively towards parent:

os.removedirs(name)

It remove every empty parent directory with self until parent which has some content

ex. os.removedirs(‘abc/xyz/pqr’) will remove the directories by order ‘abc/xyz/pqr’, ‘abc/xyz’ and ‘abc’ if they are empty.

For more info check official doc: os.unlink , os.remove, os.rmdir , shutil.rmtree, os.removedirs


回答 9

删除文件夹中的所有文件

import os
import glob

files = glob.glob(os.path.join('path/to/folder/*'))
files = glob.glob(os.path.join('path/to/folder/*.csv')) // It will give all csv files in folder
for file in files:
    os.remove(file)

删除目录中的所有文件夹

from shutil import rmtree
import os

// os.path.join()  # current working directory.

for dirct in os.listdir(os.path.join('path/to/folder')):
    rmtree(os.path.join('path/to/folder',dirct))

To remove all files in folder

import os
import glob

files = glob.glob(os.path.join('path/to/folder/*'))
files = glob.glob(os.path.join('path/to/folder/*.csv')) // It will give all csv files in folder
for file in files:
    os.remove(file)

To remove all folders in a directory

from shutil import rmtree
import os

// os.path.join()  # current working directory.

for dirct in os.listdir(os.path.join('path/to/folder')):
    rmtree(os.path.join('path/to/folder',dirct))

回答 10

为了避免ÉricAraujo 的注释突出显示的TOCTOU问题,您可以捕获异常以调用正确的方法:

def remove_file_or_dir(path: str) -> None:
    """ Remove a file or directory """
    try:
        shutil.rmtree(path)
    except NotADirectoryError:
        os.remove(path)

因为shutil.rmtree()将仅删除目录,os.remove()或者os.unlink()仅将删除文件。

To avoid the TOCTOU issue highlighted by Éric Araujo’s comment, you can catch an exception to call the correct method:

def remove_file_or_dir(path: str) -> None:
    """ Remove a file or directory """
    try:
        shutil.rmtree(path)
    except NotADirectoryError:
        os.remove(path)

Since shutil.rmtree() will only remove directories and os.remove() or os.unlink() will only remove files.


回答 11

subprocess如果您喜欢编写漂亮且易读的代码,那么我建议您使用:

import subprocess
subprocess.Popen("rm -r my_dir", shell=True)

而且,如果您不是软件工程师,那么可以考虑使用Jupyter。您可以简单地输入bash命令:

!rm -r my_dir

传统上,您使用shutil

import shutil
shutil.rmtree(my_dir) 

I recommend using subprocess if writing a beautiful and readable code is your cup of tea:

import subprocess
subprocess.Popen("rm -r my_dir", shell=True)

And if you are not a software engineer, then maybe consider using Jupyter; you can simply type bash commands:

!rm -r my_dir

Traditionally, you use shutil:

import shutil
shutil.rmtree(my_dir) 

如何查找Python中是否存在目录

问题:如何查找Python中是否存在目录

osPython模块中,有一种方法可以查找目录是否存在,例如:

>>> os.direxists(os.path.join(os.getcwd()), 'new_folder')) # in pseudocode
True/False

In the os module in Python, is there a way to find if a directory exists, something like:

>>> os.direxists(os.path.join(os.getcwd()), 'new_folder')) # in pseudocode
True/False

回答 0

您正在寻找os.path.isdir,还是os.path.exists不在乎它是文件还是目录。

例:

import os
print(os.path.isdir("/home/el"))
print(os.path.exists("/home/el/myfile.txt"))

You’re looking for os.path.isdir, or os.path.exists if you don’t care whether it’s a file or a directory.

Example:

import os
print(os.path.isdir("/home/el"))
print(os.path.exists("/home/el/myfile.txt"))

回答 1

很近!如果传入当前存在的目录名,则os.path.isdir返回True。如果不存在或不是目录,则返回False

So close! os.path.isdir returns True if you pass in the name of a directory that currently exists. If it doesn’t exist or it’s not a directory, then it returns False.


回答 2

蟒3.4引入pathlib模块到标准库,它提供了一个面向对象的方法来处理的文件系统的路径。对象的is_dir()exists()方法Path可用于回答以下问题:

In [1]: from pathlib import Path

In [2]: p = Path('/usr')

In [3]: p.exists()
Out[3]: True

In [4]: p.is_dir()
Out[4]: True

路径(和字符串)可以与/运算符连接在一起:

In [5]: q = p / 'bin' / 'vim'

In [6]: q
Out[6]: PosixPath('/usr/bin/vim') 

In [7]: q.exists()
Out[7]: True

In [8]: q.is_dir()
Out[8]: False

也可以通过PyPi上的pathlib2模块在 Python 2.7 上使用Pathlib。

Python 3.4 introduced the pathlib module into the standard library, which provides an object oriented approach to handle filesystem paths. The is_dir() and exists() methods of a Path object can be used to answer the question:

In [1]: from pathlib import Path

In [2]: p = Path('/usr')

In [3]: p.exists()
Out[3]: True

In [4]: p.is_dir()
Out[4]: True

Paths (and strings) can be joined together with the / operator:

In [5]: q = p / 'bin' / 'vim'

In [6]: q
Out[6]: PosixPath('/usr/bin/vim') 

In [7]: q.exists()
Out[7]: True

In [8]: q.is_dir()
Out[8]: False

Pathlib is also available on Python 2.7 via the pathlib2 module on PyPi.


回答 3

是的,请使用os.path.exists()


回答 4

我们可以检查2个内置函数

os.path.isdir("directory")

它将布尔值为true,指定的目录可用。

os.path.exists("directoryorfile")

如果指定的目录或文件可用,它将使boolead为true。

检查路径是否为目录;

os.path.isdir("directorypath")

如果路径为目录,则将为布尔值true

We can check with 2 built in functions

os.path.isdir("directory")

It will give boolean true the specified directory is available.

os.path.exists("directoryorfile")

It will give boolead true if specified directory or file is available.

To check whether the path is directory;

os.path.isdir("directorypath")

will give boolean true if the path is directory


回答 5

是的,使用os.path.isdir(path)


回答 6

如:

In [3]: os.path.exists('/d/temp')
Out[3]: True

可能会折腾os.path.isdir(...)以确保。

As in:

In [3]: os.path.exists('/d/temp')
Out[3]: True

Probably toss in a os.path.isdir(...) to be sure.


回答 7

仅提供os.stat版本(python 2):

import os, stat, errno
def CheckIsDir(directory):
  try:
    return stat.S_ISDIR(os.stat(directory).st_mode)
  except OSError, e:
    if e.errno == errno.ENOENT:
      return False
    raise

Just to provide the os.stat version (python 2):

import os, stat, errno
def CheckIsDir(directory):
  try:
    return stat.S_ISDIR(os.stat(directory).st_mode)
  except OSError, e:
    if e.errno == errno.ENOENT:
      return False
    raise

回答 8

os为您提供了许多以下功能:

import os
os.path.isdir(dir_in) #True/False: check if this is a directory
os.listdir(dir_in)    #gets you a list of all files and directories under dir_in

如果输入路径无效,则listdir将引发异常。

os provides you with a lot of these capabilities:

import os
os.path.isdir(dir_in) #True/False: check if this is a directory
os.listdir(dir_in)    #gets you a list of all files and directories under dir_in

the listdir will throw an exception if the input path is invalid.


回答 9

#You can also check it get help for you

if not os.path.isdir('mydir'):
    print('new directry has been created')
    os.system('mkdir mydir')
#You can also check it get help for you

if not os.path.isdir('mydir'):
    print('new directry has been created')
    os.system('mkdir mydir')

回答 10

有一个方便的Unipath模块。

>>> from unipath import Path 
>>>  
>>> Path('/var/log').exists()
True
>>> Path('/var/log').isdir()
True

您可能需要的其他相关事项:

>>> Path('/var/log/system.log').parent
Path('/var/log')
>>> Path('/var/log/system.log').ancestor(2)
Path('/var')
>>> Path('/var/log/system.log').listdir()
[Path('/var/foo'), Path('/var/bar')]
>>> (Path('/var/log') + '/system.log').isfile()
True

您可以使用pip安装它:

$ pip3 install unipath

它类似于内置的pathlib。不同之处在于,它将每个路径都视为字符串(Path是的子类str),因此,如果某些函数需要字符串,则可以轻松地将其传递给Path对象,而无需将其转换为字符串。

例如,这在Django和下非常有用settings.py

# settings.py
BASE_DIR = Path(__file__).ancestor(2)
STATIC_ROOT = BASE_DIR + '/tmp/static'

There is a convenient Unipath module.

>>> from unipath import Path 
>>>  
>>> Path('/var/log').exists()
True
>>> Path('/var/log').isdir()
True

Other related things you might need:

>>> Path('/var/log/system.log').parent
Path('/var/log')
>>> Path('/var/log/system.log').ancestor(2)
Path('/var')
>>> Path('/var/log/system.log').listdir()
[Path('/var/foo'), Path('/var/bar')]
>>> (Path('/var/log') + '/system.log').isfile()
True

You can install it using pip:

$ pip3 install unipath

It’s similar to the built-in pathlib. The difference is that it treats every path as a string (Path is a subclass of the str), so if some function expects a string, you can easily pass it a Path object without a need to convert it to a string.

For example, this works great with Django and settings.py:

# settings.py
BASE_DIR = Path(__file__).ancestor(2)
STATIC_ROOT = BASE_DIR + '/tmp/static'

回答 11

如果目录不存在,您可能还想创建该目录。

Source,如果它仍在SO上。

================================================== ===================

在Python≥3.5上,使用pathlib.Path.mkdir

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

对于旧版本的Python,我看到两个质量很好的答案,每个都有一个小缺陷,因此我将对此进行说明:

试试看os.path.exists,然后考虑os.makedirs创建。

import os
if not os.path.exists(directory):
    os.makedirs(directory)

如注释和其他地方所述,存在竞争条件–如果在os.path.existsos.makedirs调用之间创建目录,os.makedirs则将失败并显示OSError。不幸的是,毯式捕获OSError和继续操作并非万无一失,因为它将忽略由于其他因素(例如权限不足,磁盘已满等)而导致的目录创建失败。

一种选择是捕获OSError并检查嵌入式错误代码(请参阅是否存在从Python的OSError获取信息的跨平台方法):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

或者,可以有第二个os.path.exists,但是假设另一个在第一次检查后创建了目录,然后在第二个检查之前将其删除了–我们仍然可能会上当。

取决于应用程序,并发操作的危险可能比其他因素(例如文件许可权)造成的危险更大或更小。在选择实现之前,开发人员必须了解有关正在开发的特定应用程序及其预期环境的更多信息。

现代版本的Python通过暴露FileExistsError(在3.3+ 版本中)都极大地改善了此代码。

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

…并允许关键字参数os.makedirs调用exist_ok(在3.2+版本中)。

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

You may also want to create the directory if it’s not there.

Source, if it’s still there on SO.

=====================================================================

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there’s a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)…

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

…and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

回答 12

两件事情

  1. 检查目录是否存在?
  2. 如果不是,请创建目录(可选)。
import os
dirpath = "<dirpath>" # Replace the "<dirpath>" with actual directory path.

if os.path.exists(dirpath):
   print("Directory exist")
else: #this is optional if you want to create a directory if doesn't exist.
   os.mkdir(dirpath):
   print("Directory created")

Two things

  1. check if the directory exist?
  2. if not, create a directory (optional).
import os
dirpath = "<dirpath>" # Replace the "<dirpath>" with actual directory path.

if os.path.exists(dirpath):
   print("Directory exist")
else: #this is optional if you want to create a directory if doesn't exist.
   os.mkdir(dirpath):
   print("Directory created")

如何安全地创建嵌套目录?

问题:如何安全地创建嵌套目录?

检查文件目录是否存在的最优雅方法是什么,如果不存在,则使用Python创建目录?这是我尝试过的:

import os

file_path = "/my/directory/filename.txt"
directory = os.path.dirname(file_path)

try:
    os.stat(directory)
except:
    os.mkdir(directory)       

f = file(filename)

不知何故,我想念os.path.exists(感谢魔芋,布莱尔和道格拉斯)。这就是我现在所拥有的:

def ensure_dir(file_path):
    directory = os.path.dirname(file_path)
    if not os.path.exists(directory):
        os.makedirs(directory)

是否有“打开”标志,使它自动发生?

What is the most elegant way to check if the directory a file is going to be written to exists, and if not, create the directory using Python? Here is what I tried:

import os

file_path = "/my/directory/filename.txt"
directory = os.path.dirname(file_path)

try:
    os.stat(directory)
except:
    os.mkdir(directory)       

f = file(filename)

Somehow, I missed os.path.exists (thanks kanja, Blair, and Douglas). This is what I have now:

def ensure_dir(file_path):
    directory = os.path.dirname(file_path)
    if not os.path.exists(directory):
        os.makedirs(directory)

Is there a flag for “open”, that makes this happen automatically?


回答 0

在Python≥3.5上,使用pathlib.Path.mkdir

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

对于旧版本的Python,我看到两个质量很好的答案,每个都有一个小缺陷,因此我将对此进行说明:

试试看os.path.exists,然后考虑os.makedirs创建。

import os
if not os.path.exists(directory):
    os.makedirs(directory)

如注释和其他地方所述,存在竞争条件–如果在os.path.existsos.makedirs调用之间创建目录,os.makedirs则将失败并显示OSError。不幸的是,毯式捕获OSError和继续操作并非万无一失,因为它将忽略由于其他因素(例如权限不足,磁盘已满等)而导致的目录创建失败。

一种选择是捕获OSError并检查嵌入式错误代码(请参阅是否存在从Python的OSError获取信息的跨平台方法):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

或者,可以有第二个os.path.exists,但是假设另一个在第一次检查后创建了目录,然后在第二个检查之前将其删除了–我们仍然可能会上当。

取决于应用程序,并发操作的危险可能比其他因素(例如文件许可权)造成的危险更大或更小。在选择实现之前,开发人员必须了解有关正在开发的特定应用程序及其预期环境的更多信息。

现代版本的Python通过暴露FileExistsError(在3.3+ 版本中)都极大地改善了此代码。

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

…并允许关键字参数os.makedirs调用exist_ok(在3.2+版本中)。

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there’s a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)…

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

…and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

回答 1

Python 3.5以上版本:

import pathlib
pathlib.Path('/my/directory').mkdir(parents=True, exist_ok=True) 

pathlib.Path.mkdir上面使用的递归方式创建目录,并且如果目录已经存在,则不会引发异常。如果不需要或不希望创建父母,请跳过该parents参数。

Python 3.2+:

使用pathlib

如果可以,请安装pathlib名为的当前反向端口pathlib2。不要安装名为的较旧的未维护的反向端口pathlib。接下来,请参考上面的Python 3.5+部分,并对其进行相同的使用。

如果使用Python 3.4,即使它附带了pathlib,它也会丢失有用的exist_ok选项。反向端口旨在提供更新的高级实现,mkdir其中包括缺少的选项。

使用os

import os
os.makedirs(path, exist_ok=True)

os.makedirs上面使用的递归方式创建目录,并且如果目录已经存在,则不会引发异常。exist_ok仅当使用Python 3.2+时,它才具有可选参数,默认值为False。在2.7之前的Python 2.x中不存在此参数。这样,就不需要像Python 2.7那样的手动异常处理。

Python 2.7+:

使用pathlib

如果可以,请安装pathlib名为的当前反向端口pathlib2。不要安装名为的较旧的未维护的反向端口pathlib。接下来,请参考上面的Python 3.5+部分,并对其进行相同的使用。

使用os

import os
try: 
    os.makedirs(path)
except OSError:
    if not os.path.isdir(path):
        raise

虽然可能会先使用朴素的解决方案,os.path.isdir然后再使用os.makedirs,但是上述解决方案颠倒了两个操作的顺序。这样,它可以防止由于创建目录的重复尝试而导致的常见竞争情况,并且还可以消除目录中文件的歧义。

请注意,捕获异常和使用errno的作用有限,因为对于文件和目录,都会引发OSError: [Errno 17] File exists,即errno.EEXIST。仅检查目录是否存在更为可靠。

选择:

mkpath创建嵌套目录,如果目录已经存在,则不执行任何操作。这适用于Python 2和3。

import distutils.dir_util
distutils.dir_util.mkpath(path)

根据Bug 10948,此替代方案的严重局限性在于,对于给定路径,每个python进程仅工作一次。换句话说,如果您使用它来创建目录,然后从Python内部或外部删除该目录,然后mkpath再次mkpath使用它来重新创建同一目录,则将仅默默地使用其先前已创建目录的无效缓存信息,而不会实际再次创建目录。相反,os.makedirs不依赖任何此类缓存。对于某些应用程序,此限制可能是可以的。


关于目录的模式,如果您关心它,请参考文档。

Python 3.5+:

import pathlib
pathlib.Path('/my/directory').mkdir(parents=True, exist_ok=True) 

pathlib.Path.mkdir as used above recursively creates the directory and does not raise an exception if the directory already exists. If you don’t need or want the parents to be created, skip the parents argument.

Python 3.2+:

Using pathlib:

If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.

If using Python 3.4, even though it comes with pathlib, it is missing the useful exist_ok option. The backport is intended to offer a newer and superior implementation of mkdir which includes this missing option.

Using os:

import os
os.makedirs(path, exist_ok=True)

os.makedirs as used above recursively creates the directory and does not raise an exception if the directory already exists. It has the optional exist_ok argument only if using Python 3.2+, with a default value of False. This argument does not exist in Python 2.x up to 2.7. As such, there is no need for manual exception handling as with Python 2.7.

Python 2.7+:

Using pathlib:

If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.

Using os:

import os
try: 
    os.makedirs(path)
except OSError:
    if not os.path.isdir(path):
        raise

While a naive solution may first use os.path.isdir followed by os.makedirs, the solution above reverses the order of the two operations. In doing so, it prevents a common race condition having to do with a duplicated attempt at creating the directory, and also disambiguates files from directories.

Note that capturing the exception and using errno is of limited usefulness because OSError: [Errno 17] File exists, i.e. errno.EEXIST, is raised for both files and directories. It is more reliable simply to check if the directory exists.

Alternative:

mkpath creates the nested directory, and does nothing if the directory already exists. This works in both Python 2 and 3.

import distutils.dir_util
distutils.dir_util.mkpath(path)

Per Bug 10948, a severe limitation of this alternative is that it works only once per python process for a given path. In other words, if you use it to create a directory, then delete the directory from inside or outside Python, then use mkpath again to recreate the same directory, mkpath will simply silently use its invalid cached info of having previously created the directory, and will not actually make the directory again. In contrast, os.makedirs doesn’t rely on any such cache. This limitation may be okay for some applications.


With regard to the directory’s mode, please refer to the documentation if you care about it.


回答 2

使用tryexcept和来自errno模块的正确错误代码摆脱了竞争条件,并且是跨平台的:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

换句话说,我们尝试创建目录,但是如果它们已经存在,我们将忽略该错误。另一方面,将报告任何其他错误。例如,如果您预先创建目录’a’并从中删除所有权限,则会OSError引发errno.EACCES(权限被拒绝,错误13)。

Using try except and the right error code from errno module gets rid of the race condition and is cross-platform:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

In other words, we try to create the directories, but if they already exist we ignore the error. On the other hand, any other error gets reported. For example, if you create dir ‘a’ beforehand and remove all permissions from it, you will get an OSError raised with errno.EACCES (Permission denied, error 13).


回答 3

我个人建议您使用os.path.isdir()代替进行测试os.path.exists()

>>> os.path.exists('/tmp/dirname')
True
>>> os.path.exists('/tmp/dirname/filename.etc')
True
>>> os.path.isdir('/tmp/dirname/filename.etc')
False
>>> os.path.isdir('/tmp/fakedirname')
False

如果你有:

>>> dir = raw_input(":: ")

和愚蠢的用户输入:

:: /tmp/dirname/filename.etc

……你要与一个名为落得filename.etc当你传递参数os.makedirs(),如果你与测试os.path.exists()

I would personally recommend that you use os.path.isdir() to test instead of os.path.exists().

>>> os.path.exists('/tmp/dirname')
True
>>> os.path.exists('/tmp/dirname/filename.etc')
True
>>> os.path.isdir('/tmp/dirname/filename.etc')
False
>>> os.path.isdir('/tmp/fakedirname')
False

If you have:

>>> dir = raw_input(":: ")

And a foolish user input:

:: /tmp/dirname/filename.etc

… You’re going to end up with a directory named filename.etc when you pass that argument to os.makedirs() if you test with os.path.exists().


回答 4

检查os.makedirs:(确保存在完整路径。)
要处理目录可能存在的事实,请catch OSError。(如果exist_okFalse(缺省值),OSError则在目标目录已经存在时引发。)

import os
try:
    os.makedirs('./path/to/somewhere')
except OSError:
    pass

Check os.makedirs: (It makes sure the complete path exists.)
To handle the fact the directory might exist, catch OSError. (If exist_ok is False (the default), an OSError is raised if the target directory already exists.)

import os
try:
    os.makedirs('./path/to/somewhere')
except OSError:
    pass

回答 5

从Python 3.5开始,pathlib.Path.mkdir有一个exist_ok标志:

from pathlib import Path
path = Path('/my/directory/filename.txt')
path.parent.mkdir(parents=True, exist_ok=True) 
# path.parent ~ os.path.dirname(path)

这将以递归方式创建目录,并且如果目录已经存在,则不会引发异常。

(就像从python 3.2开始os.makedirsexist_ok标志一样os.makedirs(path, exist_ok=True)

Starting from Python 3.5, pathlib.Path.mkdir has an exist_ok flag:

from pathlib import Path
path = Path('/my/directory/filename.txt')
path.parent.mkdir(parents=True, exist_ok=True) 
# path.parent ~ os.path.dirname(path)

This recursively creates the directory and does not raise an exception if the directory already exists.

(just as os.makedirs got an exist_ok flag starting from python 3.2 e.g os.makedirs(path, exist_ok=True))


回答 6

对这种情况的具体见解

您在特定路径下提供特定文件,然后从文件路径中提取目录。然后,在确保您拥有目录之后,尝试打开一个文件进行读取。要对此代码发表评论:

filename = "/my/directory/filename.txt"
dir = os.path.dirname(filename)

我们要避免覆盖内置函数dir。另外,filepath或者也许fullfilepath是比它更好的语义名称,filename所以这样写会更好:

import os
filepath = '/my/directory/filename.txt'
directory = os.path.dirname(filepath)

您的最终目标是打开该文件,一开始就声明要写入,但是实际上您正在达到此目标(基于您的代码),就像这样,打开该文件进行读取

if not os.path.exists(directory):
    os.makedirs(directory)
f = file(filename)

假设开放阅读

为什么要为您希望存在并能够读取的文件创建目录?

只是尝试打开文件。

with open(filepath) as my_file:
    do_stuff(my_file)

如果目录或文件不存在,您将获得一个IOError带有相关错误代码的:errno.ENOENT无论您使用什么平台,它都将指向正确的错误代码。您可以根据需要捕获它,例如:

import errno
try:
    with open(filepath) as my_file:
        do_stuff(my_file)
except IOError as error:
    if error.errno == errno.ENOENT:
        print 'ignoring error because directory or file is not there'
    else:
        raise

假设我们正在写作

可能就是您想要的。

在这种情况下,我们可能没有面对任何比赛条件。因此,照原样进行操作,但请注意,编写时需要使用w模式打开(或a追加)。使用上下文管理器打开文件也是Python的最佳实践。

import os
if not os.path.exists(directory):
    os.makedirs(directory)
with open(filepath, 'w') as my_file:
    do_stuff(my_file)

但是,假设我们有几个Python进程试图将其所有数据放入同一目录。然后,我们可能会争执目录的创建。在这种情况下,最好将makedirs调用包装在try-except块中。

import os
import errno
if not os.path.exists(directory):
    try:
        os.makedirs(directory)
    except OSError as error:
        if error.errno != errno.EEXIST:
            raise
with open(filepath, 'w') as my_file:
    do_stuff(my_file)

Insights on the specifics of this situation

You give a particular file at a certain path and you pull the directory from the file path. Then after making sure you have the directory, you attempt to open a file for reading. To comment on this code:

filename = "/my/directory/filename.txt"
dir = os.path.dirname(filename)

We want to avoid overwriting the builtin function, dir. Also, filepath or perhaps fullfilepath is probably a better semantic name than filename so this would be better written:

import os
filepath = '/my/directory/filename.txt'
directory = os.path.dirname(filepath)

Your end goal is to open this file, you initially state, for writing, but you’re essentially approaching this goal (based on your code) like this, which opens the file for reading:

if not os.path.exists(directory):
    os.makedirs(directory)
f = file(filename)

Assuming opening for reading

Why would you make a directory for a file that you expect to be there and be able to read?

Just attempt to open the file.

with open(filepath) as my_file:
    do_stuff(my_file)

If the directory or file isn’t there, you’ll get an IOError with an associated error number: errno.ENOENT will point to the correct error number regardless of your platform. You can catch it if you want, for example:

import errno
try:
    with open(filepath) as my_file:
        do_stuff(my_file)
except IOError as error:
    if error.errno == errno.ENOENT:
        print 'ignoring error because directory or file is not there'
    else:
        raise

Assuming we’re opening for writing

This is probably what you’re wanting.

In this case, we probably aren’t facing any race conditions. So just do as you were, but note that for writing, you need to open with the w mode (or a to append). It’s also a Python best practice to use the context manager for opening files.

import os
if not os.path.exists(directory):
    os.makedirs(directory)
with open(filepath, 'w') as my_file:
    do_stuff(my_file)

However, say we have several Python processes that attempt to put all their data into the same directory. Then we may have contention over creation of the directory. In that case it’s best to wrap the makedirs call in a try-except block.

import os
import errno
if not os.path.exists(directory):
    try:
        os.makedirs(directory)
    except OSError as error:
        if error.errno != errno.EEXIST:
            raise
with open(filepath, 'w') as my_file:
    do_stuff(my_file)

回答 7

试用os.path.exists功能

if not os.path.exists(dir):
    os.mkdir(dir)

Try the os.path.exists function

if not os.path.exists(dir):
    os.mkdir(dir)

回答 8

我将以下内容放下。但是,这并非完全安全。

import os

dirname = 'create/me'

try:
    os.makedirs(dirname)
except OSError:
    if os.path.exists(dirname):
        # We are nearly safe
        pass
    else:
        # There was an error on creation, so make sure we know about it
        raise

现在,正如我所说,这并不是万无一失的,因为我们有可能无法创建目录,而在此期间可能会有另一个创建它的进程。

I have put the following down. It’s not totally foolproof though.

import os

dirname = 'create/me'

try:
    os.makedirs(dirname)
except OSError:
    if os.path.exists(dirname):
        # We are nearly safe
        pass
    else:
        # There was an error on creation, so make sure we know about it
        raise

Now as I say, this is not really foolproof, because we have the possiblity of failing to create the directory, and another process creating it during that period.


回答 9

检查目录是否存在并根据需要创建目录?

对此的直接答案是,假设有一个简单的情况,您不希望其他用户或进程弄乱您的目录:

if not os.path.exists(d):
    os.makedirs(d)

或者如果使目录符合竞争条件(即如果检查路径是否存在,则可能已经建立了其他路径),请执行以下操作:

import errno
try:
    os.makedirs(d)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise

但是,也许更好的方法是通过以下方式使用临时目录来避免资源争用问题tempfile

import tempfile

d = tempfile.mkdtemp()

以下是在线文档中的要点:

mkdtemp(suffix='', prefix='tmp', dir=None)
    User-callable function to create and return a unique temporary
    directory.  The return value is the pathname of the directory.

    The directory is readable, writable, and searchable only by the
    creating user.

    Caller is responsible for deleting the directory when done with it.

新的Python 3.5:pathlib.Pathexist_ok

有一个新的Path对象(从3.4版开始),它具有许多要与路径一起使用的方法-其中一个是mkdir

(在上下文中,我正在使用脚本跟踪我的每周代表。这是脚本中代码的相关部分,这些内容使我避免对同一数据每天多次遇到Stack Overflow。)

首先相关进口:

from pathlib import Path
import tempfile

我们现在不必处理os.path.join-只需将路径部分与结合起来即可/

directory = Path(tempfile.gettempdir()) / 'sodata'

然后,我确定地确保目录存在- exist_ok参数在Python 3.5中显示:

directory.mkdir(exist_ok=True)

这是文档的相关部分:

如果exist_ok为true,FileExistsErrorPOSIX mkdir -p仅当最后一个路径组件不是现有的非目录文件时,才会忽略异常(与命令相同的行为)。

这里还有更多脚本-就我而言,我不受竞争条件的影响,我只有一个进程希望目录(或包含的文件)存在,并且我没有任何尝试删除的过程目录。

todays_file = directory / str(datetime.datetime.utcnow().date())
if todays_file.exists():
    logger.info("todays_file exists: " + str(todays_file))
    df = pd.read_json(str(todays_file))

Path必须将对象强制转换为str其他期望str路径使用它们的API 。

也许应该更新Pandas以接受抽象基类的实例os.PathLike

Check if a directory exists and create it if necessary?

The direct answer to this is, assuming a simple situation where you don’t expect other users or processes to be messing with your directory:

if not os.path.exists(d):
    os.makedirs(d)

or if making the directory is subject to race conditions (i.e. if after checking the path exists, something else may have already made it) do this:

import errno
try:
    os.makedirs(d)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise

But perhaps an even better approach is to sidestep the resource contention issue, by using temporary directories via tempfile:

import tempfile

d = tempfile.mkdtemp()

Here’s the essentials from the online doc:

mkdtemp(suffix='', prefix='tmp', dir=None)
    User-callable function to create and return a unique temporary
    directory.  The return value is the pathname of the directory.

    The directory is readable, writable, and searchable only by the
    creating user.

    Caller is responsible for deleting the directory when done with it.

New in Python 3.5: pathlib.Path with exist_ok

There’s a new Path object (as of 3.4) with lots of methods one would want to use with paths – one of which is mkdir.

(For context, I’m tracking my weekly rep with a script. Here’s the relevant parts of code from the script that allow me to avoid hitting Stack Overflow more than once a day for the same data.)

First the relevant imports:

from pathlib import Path
import tempfile

We don’t have to deal with os.path.join now – just join path parts with a /:

directory = Path(tempfile.gettempdir()) / 'sodata'

Then I idempotently ensure the directory exists – the exist_ok argument shows up in Python 3.5:

directory.mkdir(exist_ok=True)

Here’s the relevant part of the documentation:

If exist_ok is true, FileExistsError exceptions will be ignored (same behavior as the POSIX mkdir -p command), but only if the last path component is not an existing non-directory file.

Here’s a little more of the script – in my case, I’m not subject to a race condition, I only have one process that expects the directory (or contained files) to be there, and I don’t have anything trying to remove the directory.

todays_file = directory / str(datetime.datetime.utcnow().date())
if todays_file.exists():
    logger.info("todays_file exists: " + str(todays_file))
    df = pd.read_json(str(todays_file))

Path objects have to be coerced to str before other APIs that expect str paths can use them.

Perhaps Pandas should be updated to accept instances of the abstract base class, os.PathLike.


回答 10

在Python 3.4中,您还可以使用全新的pathlib模块

from pathlib import Path
path = Path("/my/directory/filename.txt")
try:
    if not path.parent.exists():
        path.parent.mkdir(parents=True)
except OSError:
    # handle error; you can also catch specific errors like
    # FileExistsError and so on.

In Python 3.4 you can also use the brand new pathlib module:

from pathlib import Path
path = Path("/my/directory/filename.txt")
try:
    if not path.parent.exists():
        path.parent.mkdir(parents=True)
except OSError:
    # handle error; you can also catch specific errors like
    # FileExistsError and so on.

回答 11

相关的Python文档建议使用的编码风格(更容易请求原谅比许可)EAFP。这意味着代码

try:
    os.makedirs(path)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise
    else:
        print "\nBE CAREFUL! Directory %s already exists." % path

比替代品更好

if not os.path.exists(path):
    os.makedirs(path)
else:
    print "\nBE CAREFUL! Directory %s already exists." % path

该文档正是由于此问题中讨论的种族条件而提出了这一建议。此外,正如此处其他人所提到的,查询一次操作系统而不是两次查询操作系统具有性能优势。最后,在某些情况下(当开发人员知道应用程序正在运行的环境时),可能会提出支持第二个代码的参数,只有在特殊情况下才提倡该程序已为该程序建立了私有环境。本身(以及同一程序的其他实例)。

即使在这种情况下,这也是一种不好的做法,并且可能导致长时间的无用调试。例如,我们为目录设置权限的事实不应该使我们拥有为我们目的而适当设置的印象权限。可以使用其他权限挂载父目录。通常,程序应始终正常运行,并且程序员不应期望一个特定的环境。

The relevant Python documentation suggests the use of the EAFP coding style (Easier to Ask for Forgiveness than Permission). This means that the code

try:
    os.makedirs(path)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise
    else:
        print "\nBE CAREFUL! Directory %s already exists." % path

is better than the alternative

if not os.path.exists(path):
    os.makedirs(path)
else:
    print "\nBE CAREFUL! Directory %s already exists." % path

The documentation suggests this exactly because of the race condition discussed in this question. In addition, as others mention here, there is a performance advantage in querying once instead of twice the OS. Finally, the argument placed forward, potentially, in favour of the second code in some cases –when the developer knows the environment the application is running– can only be advocated in the special case that the program has set up a private environment for itself (and other instances of the same program).

Even in that case, this is a bad practice and can lead to long useless debugging. For example, the fact we set the permissions for a directory should not leave us with the impression permissions are set appropriately for our purposes. A parent directory could be mounted with other permissions. In general, a program should always work correctly and the programmer should not expect one specific environment.


回答 12

Python3中os.makedirs支持设置exist_ok。默认设置为False,这意味着OSError如果目标目录已存在,将引发。通过设置exist_okTrueOSError(目录存在)将被忽略,并且不会创建目录。

os.makedirs(path,exist_ok=True)

Python2中os.makedirs不支持设置exist_ok。您可以在heikki-toivonen的答案中使用该方法:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

In Python3, os.makedirs supports setting exist_ok. The default setting is False, which means an OSError will be raised if the target directory already exists. By setting exist_ok to True, OSError (directory exists) will be ignored and the directory will not be created.

os.makedirs(path,exist_ok=True)

In Python2, os.makedirs doesn’t support setting exist_ok. You can use the approach in heikki-toivonen’s answer:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

回答 13

对于单线解决方案,可以使用IPython.utils.path.ensure_dir_exists()

from IPython.utils.path import ensure_dir_exists
ensure_dir_exists(dir)

文档中确保目录存在。如果不存在,请尝试创建它,并在其他进程正在这样做的情况下防止出现竞争情况。

For a one-liner solution, you can use IPython.utils.path.ensure_dir_exists():

from IPython.utils.path import ensure_dir_exists
ensure_dir_exists(dir)

From the documentation: Ensure that a directory exists. If it doesn’t exist, try to create it and protect against a race condition if another process is doing the same.


回答 14

您可以使用 mkpath

# Create a directory and any missing ancestor directories. 
# If the directory already exists, do nothing.

from distutils.dir_util import mkpath
mkpath("test")    

请注意,它也会创建祖先目录。

它适用于Python 2和3。

You can use mkpath

# Create a directory and any missing ancestor directories. 
# If the directory already exists, do nothing.

from distutils.dir_util import mkpath
mkpath("test")    

Note that it will create the ancestor directories as well.

It works for Python 2 and 3.


回答 15

我使用os.path.exists()是一个Python 3脚本,可用于检查目录是否存在,如果目录不存在则创建一个,如果目录存在则将其删除(如果需要)。

它提示用户输入目录,并且可以轻松修改。

I use os.path.exists(), here is a Python 3 script that can be used to check if a directory exists, create one if it does not exist, and delete it if it does exist (if desired).

It prompts users for input of the directory and can be easily modified.


回答 16

您可以os.listdir为此使用:

import os
if 'dirName' in os.listdir('parentFolderPath')
    print('Directory Exists')

You can use os.listdir for this:

import os
if 'dirName' in os.listdir('parentFolderPath')
    print('Directory Exists')

回答 17

我找到了这个问题,起初我为自己遇到的一些失败和错误感到困惑。我正在使用Python 3(Arch Linux x86_64系统上的Anaconda虚拟环境中的v.3.5)。

考虑以下目录结构:

└── output/         ## dir
   ├── corpus       ## file
   ├── corpus2/     ## dir
   └── subdir/      ## dir

这是我的实验/注释,它们使事情变得清晰:

# ----------------------------------------------------------------------------
# [1] /programming/273192/how-can-i-create-a-directory-if-it-does-not-exist

import pathlib

""" Notes:
        1.  Include a trailing slash at the end of the directory path
            ("Method 1," below).
        2.  If a subdirectory in your intended path matches an existing file
            with same name, you will get the following error:
            "NotADirectoryError: [Errno 20] Not a directory:" ...
"""
# Uncomment and try each of these "out_dir" paths, singly:

# ----------------------------------------------------------------------------
# METHOD 1:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but no file created (os.makedirs creates dir, not files!  ;-)
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# [2] https://docs.python.org/3/library/os.html#os.makedirs

# Uncomment these to run "Method 1":

#directory = os.path.dirname(out_dir)
#os.makedirs(directory, mode=0o777, exist_ok=True)

# ----------------------------------------------------------------------------
# METHOD 2:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## works
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## works
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but creates a .../doc.txt./ dir
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# Uncomment these to run "Method 2":

#import os, errno
#try:
#       os.makedirs(out_dir)
#except OSError as e:
#       if e.errno != errno.EEXIST:
#               raise
# ----------------------------------------------------------------------------

结论:我认为“方法2”更可靠。

[1] 如果目录不存在,如何创建?

[2] https://docs.python.org/3/library/os.html#os.makedirs

I found this Q/A and I was initially puzzled by some of the failures and errors I was getting. I am working in Python 3 (v.3.5 in an Anaconda virtual environment on an Arch Linux x86_64 system).

Consider this directory structure:

└── output/         ## dir
   ├── corpus       ## file
   ├── corpus2/     ## dir
   └── subdir/      ## dir

Here are my experiments/notes, which clarifies things:

# ----------------------------------------------------------------------------
# [1] https://stackoverflow.com/questions/273192/how-can-i-create-a-directory-if-it-does-not-exist

import pathlib

""" Notes:
        1.  Include a trailing slash at the end of the directory path
            ("Method 1," below).
        2.  If a subdirectory in your intended path matches an existing file
            with same name, you will get the following error:
            "NotADirectoryError: [Errno 20] Not a directory:" ...
"""
# Uncomment and try each of these "out_dir" paths, singly:

# ----------------------------------------------------------------------------
# METHOD 1:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but no file created (os.makedirs creates dir, not files!  ;-)
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# [2] https://docs.python.org/3/library/os.html#os.makedirs

# Uncomment these to run "Method 1":

#directory = os.path.dirname(out_dir)
#os.makedirs(directory, mode=0o777, exist_ok=True)

# ----------------------------------------------------------------------------
# METHOD 2:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## works
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## works
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but creates a .../doc.txt./ dir
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# Uncomment these to run "Method 2":

#import os, errno
#try:
#       os.makedirs(out_dir)
#except OSError as e:
#       if e.errno != errno.EEXIST:
#               raise
# ----------------------------------------------------------------------------

Conclusion: in my opinion, “Method 2” is more robust.

[1] How can I create a directory if it does not exist?

[2] https://docs.python.org/3/library/os.html#os.makedirs


回答 18

我看到了Heikki ToivonenABB的答案,并想到了这种变化。

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST or not os.path.isdir(path):
            raise

I saw Heikki Toivonen and A-B-B‘s answers and thought of this variation.

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST or not os.path.isdir(path):
            raise

回答 19

使用此命令检查并创建目录

 if not os.path.isdir(test_img_dir):
     os.mkdir(test_img_dir)

Use this command check and create dir

 if not os.path.isdir(test_img_dir):
     os.mkdir(test_img_dir)

回答 20

如果在支持mkdir-p选项命令的计算机上运行,​​为什么不使用子流程模块 ?适用于python 2.7和python 3.6

from subprocess import call
call(['mkdir', '-p', 'path1/path2/path3'])

在大多数系统上都可以做到。

在可移植性无关紧要的情况下(例如,使用docker),解决方案只需2行。您也不必添加逻辑来检查目录是否存在。最后,重新运行很安全,没有任何副作用

如果您需要错误处理:

from subprocess import check_call
try:
    check_call(['mkdir', '-p', 'path1/path2/path3'])
except:
    handle...

Why not use subprocess module if running on a machine that supports command mkdir with -p option ? Works on python 2.7 and python 3.6

from subprocess import call
call(['mkdir', '-p', 'path1/path2/path3'])

Should do the trick on most systems.

In situations where portability doesn’t matter (ex, using docker) the solution is a clean 2 lines. You also don’t have to add logic to check if directories exist or not. Finally, it is safe to re-run without any side effects

If you need error handling:

from subprocess import check_call
try:
    check_call(['mkdir', '-p', 'path1/path2/path3'])
except:
    handle...

回答 21

如果考虑以下因素:

os.path.isdir('/tmp/dirname')

表示目录(路径)存在,并且是目录。所以对我来说,这种方式满足了我的需求。因此,我可以确保它是文件夹(不是文件)并且存在。

If you consider the following:

os.path.isdir('/tmp/dirname')

means a directory (path) exists AND is a directory. So for me this way does what I need. So I can make sure it is folder (not a file) and exists.


回答 22

create_dir()在程序/项目的入口点调用该函数。

import os

def create_dir(directory):
    if not os.path.exists(directory):
        print('Creating Directory '+directory)
        os.makedirs(directory)

create_dir('Project directory')

Call the function create_dir() at the entry point of your program/project.

import os

def create_dir(directory):
    if not os.path.exists(directory):
        print('Creating Directory '+directory)
        os.makedirs(directory)

create_dir('Project directory')

回答 23

您必须在创建目录之前设置完整路径:

import os,sys,inspect
import pathlib

currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
your_folder = currentdir + "/" + "your_folder"

if not os.path.exists(your_folder):
   pathlib.Path(your_folder).mkdir(parents=True, exist_ok=True)

这对我有用,希望对您也一样

You have to set the full path before creating the directory:

import os,sys,inspect
import pathlib

currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
your_folder = currentdir + "/" + "your_folder"

if not os.path.exists(your_folder):
   pathlib.Path(your_folder).mkdir(parents=True, exist_ok=True)

This works for me and hopefully, it will works for you as well


回答 24

import os
if os.path.isfile(filename):
    print "file exists"
else:
    "Your code here"

您的代码在哪里使用(touch)命令

这将检查文件是否存在,如果不存在则将创建它。

import os
if os.path.isfile(filename):
    print "file exists"
else:
    "Your code here"

Where your code here is use the (touch) command

This will check if the file is there if it is not then it will create it.


如何列出目录的所有文件?

问题:如何列出目录的所有文件?

如何在Python中列出目录的所有文件并将其添加到list

How can I list all files of a directory in Python and add them to a list?


回答 0

os.listdir()将为您提供目录中的所有内容- 文件目录

如果只需要文件,则可以使用os.path以下方法将其过滤掉:

from os import listdir
from os.path import isfile, join
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]

或者,您也可以使用os.walk()它为访问的每个目录生成两个列表 – 为您拆分为文件目录。如果只需要顶层目录,可以在第一次生成目录时中断

from os import walk

f = []
for (dirpath, dirnames, filenames) in walk(mypath):
    f.extend(filenames)
    break

os.listdir() will get you everything that’s in a directory – files and directories.

If you want just files, you could either filter this down using os.path:

from os import listdir
from os.path import isfile, join
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]

or you could use os.walk() which will yield two lists for each directory it visits – splitting into files and dirs for you. If you only want the top directory you can just break the first time it yields

from os import walk

f = []
for (dirpath, dirnames, filenames) in walk(mypath):
    f.extend(filenames)
    break

回答 1

我更喜欢使用glob模块,因为它可以进行模式匹配和扩展。

import glob
print(glob.glob("/home/adam/*.txt"))

它将返回包含查询文件的列表:

['/home/adam/file1.txt', '/home/adam/file2.txt', .... ]

I prefer using the glob module, as it does pattern matching and expansion.

import glob
print(glob.glob("/home/adam/*.txt"))

It will return a list with the queried files:

['/home/adam/file1.txt', '/home/adam/file2.txt', .... ]

回答 2

使用Python 2和3获取文件列表


os.listdir()

如何获取当前目录中的所有文件(和目录)(Python 3)

以下是在Python 3中使用oslistdir()函数仅检索当前目录中文件的简单方法。进一步的探索将演示如何返回目录中的文件夹,但是您不会在子目录中拥有该文件,因此可以使用步行-稍后讨论)。

 import os
 arr = os.listdir()
 print(arr)

 >>> ['$RECYCLE.BIN', 'work.txt', '3ebooks.txt', 'documents']

glob

我发现glob更容易选择相同类型或相同的文件。看下面的例子:

import glob

txtfiles = []
for file in glob.glob("*.txt"):
    txtfiles.append(file)

glob 具有列表理解

import glob

mylist = [f for f in glob.glob("*.txt")]

glob 具有功能

该函数在参数中返回给定扩展名(.txt,.docx等)的列表。

import glob

def filebrowser(ext=""):
    "Returns files with an extension"
    return [f for f in glob.glob(f"*{ext}")]

x = filebrowser(".txt")
print(x)

>>> ['example.txt', 'fb.txt', 'intro.txt', 'help.txt']

glob 扩展先前的代码

该函数现在返回与您作为参数传递的字符串匹配的文件列表

import glob

def filesearch(word=""):
    """Returns a list with all files with the word/extension in it"""
    file = []
    for f in glob.glob("*"):
        if word[0] == ".":
            if f.endswith(word):
                file.append(f)
                return file
        elif word in f:
            file.append(f)
            return file
    return file

lookfor = "example", ".py"
for w in lookfor:
    print(f"{w:10} found => {filesearch(w)}")

输出

example    found => []
.py        found => ['search.py']

获取完整的路径名 os.path.abspath

正如您所注意到的,上面的代码中没有文件的完整路径。如果需要绝对路径,则可以使用os.path模块的另一个函数,_getfullpathname将从os.listdir()中获取的文件作为参数。还有其他完整路径的方法,稍后我们将进行检查(如mexmex所建议,我将_getfullpathname替换为abspath)。

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 print(files_path)

 >>> ['F:\\documenti\applications.txt', 'F:\\documenti\collections.txt']

使用以下命令获取所有子目录中文件类型的完整路径名 walk

我发现这对于在许多目录中查找内容非常有用,它帮助我找到了一个我不记得其名称的文件:

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir():获取当前目录中的文件(Python 2)

在Python 2中,如果要在当前目录中列出文件列表,则必须将参数指定为“。”。或os.listdir方法中的os.getcwd()。

 import os
 arr = os.listdir('.')
 print(arr)

 >>> ['$RECYCLE.BIN', 'work.txt', '3ebooks.txt', 'documents']

进入目录树

# Method 1
x = os.listdir('..')

# Method 2
x= os.listdir('/')

获取文件:os.listdir()在特定目录中(Python 2和3)

 import os
 arr = os.listdir('F:\\python')
 print(arr)

 >>> ['$RECYCLE.BIN', 'work.txt', '3ebooks.txt', 'documents']

使用以下命令获取特定子目录的文件 os.listdir()

import os

x = os.listdir("./content")

os.walk('.') -当前目录

 import os
 arr = next(os.walk('.'))[2]
 print(arr)

 >>> ['5bs_Turismo1.pdf', '5bs_Turismo1.pptx', 'esperienza.txt']

next(os.walk('.'))os.path.join('dir', 'file')

 import os
 arr = []
 for d,r,f in next(os.walk("F:\\_python")):
     for file in f:
         arr.append(os.path.join(r,file))

 for f in arr:
     print(files)

>>> F:\\_python\\dict_class.py
>>> F:\\_python\\programmi.txt

next(os.walk('F:\\') -获取完整路径-列表理解

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\\_python")) for file in f]

 >>> ['F:\\_python\\dict_class.py', 'F:\\_python\\programmi.txt']

os.walk -获取完整路径-子目录中的所有文件**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\\_python") for file in f]
print(x)

>>> ['F:\\_python\\dict.py', 'F:\\_python\\progr.txt', 'F:\\_python\\readl.py']

os.listdir() -仅获取txt文件

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 print(arr_txt)

 >>> ['work.txt', '3ebooks.txt']

使用glob获得的文件的完整路径

如果我需要文件的绝对路径:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\\*.txt")]
for f in x:
    print(f)

>>> F:\acquistionline.txt
>>> F:\acquisti_2018.txt
>>> F:\bootstrap_jquery_ecc.txt

使用os.path.isfile列表,以避免目录

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)

>>> ['a simple game.py', 'data.txt', 'decorator.py']

使用pathlib在Python 3.4

import pathlib

flist = []
for p in pathlib.Path('.').iterdir():
    if p.is_file():
        print(p)
        flist.append(p)

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> setup.py
 >>> speak_gui2.py
 >>> thumb.PNG

list comprehension

flist = [p for p in pathlib.Path('.').iterdir() if p.is_file()]

或者,使用pathlib.Path()代替pathlib.Path(".")

在pathlib.Path()中使用glob方法

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:
    print(file)

>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py

使用os.walk获取所有文件

import os
x = [i[2] for i in os.walk('.')]
y=[]
for t in x:
    for f in t:
        y.append(f)
print(y)

>>> ['append_to_list.py', 'data.txt', 'data1.txt', 'data2.txt', 'data_180617', 'os_walk.py', 'READ2.py', 'read_data.py', 'somma_defaltdic.py', 'substitute_words.py', 'sum_data.py', 'data.txt', 'data1.txt', 'data_180617']

仅获取具有next的文件并进入目录

 import os
 x = next(os.walk('F://python'))[2]
 print(x)

 >>> ['calculator.bat','calculator.py']

仅获取具有next的目录并进入目录

 import os
 next(os.walk('F://python'))[1] # for the current dir use ('.')

 >>> ['python3','others']

使用以下命令获取所有子目录名称 walk

for r,d,f in os.walk("F:\\_python"):
    for dirs in d:
        print(dirs)

>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() 从Python 3.5及更高版本开始

import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)

>>> ['calculator.bat','calculator.py']

# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():
            print(entry.name)

>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG

例子:

例如 1:子目录中有多少个文件?

在此示例中,我们查找所有目录及其子目录中包含的文件数。

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"

print(count("F:\\python"))

>>> 'F:\\\python' : 12057 files'

例2:如何将所有文件从一个目录复制到另一个目录?

用于在计算机中排序的脚本,以查找一种类型的所有文件(默认值:pptx)并将其复制到新文件夹中。

import os
import shutil
from path import path

destination = "F:\\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype='pptx', counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\\" + f
                print(fullpath)
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print('-' * 30)
        print("\t==> Found in: `" + dir + "` : " + str(counter) + " files\n")

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == '_':
        # copyfile(dir, filetype='pdf')
        copyfile(dir, filetype='txt')


>>> _compiti18\Compito Contabilità 1\conti.txt
>>> _compiti18\Compito Contabilità 1\modula4.txt
>>> _compiti18\Compito Contabilità 1\moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

例如 3:如何获取txt文件中的所有文件

如果要使用所有文件名创建一个txt文件,请执行以下操作:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "\n"
    file.write(mylist)

示例:包含硬盘驱动器所有文件的txt

"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding='utf-8') as testo:
    for root, dirs, files in os.walk("D:\\"):
        for file in files:
            listafile.append(file)
            percorso.append(root + "\\" + file)
            testo.write(file + "\n")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "\n")

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "\n")

os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")

C:\的所有文件都在一个文本文件中

这是先前代码的简短版本。如果您需要从另一个位置开始,请更改开始查找文件的文件夹。这段代码在我的计算机上的文本文件上生成了50 mb的内容,其中包含完整路径的文件少于500.000行。

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\\"):
        for file in f:
            filewrite.write(f"{r + file}\n")

如何在一个类型的文件夹中写入所有路径的文件

使用此功能,您可以创建一个txt文件,该文件将具有要查找的文件类型的名称(例如pngfile.txt),并带有该类型所有文件的所有完整路径。我认为有时候它会很有用。

import os

def searchfiles(extension='.ttf', folder='H:\\'):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}\n")

# looking for png file (fonts) in the hard disk H:\
searchfiles('.png', 'H:\\')

>>> H:\4bs_18\Dolphins5.png
>>> H:\4bs_18\Dolphins6.png
>>> H:\4bs_18\Dolphins7.png
>>> H:\5_18\marketing html\assets\imageslogo2.png
>>> H:\7z001.png
>>> H:\7z002.png

(新)找到所有文件并使用tkinter GUI打开它们

我只是想在这个2019年添加一个小应用程序,以在目录中搜索所有文件,并能够通过双击列表中文件的名称来打开它们。 在此处输入图片说明

import tkinter as tk
import os

def searchfiles(extension='.txt', folder='H:\\'):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\\" + file)

def open_file():
    os.startfile(lb.get(lb.curselection()[0]))

root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles('.png', 'H:\\'))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()

Get a list of files with Python 2 and 3


os.listdir()

How to get all the files (and directories) in the current directory (Python 3)

Following, are simple methods to retrieve only files in the current directory, using os and the listdir() function, in Python 3. Further exploration, will demonstrate how to return folders in the directory, but you will not have the file in the subdirectory, for that you can use walk – discussed later).

 import os
 arr = os.listdir()
 print(arr)

 >>> ['$RECYCLE.BIN', 'work.txt', '3ebooks.txt', 'documents']

glob

I found glob easier to select the file of the same type or with something in common. Look at the following example:

import glob

txtfiles = []
for file in glob.glob("*.txt"):
    txtfiles.append(file)

glob with list comprehension

import glob

mylist = [f for f in glob.glob("*.txt")]

glob with a function

The function returns a list of the given extension (.txt, .docx ecc.) in the argument

import glob

def filebrowser(ext=""):
    "Returns files with an extension"
    return [f for f in glob.glob(f"*{ext}")]

x = filebrowser(".txt")
print(x)

>>> ['example.txt', 'fb.txt', 'intro.txt', 'help.txt']

glob extending the previous code

The function now returns a list of file that matched with the string you pass as argument

import glob

def filesearch(word=""):
    """Returns a list with all files with the word/extension in it"""
    file = []
    for f in glob.glob("*"):
        if word[0] == ".":
            if f.endswith(word):
                file.append(f)
                return file
        elif word in f:
            file.append(f)
            return file
    return file

lookfor = "example", ".py"
for w in lookfor:
    print(f"{w:10} found => {filesearch(w)}")

output

example    found => []
.py        found => ['search.py']

Getting the full path name with os.path.abspath

As you noticed, you don’t have the full path of the file in the code above. If you need to have the absolute path, you can use another function of the os.path module called _getfullpathname, putting the file that you get from os.listdir() as an argument. There are other ways to have the full path, as we will check later (I replaced, as suggested by mexmex, _getfullpathname with abspath).

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 print(files_path)

 >>> ['F:\\documenti\applications.txt', 'F:\\documenti\collections.txt']

Get the full path name of a type of file into all subdirectories with walk

I find this very useful to find stuff in many directories, and it helped me find a file about which I didn’t remember the name:

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir(): get files in the current directory (Python 2)

In Python 2, if you want the list of the files in the current directory, you have to give the argument as ‘.’ or os.getcwd() in the os.listdir method.

 import os
 arr = os.listdir('.')
 print(arr)

 >>> ['$RECYCLE.BIN', 'work.txt', '3ebooks.txt', 'documents']

To go up in the directory tree

# Method 1
x = os.listdir('..')

# Method 2
x= os.listdir('/')

Get files: os.listdir() in a particular directory (Python 2 and 3)

 import os
 arr = os.listdir('F:\\python')
 print(arr)

 >>> ['$RECYCLE.BIN', 'work.txt', '3ebooks.txt', 'documents']

Get files of a particular subdirectory with os.listdir()

import os

x = os.listdir("./content")

os.walk('.') – current directory

 import os
 arr = next(os.walk('.'))[2]
 print(arr)

 >>> ['5bs_Turismo1.pdf', '5bs_Turismo1.pptx', 'esperienza.txt']

next(os.walk('.')) and os.path.join('dir', 'file')

 import os
 arr = []
 for d,r,f in next(os.walk("F:\\_python")):
     for file in f:
         arr.append(os.path.join(r,file))

 for f in arr:
     print(files)

>>> F:\\_python\\dict_class.py
>>> F:\\_python\\programmi.txt

next(os.walk('F:\\') – get the full path – list comprehension

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\\_python")) for file in f]

 >>> ['F:\\_python\\dict_class.py', 'F:\\_python\\programmi.txt']

os.walk – get full path – all files in sub dirs**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\\_python") for file in f]
print(x)

>>> ['F:\\_python\\dict.py', 'F:\\_python\\progr.txt', 'F:\\_python\\readl.py']

os.listdir() – get only txt files

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 print(arr_txt)

 >>> ['work.txt', '3ebooks.txt']

Using glob to get the full path of the files

If I should need the absolute path of the files:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\\*.txt")]
for f in x:
    print(f)

>>> F:\acquistionline.txt
>>> F:\acquisti_2018.txt
>>> F:\bootstrap_jquery_ecc.txt

Using os.path.isfile to avoid directories in the list

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)

>>> ['a simple game.py', 'data.txt', 'decorator.py']

Using pathlib from Python 3.4

import pathlib

flist = []
for p in pathlib.Path('.').iterdir():
    if p.is_file():
        print(p)
        flist.append(p)

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> setup.py
 >>> speak_gui2.py
 >>> thumb.PNG

With list comprehension:

flist = [p for p in pathlib.Path('.').iterdir() if p.is_file()]

Alternatively, use pathlib.Path() instead of pathlib.Path(".")

Use glob method in pathlib.Path()

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:
    print(file)

>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py

Get all and only files with os.walk

import os
x = [i[2] for i in os.walk('.')]
y=[]
for t in x:
    for f in t:
        y.append(f)
print(y)

>>> ['append_to_list.py', 'data.txt', 'data1.txt', 'data2.txt', 'data_180617', 'os_walk.py', 'READ2.py', 'read_data.py', 'somma_defaltdic.py', 'substitute_words.py', 'sum_data.py', 'data.txt', 'data1.txt', 'data_180617']

Get only files with next and walk in a directory

 import os
 x = next(os.walk('F://python'))[2]
 print(x)

 >>> ['calculator.bat','calculator.py']

Get only directories with next and walk in a directory

 import os
 next(os.walk('F://python'))[1] # for the current dir use ('.')

 >>> ['python3','others']

Get all the subdir names with walk

for r,d,f in os.walk("F:\\_python"):
    for dirs in d:
        print(dirs)

>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() from Python 3.5 and greater

import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)

>>> ['calculator.bat','calculator.py']

# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():
            print(entry.name)

>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG

Examples:

Ex. 1: How many files are there in the subdirectories?

In this example, we look for the number of files that are included in all the directory and its subdirectories.

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"

print(count("F:\\python"))

>>> 'F:\\\python' : 12057 files'

Ex.2: How to copy all files from a directory to another?

A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

import os
import shutil
from path import path

destination = "F:\\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype='pptx', counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\\" + f
                print(fullpath)
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print('-' * 30)
        print("\t==> Found in: `" + dir + "` : " + str(counter) + " files\n")

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == '_':
        # copyfile(dir, filetype='pdf')
        copyfile(dir, filetype='txt')


>>> _compiti18\Compito Contabilità 1\conti.txt
>>> _compiti18\Compito Contabilità 1\modula4.txt
>>> _compiti18\Compito Contabilità 1\moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

Ex. 3: How to get all the files in a txt file

In case you want to create a txt file with all the file names:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "\n"
    file.write(mylist)

Example: txt with all the files of an hard drive

"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding='utf-8') as testo:
    for root, dirs, files in os.walk("D:\\"):
        for file in files:
            listafile.append(file)
            percorso.append(root + "\\" + file)
            testo.write(file + "\n")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "\n")

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "\n")

os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")

All the file of C:\ in one text file

This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\\"):
        for file in f:
            filewrite.write(f"{r + file}\n")

How to write a file with all paths in a folder of a type

With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

import os

def searchfiles(extension='.ttf', folder='H:\\'):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}\n")

# looking for png file (fonts) in the hard disk H:\
searchfiles('.png', 'H:\\')

>>> H:\4bs_18\Dolphins5.png
>>> H:\4bs_18\Dolphins6.png
>>> H:\4bs_18\Dolphins7.png
>>> H:\5_18\marketing html\assets\imageslogo2.png
>>> H:\7z001.png
>>> H:\7z002.png

(New) Find all files and open them with tkinter GUI

I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list. enter image description here

import tkinter as tk
import os

def searchfiles(extension='.txt', folder='H:\\'):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\\" + file)

def open_file():
    os.startfile(lb.get(lb.curselection()[0]))

root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles('.png', 'H:\\'))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()

回答 3

import os
os.listdir("somedirectory")

将返回“ somedirectory”中所有文件和目录的列表。

import os
os.listdir("somedirectory")

will return a list of all files and directories in “somedirectory”.


回答 4

一种获取文件列表(不包含子目录)的单行解决方案:

filenames = next(os.walk(path))[2]

或绝对路径名:

paths = [os.path.join(path, fn) for fn in next(os.walk(path))[2]]

A one-line solution to get only list of files (no subdirectories):

filenames = next(os.walk(path))[2]

or absolute pathnames:

paths = [os.path.join(path, fn) for fn in next(os.walk(path))[2]]

回答 5

从目录及其所有子目录获取完整的文件路径

import os

def get_filepaths(directory):
    """
    This function will generate the file names in a directory 
    tree by walking the tree either top-down or bottom-up. For each 
    directory in the tree rooted at directory top (including top itself), 
    it yields a 3-tuple (dirpath, dirnames, filenames).
    """
    file_paths = []  # List which will store all of the full filepaths.

    # Walk the tree.
    for root, directories, files in os.walk(directory):
        for filename in files:
            # Join the two strings in order to form the full filepath.
            filepath = os.path.join(root, filename)
            file_paths.append(filepath)  # Add it to the list.

    return file_paths  # Self-explanatory.

# Run the above function and store its results in a variable.   
full_file_paths = get_filepaths("/Users/johnny/Desktop/TEST")

  • 我在上述函数中提供的路径包含3个文件-其中两个在根目录中,另一个在子文件夹“ SUBFOLDER”中。您现在可以执行以下操作:
  • print full_file_paths 这将打印列表:

    • ['/Users/johnny/Desktop/TEST/file1.txt', '/Users/johnny/Desktop/TEST/file2.txt', '/Users/johnny/Desktop/TEST/SUBFOLDER/file3.dat']

如果愿意,您可以打开和阅读内容,或仅关注扩展名为“ .dat”的文件,如以下代码所示:

for f in full_file_paths:
  if f.endswith(".dat"):
    print f

/Users/johnny/Desktop/TEST/SUBFOLDER/file3.dat

Getting Full File Paths From a Directory and All Its Subdirectories

import os

def get_filepaths(directory):
    """
    This function will generate the file names in a directory 
    tree by walking the tree either top-down or bottom-up. For each 
    directory in the tree rooted at directory top (including top itself), 
    it yields a 3-tuple (dirpath, dirnames, filenames).
    """
    file_paths = []  # List which will store all of the full filepaths.

    # Walk the tree.
    for root, directories, files in os.walk(directory):
        for filename in files:
            # Join the two strings in order to form the full filepath.
            filepath = os.path.join(root, filename)
            file_paths.append(filepath)  # Add it to the list.

    return file_paths  # Self-explanatory.

# Run the above function and store its results in a variable.   
full_file_paths = get_filepaths("/Users/johnny/Desktop/TEST")

  • The path I provided in the above function contained 3 files— two of them in the root directory, and another in a subfolder called “SUBFOLDER.” You can now do things like:
  • print full_file_paths which will print the list:

    • ['/Users/johnny/Desktop/TEST/file1.txt', '/Users/johnny/Desktop/TEST/file2.txt', '/Users/johnny/Desktop/TEST/SUBFOLDER/file3.dat']

If you’d like, you can open and read the contents, or focus only on files with the extension “.dat” like in the code below:

for f in full_file_paths:
  if f.endswith(".dat"):
    print f

/Users/johnny/Desktop/TEST/SUBFOLDER/file3.dat


回答 6

从3.4版开始,有内置的迭代器,它比os.listdir()

pathlib版本3.4中的新功能。

>>> import pathlib
>>> [p for p in pathlib.Path('.').iterdir() if p.is_file()]

根据PEP 428,该pathlib库的目的是提供一个简单的类层次结构,以处理文件系统路径以及用户对其进行的常见操作。

os.scandir()3.5版中的新功能。

>>> import os
>>> [entry for entry in os.scandir('.') if entry.is_file()]

请注意,os.walk()使用os.scandir()代替了os.listdir()3.5版,并且根据PEP 471将其速度提高了2-20倍。

我也建议您阅读下面的ShadowRanger评论。

Since version 3.4 there are builtin iterators for this which are a lot more efficient than os.listdir():

pathlib: New in version 3.4.

>>> import pathlib
>>> [p for p in pathlib.Path('.').iterdir() if p.is_file()]

According to PEP 428, the aim of the pathlib library is to provide a simple hierarchy of classes to handle filesystem paths and the common operations users do over them.

os.scandir(): New in version 3.5.

>>> import os
>>> [entry for entry in os.scandir('.') if entry.is_file()]

Note that os.walk() uses os.scandir() instead of os.listdir() from version 3.5, and its speed got increased by 2-20 times according to PEP 471.

Let me also recommend reading ShadowRanger’s comment below.


回答 7

初步说明

  • 尽管文件目录问题文本中的术语,但有些人可能会认为目录实际上是特殊文件
  • 该声明: ”目录的所有文件 ”可以用两种方式解释:
    1. 所有直接(或1级)后代
    2. 整个目录树中的所有子代(包括子目录中的子代)
  • 提出问题时,我认为Python 2LTS版本,但是代码示例将由Python 3.5)运行(我将使其尽可能与Python 2兼容;此外,属于我要发布的Python来自v3.5.4-除非另有说明)。结果与问题中的另一个关键字相关:“ 将它们添加到列表中 ”:

    • Python 2.2之前的版本中版本中,序列(可迭代)主要由列表(元组,集合等)表示。
    • Python 2.2中引入生成器[Python.Wiki]:Generators)的概念-由[Python 3]:yield语句提供。随着时间的流逝,对于返回/使用列表的函数,生成器对应对象开始出现
    • Python 3中,generator是默认行为
    • 不知道返回列表是否仍然是强制性的(或者生成器也可以执行),但是将生成器传递给列表构造函数会从列表构造器中创建列表(并消耗列表)。以下示例说明了[Python 3]的区别mapfunction,iterable,…
    >>> import sys
    >>> sys.version
    '2.7.10 (default, Mar  8 2016, 15:02:46) [MSC v.1600 64 bit (AMD64)]'
    >>> m = map(lambda x: x, [1, 2, 3])  # Just a dummy lambda function
    >>> m, type(m)
    ([1, 2, 3], <type 'list'>)
    >>> len(m)
    3


    >>> import sys
    >>> sys.version
    '3.5.4 (v3.5.4:3f56838, Aug  8 2017, 02:17:05) [MSC v.1900 64 bit (AMD64)]'
    >>> m = map(lambda x: x, [1, 2, 3])
    >>> m, type(m)
    (<map object at 0x000001B4257342B0>, <class 'map'>)
    >>> len(m)
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
    TypeError: object of type 'map' has no len()
    >>> lm0 = list(m)  # Build a list from the generator
    >>> lm0, type(lm0)
    ([1, 2, 3], <class 'list'>)
    >>>
    >>> lm1 = list(m)  # Build a list from the same generator
    >>> lm1, type(lm1)  # Empty list now - generator already consumed
    ([], <class 'list'>)
  • 这些示例将基于具有以下结构的名为root_dir的目录(此示例适用于Win,但我也在Lnx上使用同一棵树):

    E:\Work\Dev\StackOverflow\q003207219>tree /f "root_dir"
    Folder PATH listing for volume Work
    Volume serial number is 00000029 3655:6FED
    E:\WORK\DEV\STACKOVERFLOW\Q003207219\ROOT_DIR
    ¦   file0
    ¦   file1
    ¦
    +---dir0
    ¦   +---dir00
    ¦   ¦   ¦   file000
    ¦   ¦   ¦
    ¦   ¦   +---dir000
    ¦   ¦           file0000
    ¦   ¦
    ¦   +---dir01
    ¦   ¦       file010
    ¦   ¦       file011
    ¦   ¦
    ¦   +---dir02
    ¦       +---dir020
    ¦           +---dir0200
    +---dir1
    ¦       file10
    ¦       file11
    ¦       file12
    ¦
    +---dir2
    ¦   ¦   file20
    ¦   ¦
    ¦   +---dir20
    ¦           file200
    ¦
    +---dir3


解决方案

程序化方法:

  1. [Python 3]:操作系统。listdirpath =’。’

    返回一个列表,其中包含由path给出的目录中条目的名称。列表按任意顺序排列,不包括特殊条目'.''..'


    >>> import os
    >>> root_dir = "root_dir"  # Path relative to current dir (os.getcwd())
    >>>
    >>> os.listdir(root_dir)  # List all the items in root_dir
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [item for item in os.listdir(root_dir) if os.path.isfile(os.path.join(root_dir, item))]  # Filter items and only keep files (strip out directories)
    ['file0', 'file1']

    一个更详细的示例(code_os_listdir.py):

    import os
    from pprint import pformat
    
    
    def _get_dir_content(path, include_folders, recursive):
        entries = os.listdir(path)
        for entry in entries:
            entry_with_path = os.path.join(path, entry)
            if os.path.isdir(entry_with_path):
                if include_folders:
                    yield entry_with_path
                if recursive:
                    for sub_entry in _get_dir_content(entry_with_path, include_folders, recursive):
                        yield sub_entry
            else:
                yield entry_with_path
    
    
    def get_dir_content(path, include_folders=True, recursive=True, prepend_folder_name=True):
        path_len = len(path) + len(os.path.sep)
        for item in _get_dir_content(path, include_folders, recursive):
            yield item if prepend_folder_name else item[path_len:]
    
    
    def _get_dir_content_old(path, include_folders, recursive):
        entries = os.listdir(path)
        ret = list()
        for entry in entries:
            entry_with_path = os.path.join(path, entry)
            if os.path.isdir(entry_with_path):
                if include_folders:
                    ret.append(entry_with_path)
                if recursive:
                    ret.extend(_get_dir_content_old(entry_with_path, include_folders, recursive))
            else:
                ret.append(entry_with_path)
        return ret
    
    
    def get_dir_content_old(path, include_folders=True, recursive=True, prepend_folder_name=True):
        path_len = len(path) + len(os.path.sep)
        return [item if prepend_folder_name else item[path_len:] for item in _get_dir_content_old(path, include_folders, recursive)]
    
    
    def main():
        root_dir = "root_dir"
        ret0 = get_dir_content(root_dir, include_folders=True, recursive=True, prepend_folder_name=True)
        lret0 = list(ret0)
        print(ret0, len(lret0), pformat(lret0))
        ret1 = get_dir_content_old(root_dir, include_folders=False, recursive=True, prepend_folder_name=False)
        print(len(ret1), pformat(ret1))
    
    
    if __name__ == "__main__":
        main()

    注意事项

    • 有两种实现:
      • 使用生成器的生成器(当然这里似乎没有用,因为我立即将结果转换为列表)
      • 经典的(函数名称以 _old
    • 使用递归(进入子目录)
    • 对于每种实现,都有两个功能:
      • 下划线 _):“ private”(不应直接调用)-完成所有工作
      • 公共的(包装上一个):它只是从返回的条目中剥离出初始路径(如果需要)。这是一个丑陋的实现,但这是我目前唯一能想到的想法
    • 在性能方面,生成器通常要快一些(考虑这两个创建迭代时间),但是我没有在递归函数中对其进行测试,而且我还在内部生成器上迭代了函数-不知道性能如何友好的是
    • 玩弄参数以获得不同的结果


    输出

    (py35x64_test) E:\Work\Dev\StackOverflow\q003207219>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" "code_os_listdir.py"
    <generator object get_dir_content at 0x000001BDDBB3DF10> 22 ['root_dir\\dir0',
     'root_dir\\dir0\\dir00',
     'root_dir\\dir0\\dir00\\dir000',
     'root_dir\\dir0\\dir00\\dir000\\file0000',
     'root_dir\\dir0\\dir00\\file000',
     'root_dir\\dir0\\dir01',
     'root_dir\\dir0\\dir01\\file010',
     'root_dir\\dir0\\dir01\\file011',
     'root_dir\\dir0\\dir02',
     'root_dir\\dir0\\dir02\\dir020',
     'root_dir\\dir0\\dir02\\dir020\\dir0200',
     'root_dir\\dir1',
     'root_dir\\dir1\\file10',
     'root_dir\\dir1\\file11',
     'root_dir\\dir1\\file12',
     'root_dir\\dir2',
     'root_dir\\dir2\\dir20',
     'root_dir\\dir2\\dir20\\file200',
     'root_dir\\dir2\\file20',
     'root_dir\\dir3',
     'root_dir\\file0',
     'root_dir\\file1']
    11 ['dir0\\dir00\\dir000\\file0000',
     'dir0\\dir00\\file000',
     'dir0\\dir01\\file010',
     'dir0\\dir01\\file011',
     'dir1\\file10',
     'dir1\\file11',
     'dir1\\file12',
     'dir2\\dir20\\file200',
     'dir2\\file20',
     'file0',
     'file1']


  1. [Python 3]:操作系统。scandirpath =’。’Python 3.5 +,反向移植:[PyPI]:scandir

    返回与path指定的目录中的条目相对应的os.DirEntry对象的迭代器。条目以任意顺序产生,特殊条目'.''..'不包括在内。

    使用scandir()而不是listdir()可以显着提高还需要文件类型或文件属性信息的代码的性能,因为如果操作系统在扫描目录时提供了os.DirEntry对象,则该信息会公开。所有的os.DirEntry方法都可以执行系统调用,但是is_dir()is_file()通常只需要系统调用即可进行符号链接。os.DirEntry.stat()在Unix上始终需要系统调用,而在Windows上只需要一个系统调用即可。


    >>> import os
    >>> root_dir = os.path.join(".", "root_dir")  # Explicitly prepending current directory
    >>> root_dir
    '.\\root_dir'
    >>>
    >>> scandir_iterator = os.scandir(root_dir)
    >>> scandir_iterator
    <nt.ScandirIterator object at 0x00000268CF4BC140>
    >>> [item.path for item in scandir_iterator]
    ['.\\root_dir\\dir0', '.\\root_dir\\dir1', '.\\root_dir\\dir2', '.\\root_dir\\dir3', '.\\root_dir\\file0', '.\\root_dir\\file1']
    >>>
    >>> [item.path for item in scandir_iterator]  # Will yield an empty list as it was consumed by previous iteration (automatically performed by the list comprehension)
    []
    >>>
    >>> scandir_iterator = os.scandir(root_dir)  # Reinitialize the generator
    >>> for item in scandir_iterator :
    ...     if os.path.isfile(item.path):
    ...             print(item.name)
    ...
    file0
    file1

    注意事项

    • 类似于 os.listdir
    • 但是它也更灵活(并提供更多功能),更多Python ic(在某些情况下更快)


  1. [Python 3]:操作系统。步行top,topdown = True,onerror = None,followlinks = False

    通过自上而下或自下而上移动目录树来生成文件名。对于在目录为根的树中的每个目录顶部(包括顶部本身),它产生一个3元组(dirpathdirnamesfilenames)。


    >>> import os
    >>> root_dir = os.path.join(os.getcwd(), "root_dir")  # Specify the full path
    >>> root_dir
    'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir'
    >>>
    >>> walk_generator = os.walk(root_dir)
    >>> root_dir_entry = next(walk_generator)  # First entry corresponds to the root dir (passed as an argument)
    >>> root_dir_entry
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir', ['dir0', 'dir1', 'dir2', 'dir3'], ['file0', 'file1'])
    >>>
    >>> root_dir_entry[1] + root_dir_entry[2]  # Display dirs and files (direct descendants) in a single list
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [os.path.join(root_dir_entry[0], item) for item in root_dir_entry[1] + root_dir_entry[2]]  # Display all the entries in the previous list by their full path
    ['E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir1', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir2', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir3', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\file0', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\file1']
    >>>
    >>> for entry in walk_generator:  # Display the rest of the elements (corresponding to every subdir)
    ...     print(entry)
    ...
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0', ['dir00', 'dir01', 'dir02'], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir00', ['dir000'], ['file000'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir00\\dir000', [], ['file0000'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir01', [], ['file010', 'file011'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir02', ['dir020'], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir02\\dir020', ['dir0200'], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir02\\dir020\\dir0200', [], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir1', [], ['file10', 'file11', 'file12'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir2', ['dir20'], ['file20'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir2\\dir20', [], ['file200'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir3', [], [])

    注意事项

    • 在场景下,它使用os.scandiros.listdir在旧版本上)
    • 它通过重复子文件夹来完成繁重的工作


  1. [Python 3]:glob。globpathname,*,recursive = False[Python 3]:glob。iglobpathname,*,recursive = False

    返回与pathname匹配的路径名的可能为空的列表,该列表必须是包含路径说明的字符串。路径名可以是绝对的(如/usr/src/Python-1.5/Makefile)或相对的(如../../Tools/*/*.gif),并且可以包含壳式通配符。损坏的符号链接包含在结果中(如在shell中)。

    在版本3.5中更改:使用“ **” 支持递归glob 。


    >>> import glob, os
    >>> wildcard_pattern = "*"
    >>> root_dir = os.path.join("root_dir", wildcard_pattern)  # Match every file/dir name
    >>> root_dir
    'root_dir\\*'
    >>>
    >>> glob_list = glob.glob(root_dir)
    >>> glob_list
    ['root_dir\\dir0', 'root_dir\\dir1', 'root_dir\\dir2', 'root_dir\\dir3', 'root_dir\\file0', 'root_dir\\file1']
    >>>
    >>> [item.replace("root_dir" + os.path.sep, "") for item in glob_list]  # Strip the dir name and the path separator from begining
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> for entry in glob.iglob(root_dir + "*", recursive=True):
    ...     print(entry)
    ...
    root_dir\
    root_dir\dir0
    root_dir\dir0\dir00
    root_dir\dir0\dir00\dir000
    root_dir\dir0\dir00\dir000\file0000
    root_dir\dir0\dir00\file000
    root_dir\dir0\dir01
    root_dir\dir0\dir01\file010
    root_dir\dir0\dir01\file011
    root_dir\dir0\dir02
    root_dir\dir0\dir02\dir020
    root_dir\dir0\dir02\dir020\dir0200
    root_dir\dir1
    root_dir\dir1\file10
    root_dir\dir1\file11
    root_dir\dir1\file12
    root_dir\dir2
    root_dir\dir2\dir20
    root_dir\dir2\dir20\file200
    root_dir\dir2\file20
    root_dir\dir3
    root_dir\file0
    root_dir\file1

    注意事项

    • 用途 os.listdir
    • 对于大树(尤其是在启用递归的情况下),iglob首选
    • 允许基于名称进行高级过滤(由于通配符)


  1. [Python 3]:类pathlib。路径* pathsegmentsPython 3.4 +,backport:[PyPI]:pathlib2

    >>> import pathlib
    >>> root_dir = "root_dir"
    >>> root_dir_instance = pathlib.Path(root_dir)
    >>> root_dir_instance
    WindowsPath('root_dir')
    >>> root_dir_instance.name
    'root_dir'
    >>> root_dir_instance.is_dir()
    True
    >>>
    >>> [item.name for item in root_dir_instance.glob("*")]  # Wildcard searching for all direct descendants
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [os.path.join(item.parent.name, item.name) for item in root_dir_instance.glob("*") if not item.is_dir()]  # Display paths (including parent) for files only
    ['root_dir\\file0', 'root_dir\\file1']

    注意事项

    • 这是实现我们目标的一种方式
    • 这是处理路径的OOP风格
    • 提供许多功能


  1. [Python 2]:dircache.listdir(path)(仅Python 2


    def listdir(path):
        """List directory contents, using cache."""
        try:
            cached_mtime, list = cache[path]
            del cache[path]
        except KeyError:
            cached_mtime, list = -1, []
        mtime = os.stat(path).st_mtime
        if mtime != cached_mtime:
            list = os.listdir(path)
            list.sort()
        cache[path] = mtime, list
        return list


  1. [man7]:OPENDIR(3) / [man7]:READDIR(3) / [man7]:CLOSEDIR(3)通过[Python 3]:ctypes-Python的外部函数库(特定POSIX

    ctypes是Python的外部函数库。它提供C兼容的数据类型,并允许在DLL或共享库中调用函数。它可以用于将这些库包装在纯Python中。

    code_ctypes.py

    #!/usr/bin/env python3
    
    import sys
    from ctypes import Structure, \
        c_ulonglong, c_longlong, c_ushort, c_ubyte, c_char, c_int, \
        CDLL, POINTER, \
        create_string_buffer, get_errno, set_errno, cast
    
    
    DT_DIR = 4
    DT_REG = 8
    
    char256 = c_char * 256
    
    
    class LinuxDirent64(Structure):
        _fields_ = [
            ("d_ino", c_ulonglong),
            ("d_off", c_longlong),
            ("d_reclen", c_ushort),
            ("d_type", c_ubyte),
            ("d_name", char256),
        ]
    
    LinuxDirent64Ptr = POINTER(LinuxDirent64)
    
    libc_dll = this_process = CDLL(None, use_errno=True)
    # ALWAYS set argtypes and restype for functions, otherwise it's UB!!!
    opendir = libc_dll.opendir
    readdir = libc_dll.readdir
    closedir = libc_dll.closedir
    
    
    def get_dir_content(path):
        ret = [path, list(), list()]
        dir_stream = opendir(create_string_buffer(path.encode()))
        if (dir_stream == 0):
            print("opendir returned NULL (errno: {:d})".format(get_errno()))
            return ret
        set_errno(0)
        dirent_addr = readdir(dir_stream)
        while dirent_addr:
            dirent_ptr = cast(dirent_addr, LinuxDirent64Ptr)
            dirent = dirent_ptr.contents
            name = dirent.d_name.decode()
            if dirent.d_type & DT_DIR:
                if name not in (".", ".."):
                    ret[1].append(name)
            elif dirent.d_type & DT_REG:
                ret[2].append(name)
            dirent_addr = readdir(dir_stream)
        if get_errno():
            print("readdir returned NULL (errno: {:d})".format(get_errno()))
        closedir(dir_stream)
        return ret
    
    
    def main():
        print("{:s} on {:s}\n".format(sys.version, sys.platform))
        root_dir = "root_dir"
        entries = get_dir_content(root_dir)
        print(entries)
    
    
    if __name__ == "__main__":
        main()

    注意事项

    • 它从libc加载三个函数(在当前进程中加载​​)并调用它们(有关更多详细信息,请检查[SO]:如何检查文件是否存在无异常?(@ CristiFati的回答))第4项的最后注释)。这将使这种方法非常接近Python / C边缘
    • LinuxDirent64ctypes的代表性结构dirent64[man7]:dirent.h(0P) (因此是DT_从我的机器常量):Ubtu 16 644.10.0-40泛型libc6的-dev的:AMD64)。在其他口味/版本上,结构定义可能会有所不同,如果是,则应更新ctypes别名,否则将产生未定义行为
    • 它以os.walk格式返回数据。我没有麻烦使其递归,但是从现有代码开始,这将是一件相当琐碎的任务
    • 一切在Win上也是可行的,数据(库,函数,结构,常量等)不同


    输出

    [cfati@cfati-ubtu16x64-0:~/Work/Dev/StackOverflow/q003207219]> ./code_ctypes.py
    3.5.2 (default, Nov 12 2018, 13:43:14)
    [GCC 5.4.0 20160609] on linux
    
    ['root_dir', ['dir2', 'dir1', 'dir3', 'dir0'], ['file1', 'file0']]


  1. [ActiveState.Docs]:win32file.FindFilesW(特定Win

    使用Windows Unicode API检索匹配文件名的列表。API FindFirstFileW / FindNextFileW / Find关闭函数的接口。


    >>> import os, win32file, win32con
    >>> root_dir = "root_dir"
    >>> wildcard = "*"
    >>> root_dir_wildcard = os.path.join(root_dir, wildcard)
    >>> entry_list = win32file.FindFilesW(root_dir_wildcard)
    >>> len(entry_list)  # Don't display the whole content as it's too long
    8
    >>> [entry[-2] for entry in entry_list]  # Only display the entry names
    ['.', '..', 'dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [entry[-2] for entry in entry_list if entry[0] & win32con.FILE_ATTRIBUTE_DIRECTORY and entry[-2] not in (".", "..")]  # Filter entries and only display dir names (except self and parent)
    ['dir0', 'dir1', 'dir2', 'dir3']
    >>>
    >>> [os.path.join(root_dir, entry[-2]) for entry in entry_list if entry[0] & (win32con.FILE_ATTRIBUTE_NORMAL | win32con.FILE_ATTRIBUTE_ARCHIVE)]  # Only display file "full" names
    ['root_dir\\file0', 'root_dir\\file1']

    注意事项


  1. 安装一些(其他)第三方软件包即可
    • 最有可能会依赖于上述一项(或多项)(可能需要进行一些自定义)


注意事项

  • 代码应具有可移植性(针对特定区域的地方-带有标记的地方除外)或交叉的:

    • 平台(NixWin,)
    • Python版本(2、3,)
  • 在上述变体中使用了多种路径样式(绝对路径,相对路径),以说明以下事实:所使用的“工具”在此方向上是灵活的

  • os.listdiros.scandir使用opendir / readdir / closedir[MS.Docs]:FindFirstFileW函数 / [MS.Docs]:FindNextFileW函数 / [MS.Docs]:FindClose函数)(通过[GitHub]:python / cpython-(master)cpython /模块/posixmodule.c

  • win32file.FindFilesW也使用那些(特定于Win的)函数(通过[GitHub]:mhammond / pywin32-(主)pywin32 / win32 / src / win32file.i

  • _get_dir_content(从第1点开始)可以使用以下任何一种方法来实现(有些需要更多的工作,有些需要更少的工作)

    • 可以执行一些高级过滤(而不只是文件目录):例如,include_folders参数可以被另一个参数(例如filter_func)代替,该函数将以路径作为参数:(filter_func=lambda x: True这不会删除)内容)和_get_dir_content内的内容类似:(if not filter_func(entry_with_path): continue如果该函数对一项失败,则将被跳过),但是代码越复杂,执行所花费的时间就越长
  • 诺娜·贝恩!由于使用了递归,因此我必须提到我在笔记本电脑(Win 10 x64)上进行了一些测试,与该问题完全无关,并且当递归级别达到(990 .. 1000)范围内的某个值时(recursionlimit -1000 (默认)),我得到了StackOverflow :)。如果目录树超过该限制(我不是FS专家,那么我什至不知道那是否可能),那可能是个问题。
    我还必须提到,我没有尝试增加递归限制,因为我在该领域没有经验(必须增加OS的堆栈之前,我可以增加多少?级别),但从理论上讲,如果目录深度大于可能的最高递归限制(在该计算机上),则总有失败的可能性

  • 代码示例仅用于说明目的。这意味着我没有考虑错误处理(我不认为有任何尝试 / 除外 / else / finally块),因此代码并不健壮(原因是:使其尽可能简单和简短) )。对于生产,还应添加错误处理

其他方法:

  1. 仅将Python用作包装器

    • 一切都使用另一种技术完成
    • 该技术是从Python调用的
    • 我所知道的最著名的味道是我所说的系统管理员方法:

      • 采用 Python(或与此相关的任何编程语言)执行Shell命令(并解析其输出)
      • 有人认为这是一个很好的技巧
      • 我认为这更像是一种la脚的解决方法(gainarie),因为操作本身是从shell在这种情况下为cmd)执行的,因此与Python无关。
      • 过滤(grep/ findstr)或输出格式化都可以在两面进行,但我不会坚持这样做。另外,我故意使用os.system代替subprocess.Popen
      (py35x64_test) E:\Work\Dev\StackOverflow\q003207219>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" -c "import os;os.system(\"dir /b root_dir\")"
      dir0
      dir1
      dir2
      dir3
      file0
      file1

    通常应避免这种方法,因为如果某些命令输出格式在OS版本/风格之间略有不同,则解析代码也应进行调整;例如,更不用说语言环境之间的差异了)。

Preliminary notes

  • Although there’s a clear differentiation between file and directory terms in the question text, some may argue that directories are actually special files
  • The statement: “all files of a directory” can be interpreted in two ways:
    1. All direct (or level 1) descendants only
    2. All descendants in the whole directory tree (including the ones in sub-directories)
  • When the question was asked, I imagine that Python 2, was the LTS version, however the code samples will be run by Python 3(.5) (I’ll keep them as Python 2 compliant as possible; also, any code belonging to Python that I’m going to post, is from v3.5.4 – unless otherwise specified). That has consequences related to another keyword in the question: “add them into a list“:

    • In pre Python 2.2 versions, sequences (iterables) were mostly represented by lists (tuples, sets, …)
    • In Python 2.2, the concept of generator ([Python.Wiki]: Generators) – courtesy of [Python 3]: The yield statement) – was introduced. As time passed, generator counterparts started to appear for functions that returned/worked with lists
    • In Python 3, generator is the default behavior
    • Not sure if returning a list is still mandatory (or a generator would do as well), but passing a generator to the list constructor, will create a list out of it (and also consume it). The example below illustrates the differences on [Python 3]: map(function, iterable, …)
    >>> import sys
    >>> sys.version
    '2.7.10 (default, Mar  8 2016, 15:02:46) [MSC v.1600 64 bit (AMD64)]'
    >>> m = map(lambda x: x, [1, 2, 3])  # Just a dummy lambda function
    >>> m, type(m)
    ([1, 2, 3], <type 'list'>)
    >>> len(m)
    3
    


    >>> import sys
    >>> sys.version
    '3.5.4 (v3.5.4:3f56838, Aug  8 2017, 02:17:05) [MSC v.1900 64 bit (AMD64)]'
    >>> m = map(lambda x: x, [1, 2, 3])
    >>> m, type(m)
    (<map object at 0x000001B4257342B0>, <class 'map'>)
    >>> len(m)
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
    TypeError: object of type 'map' has no len()
    >>> lm0 = list(m)  # Build a list from the generator
    >>> lm0, type(lm0)
    ([1, 2, 3], <class 'list'>)
    >>>
    >>> lm1 = list(m)  # Build a list from the same generator
    >>> lm1, type(lm1)  # Empty list now - generator already consumed
    ([], <class 'list'>)
    
  • The examples will be based on a directory called root_dir with the following structure (this example is for Win, but I’m using the same tree on Lnx as well):

    E:\Work\Dev\StackOverflow\q003207219>tree /f "root_dir"
    Folder PATH listing for volume Work
    Volume serial number is 00000029 3655:6FED
    E:\WORK\DEV\STACKOVERFLOW\Q003207219\ROOT_DIR
    ¦   file0
    ¦   file1
    ¦
    +---dir0
    ¦   +---dir00
    ¦   ¦   ¦   file000
    ¦   ¦   ¦
    ¦   ¦   +---dir000
    ¦   ¦           file0000
    ¦   ¦
    ¦   +---dir01
    ¦   ¦       file010
    ¦   ¦       file011
    ¦   ¦
    ¦   +---dir02
    ¦       +---dir020
    ¦           +---dir0200
    +---dir1
    ¦       file10
    ¦       file11
    ¦       file12
    ¦
    +---dir2
    ¦   ¦   file20
    ¦   ¦
    ¦   +---dir20
    ¦           file200
    ¦
    +---dir3
    


Solutions

Programmatic approaches:

  1. [Python 3]: os.listdir(path=’.’)

    Return a list containing the names of the entries in the directory given by path. The list is in arbitrary order, and does not include the special entries '.' and '..'


    >>> import os
    >>> root_dir = "root_dir"  # Path relative to current dir (os.getcwd())
    >>>
    >>> os.listdir(root_dir)  # List all the items in root_dir
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [item for item in os.listdir(root_dir) if os.path.isfile(os.path.join(root_dir, item))]  # Filter items and only keep files (strip out directories)
    ['file0', 'file1']
    

    A more elaborate example (code_os_listdir.py):

    import os
    from pprint import pformat
    
    
    def _get_dir_content(path, include_folders, recursive):
        entries = os.listdir(path)
        for entry in entries:
            entry_with_path = os.path.join(path, entry)
            if os.path.isdir(entry_with_path):
                if include_folders:
                    yield entry_with_path
                if recursive:
                    for sub_entry in _get_dir_content(entry_with_path, include_folders, recursive):
                        yield sub_entry
            else:
                yield entry_with_path
    
    
    def get_dir_content(path, include_folders=True, recursive=True, prepend_folder_name=True):
        path_len = len(path) + len(os.path.sep)
        for item in _get_dir_content(path, include_folders, recursive):
            yield item if prepend_folder_name else item[path_len:]
    
    
    def _get_dir_content_old(path, include_folders, recursive):
        entries = os.listdir(path)
        ret = list()
        for entry in entries:
            entry_with_path = os.path.join(path, entry)
            if os.path.isdir(entry_with_path):
                if include_folders:
                    ret.append(entry_with_path)
                if recursive:
                    ret.extend(_get_dir_content_old(entry_with_path, include_folders, recursive))
            else:
                ret.append(entry_with_path)
        return ret
    
    
    def get_dir_content_old(path, include_folders=True, recursive=True, prepend_folder_name=True):
        path_len = len(path) + len(os.path.sep)
        return [item if prepend_folder_name else item[path_len:] for item in _get_dir_content_old(path, include_folders, recursive)]
    
    
    def main():
        root_dir = "root_dir"
        ret0 = get_dir_content(root_dir, include_folders=True, recursive=True, prepend_folder_name=True)
        lret0 = list(ret0)
        print(ret0, len(lret0), pformat(lret0))
        ret1 = get_dir_content_old(root_dir, include_folders=False, recursive=True, prepend_folder_name=False)
        print(len(ret1), pformat(ret1))
    
    
    if __name__ == "__main__":
        main()
    

    Notes:

    • There are two implementations:
      • One that uses generators (of course here it seems useless, since I immediately convert the result to a list)
      • The classic one (function names ending in _old)
    • Recursion is used (to get into subdirectories)
    • For each implementation there are two functions:
      • One that starts with an underscore (_): “private” (should not be called directly) – that does all the work
      • The public one (wrapper over previous): it just strips off the initial path (if required) from the returned entries. It’s an ugly implementation, but it’s the only idea that I could come with at this point
    • In terms of performance, generators are generally a little bit faster (considering both creation and iteration times), but I didn’t test them in recursive functions, and also I am iterating inside the function over inner generators – don’t know how performance friendly is that
    • Play with the arguments to get different results


    Output:

    (py35x64_test) E:\Work\Dev\StackOverflow\q003207219>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" "code_os_listdir.py"
    <generator object get_dir_content at 0x000001BDDBB3DF10> 22 ['root_dir\\dir0',
     'root_dir\\dir0\\dir00',
     'root_dir\\dir0\\dir00\\dir000',
     'root_dir\\dir0\\dir00\\dir000\\file0000',
     'root_dir\\dir0\\dir00\\file000',
     'root_dir\\dir0\\dir01',
     'root_dir\\dir0\\dir01\\file010',
     'root_dir\\dir0\\dir01\\file011',
     'root_dir\\dir0\\dir02',
     'root_dir\\dir0\\dir02\\dir020',
     'root_dir\\dir0\\dir02\\dir020\\dir0200',
     'root_dir\\dir1',
     'root_dir\\dir1\\file10',
     'root_dir\\dir1\\file11',
     'root_dir\\dir1\\file12',
     'root_dir\\dir2',
     'root_dir\\dir2\\dir20',
     'root_dir\\dir2\\dir20\\file200',
     'root_dir\\dir2\\file20',
     'root_dir\\dir3',
     'root_dir\\file0',
     'root_dir\\file1']
    11 ['dir0\\dir00\\dir000\\file0000',
     'dir0\\dir00\\file000',
     'dir0\\dir01\\file010',
     'dir0\\dir01\\file011',
     'dir1\\file10',
     'dir1\\file11',
     'dir1\\file12',
     'dir2\\dir20\\file200',
     'dir2\\file20',
     'file0',
     'file1']
    


  1. [Python 3]: os.scandir(path=’.’) (Python 3.5+, backport: [PyPI]: scandir)

    Return an iterator of os.DirEntry objects corresponding to the entries in the directory given by path. The entries are yielded in arbitrary order, and the special entries '.' and '..' are not included.

    Using scandir() instead of listdir() can significantly increase the performance of code that also needs file type or file attribute information, because os.DirEntry objects expose this information if the operating system provides it when scanning a directory. All os.DirEntry methods may perform a system call, but is_dir() and is_file() usually only require a system call for symbolic links; os.DirEntry.stat() always requires a system call on Unix but only requires one for symbolic links on Windows.


    >>> import os
    >>> root_dir = os.path.join(".", "root_dir")  # Explicitly prepending current directory
    >>> root_dir
    '.\\root_dir'
    >>>
    >>> scandir_iterator = os.scandir(root_dir)
    >>> scandir_iterator
    <nt.ScandirIterator object at 0x00000268CF4BC140>
    >>> [item.path for item in scandir_iterator]
    ['.\\root_dir\\dir0', '.\\root_dir\\dir1', '.\\root_dir\\dir2', '.\\root_dir\\dir3', '.\\root_dir\\file0', '.\\root_dir\\file1']
    >>>
    >>> [item.path for item in scandir_iterator]  # Will yield an empty list as it was consumed by previous iteration (automatically performed by the list comprehension)
    []
    >>>
    >>> scandir_iterator = os.scandir(root_dir)  # Reinitialize the generator
    >>> for item in scandir_iterator :
    ...     if os.path.isfile(item.path):
    ...             print(item.name)
    ...
    file0
    file1
    

    Notes:

    • It’s similar to os.listdir
    • But it’s also more flexible (and offers more functionality), more Pythonic (and in some cases, faster)


  1. [Python 3]: os.walk(top, topdown=True, onerror=None, followlinks=False)

    Generate the file names in a directory tree by walking the tree either top-down or bottom-up. For each directory in the tree rooted at directory top (including top itself), it yields a 3-tuple (dirpath, dirnames, filenames).


    >>> import os
    >>> root_dir = os.path.join(os.getcwd(), "root_dir")  # Specify the full path
    >>> root_dir
    'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir'
    >>>
    >>> walk_generator = os.walk(root_dir)
    >>> root_dir_entry = next(walk_generator)  # First entry corresponds to the root dir (passed as an argument)
    >>> root_dir_entry
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir', ['dir0', 'dir1', 'dir2', 'dir3'], ['file0', 'file1'])
    >>>
    >>> root_dir_entry[1] + root_dir_entry[2]  # Display dirs and files (direct descendants) in a single list
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [os.path.join(root_dir_entry[0], item) for item in root_dir_entry[1] + root_dir_entry[2]]  # Display all the entries in the previous list by their full path
    ['E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir1', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir2', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir3', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\file0', 'E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\file1']
    >>>
    >>> for entry in walk_generator:  # Display the rest of the elements (corresponding to every subdir)
    ...     print(entry)
    ...
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0', ['dir00', 'dir01', 'dir02'], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir00', ['dir000'], ['file000'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir00\\dir000', [], ['file0000'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir01', [], ['file010', 'file011'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir02', ['dir020'], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir02\\dir020', ['dir0200'], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir0\\dir02\\dir020\\dir0200', [], [])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir1', [], ['file10', 'file11', 'file12'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir2', ['dir20'], ['file20'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir2\\dir20', [], ['file200'])
    ('E:\\Work\\Dev\\StackOverflow\\q003207219\\root_dir\\dir3', [], [])
    

    Notes:

    • Under the scenes, it uses os.scandir (os.listdir on older versions)
    • It does the heavy lifting by recurring in subfolders


  1. [Python 3]: glob.glob(pathname, *, recursive=False) ([Python 3]: glob.iglob(pathname, *, recursive=False))

    Return a possibly-empty list of path names that match pathname, which must be a string containing a path specification. pathname can be either absolute (like /usr/src/Python-1.5/Makefile) or relative (like ../../Tools/*/*.gif), and can contain shell-style wildcards. Broken symlinks are included in the results (as in the shell).

    Changed in version 3.5: Support for recursive globs using “**”.


    >>> import glob, os
    >>> wildcard_pattern = "*"
    >>> root_dir = os.path.join("root_dir", wildcard_pattern)  # Match every file/dir name
    >>> root_dir
    'root_dir\\*'
    >>>
    >>> glob_list = glob.glob(root_dir)
    >>> glob_list
    ['root_dir\\dir0', 'root_dir\\dir1', 'root_dir\\dir2', 'root_dir\\dir3', 'root_dir\\file0', 'root_dir\\file1']
    >>>
    >>> [item.replace("root_dir" + os.path.sep, "") for item in glob_list]  # Strip the dir name and the path separator from begining
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> for entry in glob.iglob(root_dir + "*", recursive=True):
    ...     print(entry)
    ...
    root_dir\
    root_dir\dir0
    root_dir\dir0\dir00
    root_dir\dir0\dir00\dir000
    root_dir\dir0\dir00\dir000\file0000
    root_dir\dir0\dir00\file000
    root_dir\dir0\dir01
    root_dir\dir0\dir01\file010
    root_dir\dir0\dir01\file011
    root_dir\dir0\dir02
    root_dir\dir0\dir02\dir020
    root_dir\dir0\dir02\dir020\dir0200
    root_dir\dir1
    root_dir\dir1\file10
    root_dir\dir1\file11
    root_dir\dir1\file12
    root_dir\dir2
    root_dir\dir2\dir20
    root_dir\dir2\dir20\file200
    root_dir\dir2\file20
    root_dir\dir3
    root_dir\file0
    root_dir\file1
    

    Notes:

    • Uses os.listdir
    • For large trees (especially if recursive is on), iglob is preferred
    • Allows advanced filtering based on name (due to the wildcard)


  1. [Python 3]: class pathlib.Path(*pathsegments) (Python 3.4+, backport: [PyPI]: pathlib2)

    >>> import pathlib
    >>> root_dir = "root_dir"
    >>> root_dir_instance = pathlib.Path(root_dir)
    >>> root_dir_instance
    WindowsPath('root_dir')
    >>> root_dir_instance.name
    'root_dir'
    >>> root_dir_instance.is_dir()
    True
    >>>
    >>> [item.name for item in root_dir_instance.glob("*")]  # Wildcard searching for all direct descendants
    ['dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [os.path.join(item.parent.name, item.name) for item in root_dir_instance.glob("*") if not item.is_dir()]  # Display paths (including parent) for files only
    ['root_dir\\file0', 'root_dir\\file1']
    

    Notes:

    • This is one way of achieving our goal
    • It’s the OOP style of handling paths
    • Offers lots of functionalities


  1. [Python 2]: dircache.listdir(path) (Python 2 only)


    def listdir(path):
        """List directory contents, using cache."""
        try:
            cached_mtime, list = cache[path]
            del cache[path]
        except KeyError:
            cached_mtime, list = -1, []
        mtime = os.stat(path).st_mtime
        if mtime != cached_mtime:
            list = os.listdir(path)
            list.sort()
        cache[path] = mtime, list
        return list
    


  1. [man7]: OPENDIR(3) / [man7]: READDIR(3) / [man7]: CLOSEDIR(3) via [Python 3]: ctypes – A foreign function library for Python (POSIX specific)

    ctypes is a foreign function library for Python. It provides C compatible data types, and allows calling functions in DLLs or shared libraries. It can be used to wrap these libraries in pure Python.

    code_ctypes.py:

    #!/usr/bin/env python3
    
    import sys
    from ctypes import Structure, \
        c_ulonglong, c_longlong, c_ushort, c_ubyte, c_char, c_int, \
        CDLL, POINTER, \
        create_string_buffer, get_errno, set_errno, cast
    
    
    DT_DIR = 4
    DT_REG = 8
    
    char256 = c_char * 256
    
    
    class LinuxDirent64(Structure):
        _fields_ = [
            ("d_ino", c_ulonglong),
            ("d_off", c_longlong),
            ("d_reclen", c_ushort),
            ("d_type", c_ubyte),
            ("d_name", char256),
        ]
    
    LinuxDirent64Ptr = POINTER(LinuxDirent64)
    
    libc_dll = this_process = CDLL(None, use_errno=True)
    # ALWAYS set argtypes and restype for functions, otherwise it's UB!!!
    opendir = libc_dll.opendir
    readdir = libc_dll.readdir
    closedir = libc_dll.closedir
    
    
    def get_dir_content(path):
        ret = [path, list(), list()]
        dir_stream = opendir(create_string_buffer(path.encode()))
        if (dir_stream == 0):
            print("opendir returned NULL (errno: {:d})".format(get_errno()))
            return ret
        set_errno(0)
        dirent_addr = readdir(dir_stream)
        while dirent_addr:
            dirent_ptr = cast(dirent_addr, LinuxDirent64Ptr)
            dirent = dirent_ptr.contents
            name = dirent.d_name.decode()
            if dirent.d_type & DT_DIR:
                if name not in (".", ".."):
                    ret[1].append(name)
            elif dirent.d_type & DT_REG:
                ret[2].append(name)
            dirent_addr = readdir(dir_stream)
        if get_errno():
            print("readdir returned NULL (errno: {:d})".format(get_errno()))
        closedir(dir_stream)
        return ret
    
    
    def main():
        print("{:s} on {:s}\n".format(sys.version, sys.platform))
        root_dir = "root_dir"
        entries = get_dir_content(root_dir)
        print(entries)
    
    
    if __name__ == "__main__":
        main()
    

    Notes:

    • It loads the three functions from libc (loaded in the current process) and calls them (for more details check [SO]: How do I check whether a file exists without exceptions? (@CristiFati’s answer) – last notes from item #4.). That would place this approach very close to the Python / C edge
    • LinuxDirent64 is the ctypes representation of struct dirent64 from [man7]: dirent.h(0P) (so are the DT_ constants) from my machine: Ubtu 16 x64 (4.10.0-40-generic and libc6-dev:amd64). On other flavors/versions, the struct definition might differ, and if so, the ctypes alias should be updated, otherwise it will yield Undefined Behavior
    • It returns data in the os.walk‘s format. I didn’t bother to make it recursive, but starting from the existing code, that would be a fairly trivial task
    • Everything is doable on Win as well, the data (libraries, functions, structs, constants, …) differ


    Output:

    [cfati@cfati-ubtu16x64-0:~/Work/Dev/StackOverflow/q003207219]> ./code_ctypes.py
    3.5.2 (default, Nov 12 2018, 13:43:14)
    [GCC 5.4.0 20160609] on linux
    
    ['root_dir', ['dir2', 'dir1', 'dir3', 'dir0'], ['file1', 'file0']]
    


  1. [ActiveState.Docs]: win32file.FindFilesW (Win specific)

    Retrieves a list of matching filenames, using the Windows Unicode API. An interface to the API FindFirstFileW/FindNextFileW/Find close functions.


    >>> import os, win32file, win32con
    >>> root_dir = "root_dir"
    >>> wildcard = "*"
    >>> root_dir_wildcard = os.path.join(root_dir, wildcard)
    >>> entry_list = win32file.FindFilesW(root_dir_wildcard)
    >>> len(entry_list)  # Don't display the whole content as it's too long
    8
    >>> [entry[-2] for entry in entry_list]  # Only display the entry names
    ['.', '..', 'dir0', 'dir1', 'dir2', 'dir3', 'file0', 'file1']
    >>>
    >>> [entry[-2] for entry in entry_list if entry[0] & win32con.FILE_ATTRIBUTE_DIRECTORY and entry[-2] not in (".", "..")]  # Filter entries and only display dir names (except self and parent)
    ['dir0', 'dir1', 'dir2', 'dir3']
    >>>
    >>> [os.path.join(root_dir, entry[-2]) for entry in entry_list if entry[0] & (win32con.FILE_ATTRIBUTE_NORMAL | win32con.FILE_ATTRIBUTE_ARCHIVE)]  # Only display file "full" names
    ['root_dir\\file0', 'root_dir\\file1']
    

    Notes:


  1. Install some (other) third-party package that does the trick
    • Most likely, will rely on one (or more) of the above (maybe with slight customizations)


Notes:

  • Code is meant to be portable (except places that target a specific area – which are marked) or cross:

    • platform (Nix, Win, )
    • Python version (2, 3, )
  • Multiple path styles (absolute, relatives) were used across the above variants, to illustrate the fact that the “tools” used are flexible in this direction

  • os.listdir and os.scandir use opendir / readdir / closedir ([MS.Docs]: FindFirstFileW function / [MS.Docs]: FindNextFileW function / [MS.Docs]: FindClose function) (via [GitHub]: python/cpython – (master) cpython/Modules/posixmodule.c)

  • win32file.FindFilesW uses those (Win specific) functions as well (via [GitHub]: mhammond/pywin32 – (master) pywin32/win32/src/win32file.i)

  • _get_dir_content (from point #1.) can be implemented using any of these approaches (some will require more work and some less)

    • Some advanced filtering (instead of just file vs. dir) could be done: e.g. the include_folders argument could be replaced by another one (e.g. filter_func) which would be a function that takes a path as an argument: filter_func=lambda x: True (this doesn’t strip out anything) and inside _get_dir_content something like: if not filter_func(entry_with_path): continue (if the function fails for one entry, it will be skipped), but the more complex the code becomes, the longer it will take to execute
  • Nota bene! Since recursion is used, I must mention that I did some tests on my laptop (Win 10 x64), totally unrelated to this problem, and when the recursion level was reaching values somewhere in the (990 .. 1000) range (recursionlimit – 1000 (default)), I got StackOverflow :). If the directory tree exceeds that limit (I am not an FS expert, so I don’t know if that is even possible), that could be a problem.
    I must also mention that I didn’t try to increase recursionlimit because I have no experience in the area (how much can I increase it before having to also increase the stack at OS level), but in theory there will always be the possibility for failure, if the dir depth is larger than the highest possible recursionlimit (on that machine)

  • The code samples are for demonstrative purposes only. That means that I didn’t take into account error handling (I don’t think there’s any try / except / else / finally block), so the code is not robust (the reason is: to keep it as simple and short as possible). For production, error handling should be added as well

Other approaches:

  1. Use Python only as a wrapper

    • Everything is done using another technology
    • That technology is invoked from Python
    • The most famous flavor that I know is what I call the system administrator approach:

      • Use Python (or any programming language for that matter) in order to execute shell commands (and parse their outputs)
      • Some consider this a neat hack
      • I consider it more like a lame workaround (gainarie), as the action per se is performed from shell (cmd in this case), and thus doesn’t have anything to do with Python.
      • Filtering (grep / findstr) or output formatting could be done on both sides, but I’m not going to insist on it. Also, I deliberately used os.system instead of subprocess.Popen.
      (py35x64_test) E:\Work\Dev\StackOverflow\q003207219>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" -c "import os;os.system(\"dir /b root_dir\")"
      dir0
      dir1
      dir2
      dir3
      file0
      file1
      

    In general this approach is to be avoided, since if some command output format slightly differs between OS versions/flavors, the parsing code should be adapted as well; not to mention differences between locales).


回答 8

我真的很喜欢adamk的答案,建议您使用glob()同名模块中的。这使您可以与进行模式匹配*

但是,正如其他人在评论中指出的那样,它们glob()可能会因不一致的斜线方向而绊倒。为了解决这个问题,建议您使用模块中的join()expanduser()函数,也可以使用os.path模块中的getcwd()函数os

例如:

from glob import glob

# Return everything under C:\Users\admin that contains a folder called wlp.
glob('C:\Users\admin\*\wlp')

上面的代码很糟糕-路径已经过硬编码,并且只能在Windows上的驱动器名称和\s硬编码到路径之间使用。

from glob    import glob
from os.path import join

# Return everything under Users, admin, that contains a folder called wlp.
glob(join('Users', 'admin', '*', 'wlp'))

上面的方法效果更好,但是它依赖Users于Windows上经常发现的文件夹名称,而在其他OS上却很少见。它还依赖于具有特定名称的用户admin

from glob    import glob
from os.path import expanduser, join

# Return everything under the user directory that contains a folder called wlp.
glob(join(expanduser('~'), '*', 'wlp'))

这可以在所有平台上完美运行。

另一个很好的示例,它可以在各种平台上完美运行,并且有所不同:

from glob    import glob
from os      import getcwd
from os.path import join

# Return everything under the current directory that contains a folder called wlp.
glob(join(getcwd(), '*', 'wlp'))

希望这些示例可以帮助您了解在标准Python库模块中可以找到的一些功能的强大功能。

I really liked adamk’s answer, suggesting that you use glob(), from the module of the same name. This allows you to have pattern matching with *s.

But as other people pointed out in the comments, glob() can get tripped up over inconsistent slash directions. To help with that, I suggest you use the join() and expanduser() functions in the os.path module, and perhaps the getcwd() function in the os module, as well.

As examples:

from glob import glob

# Return everything under C:\Users\admin that contains a folder called wlp.
glob('C:\Users\admin\*\wlp')

The above is terrible – the path has been hardcoded and will only ever work on Windows between the drive name and the \s being hardcoded into the path.

from glob    import glob
from os.path import join

# Return everything under Users, admin, that contains a folder called wlp.
glob(join('Users', 'admin', '*', 'wlp'))

The above works better, but it relies on the folder name Users which is often found on Windows and not so often found on other OSs. It also relies on the user having a specific name, admin.

from glob    import glob
from os.path import expanduser, join

# Return everything under the user directory that contains a folder called wlp.
glob(join(expanduser('~'), '*', 'wlp'))

This works perfectly across all platforms.

Another great example that works perfectly across platforms and does something a bit different:

from glob    import glob
from os      import getcwd
from os.path import join

# Return everything under the current directory that contains a folder called wlp.
glob(join(getcwd(), '*', 'wlp'))

Hope these examples help you see the power of a few of the functions you can find in the standard Python library modules.


回答 9

def list_files(path):
    # returns a list of names (with extension, without full path) of all files 
    # in folder path
    files = []
    for name in os.listdir(path):
        if os.path.isfile(os.path.join(path, name)):
            files.append(name)
    return files 
def list_files(path):
    # returns a list of names (with extension, without full path) of all files 
    # in folder path
    files = []
    for name in os.listdir(path):
        if os.path.isfile(os.path.join(path, name)):
            files.append(name)
    return files 

回答 10

如果您正在寻找find的Python实现,这是我经常使用的食谱:

from findtools.find_files import (find_files, Match)

# Recursively find all *.sh files in **/usr/bin**
sh_files_pattern = Match(filetype='f', name='*.sh')
found_files = find_files(path='/usr/bin', match=sh_files_pattern)

for found_file in found_files:
    print found_file

因此,我用它制作了一个PyPI 软件包,并且还有一个GitHub存储库。我希望有人发现它可能对该代码有用。

If you are looking for a Python implementation of find, this is a recipe I use rather frequently:

from findtools.find_files import (find_files, Match)

# Recursively find all *.sh files in **/usr/bin**
sh_files_pattern = Match(filetype='f', name='*.sh')
found_files = find_files(path='/usr/bin', match=sh_files_pattern)

for found_file in found_files:
    print found_file

So I made a PyPI package out of it and there is also a GitHub repository. I hope that someone finds it potentially useful for this code.


回答 11

为了获得更好的结果,您可以listdir()os模块的方法与生成器一起使用(生成器是保持其状态的强大迭代器,还记得吗?)。以下代码在这两个版本上均可正常使用:Python 2和Python 3。

这是一个代码:

import os

def files(path):  
    for file in os.listdir(path):
        if os.path.isfile(os.path.join(path, file)):
            yield file

for file in files("."):  
    print (file)

listdir()方法返回给定目录的条目列表。如果给定的条目是文件,则该方法os.path.isfile()返回True。并且yield操作员退出功能但保持其当前状态,并且仅返回检测为文件的条目的名称。以上所有内容使我们可以循环生成器功能。

For greater results, you can use listdir() method of the os module along with a generator (a generator is a powerful iterator that keeps its state, remember?). The following code works fine with both versions: Python 2 and Python 3.

Here’s a code:

import os

def files(path):  
    for file in os.listdir(path):
        if os.path.isfile(os.path.join(path, file)):
            yield file

for file in files("."):  
    print (file)

The listdir() method returns the list of entries for the given directory. The method os.path.isfile() returns True if the given entry is a file. And the yield operator quits the func but keeps its current state, and it returns only the name of the entry detected as a file. All the above allows us to loop over the generator function.


回答 12

返回绝对文件路径的列表,不会递归到子目录中

L = [os.path.join(os.getcwd(),f) for f in os.listdir('.') if os.path.isfile(os.path.join(os.getcwd(),f))]

Returning a list of absolute filepaths, does not recurse into subdirectories

L = [os.path.join(os.getcwd(),f) for f in os.listdir('.') if os.path.isfile(os.path.join(os.getcwd(),f))]

回答 13

import os
import os.path


def get_files(target_dir):
    item_list = os.listdir(target_dir)

    file_list = list()
    for item in item_list:
        item_dir = os.path.join(target_dir,item)
        if os.path.isdir(item_dir):
            file_list += get_files(item_dir)
        else:
            file_list.append(item_dir)
    return file_list

在这里,我使用递归结构。

import os
import os.path


def get_files(target_dir):
    item_list = os.listdir(target_dir)

    file_list = list()
    for item in item_list:
        item_dir = os.path.join(target_dir,item)
        if os.path.isdir(item_dir):
            file_list += get_files(item_dir)
        else:
            file_list.append(item_dir)
    return file_list

Here I use a recursive structure.


回答 14

一位聪明的老师曾经告诉我:

当有几种确定的方法可以做某事时,没有一种方法适合所有情况。

因此,我将为问题的一个子集添加一个解决方案:很多时候,我们只想检查文件是否匹配开始字符串和结束字符串,而无需进入子目录。因此,我们想要一个返回文件名列表的函数,例如:

filenames = dir_filter('foo/baz', radical='radical', extension='.txt')

如果您想先声明两个函数,可以这样做:

def file_filter(filename, radical='', extension=''):
    "Check if a filename matches a radical and extension"
    if not filename:
        return False
    filename = filename.strip()
    return(filename.startswith(radical) and filename.endswith(extension))

def dir_filter(dirname='', radical='', extension=''):
    "Filter filenames in directory according to radical and extension"
    if not dirname:
        dirname = '.'
    return [filename for filename in os.listdir(dirname)
                if file_filter(filename, radical, extension)]

此解决方案可以使用正则表达式轻松进行一般化(pattern如果您不希望模式始终坚持文件名的开头或结尾,则可能需要添加一个参数)。

A wise teacher told me once that:

When there are several established ways to do something, none of them is good for all cases.

I will thus add a solution for a subset of the problem: quite often, we only want to check whether a file matches a start string and an end string, without going into subdirectories. We would thus like a function that returns a list of filenames, like:

filenames = dir_filter('foo/baz', radical='radical', extension='.txt')

If you care to first declare two functions, this can be done:

def file_filter(filename, radical='', extension=''):
    "Check if a filename matches a radical and extension"
    if not filename:
        return False
    filename = filename.strip()
    return(filename.startswith(radical) and filename.endswith(extension))

def dir_filter(dirname='', radical='', extension=''):
    "Filter filenames in directory according to radical and extension"
    if not dirname:
        dirname = '.'
    return [filename for filename in os.listdir(dirname)
                if file_filter(filename, radical, extension)]

This solution could be easily generalized with regular expressions (and you might want to add a pattern argument, if you do not want your patterns to always stick to the start or end of the filename).


回答 15

使用生成器

import os
def get_files(search_path):
     for (dirpath, _, filenames) in os.walk(search_path):
         for filename in filenames:
             yield os.path.join(dirpath, filename)
list_files = get_files('.')
for filename in list_files:
    print(filename)

Using generators

import os
def get_files(search_path):
     for (dirpath, _, filenames) in os.walk(search_path):
         for filename in filenames:
             yield os.path.join(dirpath, filename)
list_files = get_files('.')
for filename in list_files:
    print(filename)

回答 16

Python 3.4+的另一个非常易读的变体是使用pathlib.Path.glob:

from pathlib import Path
folder = '/foo'
[f for f in Path(folder).glob('*') if f.is_file()]

进行更具体的说明很简单,例如,在所有子目录中仅查找不是符号链接的Python源文件:

[f for f in Path(folder).glob('**/*.py') if not f.is_symlink()]

Another very readable variant for Python 3.4+ is using pathlib.Path.glob:

from pathlib import Path
folder = '/foo'
[f for f in Path(folder).glob('*') if f.is_file()]

It is simple to make more specific, e.g. only look for Python source files which are not symbolic links, also in all subdirectories:

[f for f in Path(folder).glob('**/*.py') if not f.is_symlink()]

回答 17

这是我的通用功能。它返回文件路径而不是文件名的列表,因为我发现它更有用。它具有一些可选参数,使其具有通用性。例如,我经常将其与诸如pattern='*.txt'或那样的参数一起使用subfolders=True

import os
import fnmatch

def list_paths(folder='.', pattern='*', case_sensitive=False, subfolders=False):
    """Return a list of the file paths matching the pattern in the specified 
    folder, optionally including files inside subfolders.
    """
    match = fnmatch.fnmatchcase if case_sensitive else fnmatch.fnmatch
    walked = os.walk(folder) if subfolders else [next(os.walk(folder))]
    return [os.path.join(root, f)
            for root, dirnames, filenames in walked
            for f in filenames if match(f, pattern)]

Here’s my general-purpose function for this. It returns a list of file paths rather than filenames since I found that to be more useful. It has a few optional arguments that make it versatile. For instance, I often use it with arguments like pattern='*.txt' or subfolders=True.

import os
import fnmatch

def list_paths(folder='.', pattern='*', case_sensitive=False, subfolders=False):
    """Return a list of the file paths matching the pattern in the specified 
    folder, optionally including files inside subfolders.
    """
    match = fnmatch.fnmatchcase if case_sensitive else fnmatch.fnmatch
    walked = os.walk(folder) if subfolders else [next(os.walk(folder))]
    return [os.path.join(root, f)
            for root, dirnames, filenames in walked
            for f in filenames if match(f, pattern)]

回答 18

我将提供一个示例liner,其中可以提供sourcepath和文件类型作为输入。该代码返回带有csv扩展名的文件名列表。使用万一需要返回所有文件。这还将递归扫描子目录。

[y for x in os.walk(sourcePath) for y in glob(os.path.join(x[0], '*.csv'))]

根据需要修改文件扩展名和源路径。

I will provide a sample one liner where sourcepath and file type can be provided as input. The code returns a list of filenames with csv extension. Use . in case all files needs to be returned. This will also recursively scans the subdirectories.

[y for x in os.walk(sourcePath) for y in glob(os.path.join(x[0], '*.csv'))]

Modify file extensions and source path as needed.


回答 19

对于python2:pip install rglob

import rglob
file_list=rglob.rglob("/home/base/dir/", "*")
print file_list

For python2: pip install rglob

import rglob
file_list=rglob.rglob("/home/base/dir/", "*")
print file_list

回答 20

dircache是“自2.6版起不推荐使用:dircache模块已在Python 3.0中删除。”

import dircache
list = dircache.listdir(pathname)
i = 0
check = len(list[0])
temp = []
count = len(list)
while count != 0:
  if len(list[i]) != check:
     temp.append(list[i-1])
     check = len(list[i])
  else:
    i = i + 1
    count = count - 1

print temp

dircache is “Deprecated since version 2.6: The dircache module has been removed in Python 3.0.”

import dircache
list = dircache.listdir(pathname)
i = 0
check = len(list[0])
temp = []
count = len(list)
while count != 0:
  if len(list[i]) != check:
     temp.append(list[i-1])
     check = len(list[i])
  else:
    i = i + 1
    count = count - 1

print temp