获取目录中文件的过滤列表

问题:获取目录中文件的过滤列表

我正在尝试使用Python获取目录中的文件列表,但是我不想要所有文件的列表。

我本质上想要的是能够执行以下操作但使用Python而不执行ls的功能。

ls 145592*.jpg

如果没有内置方法,我目前正在考虑编写一个for循环以遍历an的结果。 os.listdir()并将所有匹配的文件附加到新列表中。

但是,该目录中有很多文件,因此我希望有一种更有效的方法(或内置方法)。

I am trying to get a list of files in a directory using Python, but I do not want a list of ALL the files.

What I essentially want is the ability to do something like the following but using Python and not executing ls.

ls 145592*.jpg

If there is no built-in method for this, I am currently thinking of writing a for loop to iterate through the results of an os.listdir() and to append all the matching files to a new list.

However, there are a lot of files in that directory and therefore I am hoping there is a more efficient method (or a built-in method).


回答 0

import glob

jpgFilenamesList = glob.glob('145592*.jpg')

See glob in python documenttion


回答 1

glob.glob()绝对是做到这一点的方式(根据Ignacio)。但是,如果您确实需要更复杂的匹配,则可以使用列表理解和来完成re.match(),例如:

files = [f for f in os.listdir('.') if re.match(r'[0-9]+.*\.jpg', f)]

更加灵活,但是您注意到效率更低。

glob.glob() is definitely the way to do it (as per Ignacio). However, if you do need more complicated matching, you can do it with a list comprehension and re.match(), something like so:

files = [f for f in os.listdir('.') if re.match(r'[0-9]+.*\.jpg', f)]

More flexible, but as you note, less efficient.


回答 2

把事情简单化:

import os
relevant_path = "[path to folder]"
included_extensions = ['jpg','jpeg', 'bmp', 'png', 'gif']
file_names = [fn for fn in os.listdir(relevant_path)
              if any(fn.endswith(ext) for ext in included_extensions)]

我更喜欢这种形式的列表理解,因为它的英文读起来很好。

我将第四行读为:对于os.listdir中路径的每个fn,请仅提供与我包含的任何扩展名匹配的那些fn。

对于新手python程序员来说,可能很难真正习惯于使用列表推导进行过滤,并且对于非常大的数据集,它可能会有一些内存开销,但是对于列出目录和其他简单的字符串过滤任务,列表推导会导致更干净可记录的代码。

这种设计的唯一之处在于,它不能保护您避免犯错误,而不是传递字符串而不是列表。例如,如果您不小心将字符串转换为列表,并最终检查了字符串的所有字符,则可能最终会得到一系列误报。

但是,拥有一个易于解决的问题比解决一个难以理解的解决方案要好。

Keep it simple:

import os
relevant_path = "[path to folder]"
included_extensions = ['jpg','jpeg', 'bmp', 'png', 'gif']
file_names = [fn for fn in os.listdir(relevant_path)
              if any(fn.endswith(ext) for ext in included_extensions)]

I prefer this form of list comprehensions because it reads well in English.

I read the fourth line as: For each fn in os.listdir for my path, give me only the ones that match any one of my included extensions.

It may be hard for novice python programmers to really get used to using list comprehensions for filtering, and it can have some memory overhead for very large data sets, but for listing a directory and other simple string filtering tasks, list comprehensions lead to more clean documentable code.

The only thing about this design is that it doesn’t protect you against making the mistake of passing a string instead of a list. For example if you accidentally convert a string to a list and end up checking against all the characters of a string, you could end up getting a slew of false positives.

But it’s better to have a problem that’s easy to fix than a solution that’s hard to understand.


回答 3

另外的选择:

>>> import os, fnmatch
>>> fnmatch.filter(os.listdir('.'), '*.py')
['manage.py']

https://docs.python.org/3/library/fnmatch.html

Another option:

>>> import os, fnmatch
>>> fnmatch.filter(os.listdir('.'), '*.py')
['manage.py']

https://docs.python.org/3/library/fnmatch.html


回答 4

过滤glob模块:

导入球

import glob

通配符:

files=glob.glob("data/*")
print(files)

Out:

['data/ks_10000_0', 'data/ks_1000_0', 'data/ks_100_0', 'data/ks_100_1',
'data/ks_100_2', 'data/ks_106_0', 'data/ks_19_0', 'data/ks_200_0', 'data/ks_200_1', 
'data/ks_300_0', 'data/ks_30_0', 'data/ks_400_0', 'data/ks_40_0', 'data/ks_45_0', 
'data/ks_4_0', 'data/ks_500_0', 'data/ks_50_0', 'data/ks_50_1', 'data/ks_60_0', 
'data/ks_82_0', 'data/ks_lecture_dp_1', 'data/ks_lecture_dp_2']

接头扩展.txt

files = glob.glob("/home/ach/*/*.txt")

一个字符

glob.glob("/home/ach/file?.txt")

编号范围

glob.glob("/home/ach/*[0-9]*")

字母范围

glob.glob("/home/ach/[a-c]*")

Filter with glob module:

Import glob

import glob

Wild Cards:

files=glob.glob("data/*")
print(files)

Out:

['data/ks_10000_0', 'data/ks_1000_0', 'data/ks_100_0', 'data/ks_100_1',
'data/ks_100_2', 'data/ks_106_0', 'data/ks_19_0', 'data/ks_200_0', 'data/ks_200_1', 
'data/ks_300_0', 'data/ks_30_0', 'data/ks_400_0', 'data/ks_40_0', 'data/ks_45_0', 
'data/ks_4_0', 'data/ks_500_0', 'data/ks_50_0', 'data/ks_50_1', 'data/ks_60_0', 
'data/ks_82_0', 'data/ks_lecture_dp_1', 'data/ks_lecture_dp_2']

Fiter extension .txt:

files = glob.glob("/home/ach/*/*.txt")

A single character

glob.glob("/home/ach/file?.txt")

Number Ranges

glob.glob("/home/ach/*[0-9]*")

Alphabet Ranges

glob.glob("/home/ach/[a-c]*")

回答 5

初步代码

import glob
import fnmatch
import pathlib
import os

pattern = '*.py'
path = '.'

解决方案1-使用“ glob”

# lookup in current dir
glob.glob(pattern)

In [2]: glob.glob(pattern)
Out[2]: ['wsgi.py', 'manage.py', 'tasks.py']

解决方案2-使用“操作系统” +“ fnmatch”

版本2.1-在当前目录中查找

# lookup in current dir
fnmatch.filter(os.listdir(path), pattern)

In [3]: fnmatch.filter(os.listdir(path), pattern)
Out[3]: ['wsgi.py', 'manage.py', 'tasks.py']

版本2.2-递归查找

# lookup recursive
for dirpath, dirnames, filenames in os.walk(path):

    if not filenames:
        continue

    pythonic_files = fnmatch.filter(filenames, pattern)
    if pythonic_files:
        for file in pythonic_files:
            print('{}/{}'.format(dirpath, file))

结果

./wsgi.py
./manage.py
./tasks.py
./temp/temp.py
./apps/diaries/urls.py
./apps/diaries/signals.py
./apps/diaries/actions.py
./apps/diaries/querysets.py
./apps/library/tests/test_forms.py
./apps/library/migrations/0001_initial.py
./apps/polls/views.py
./apps/polls/formsets.py
./apps/polls/reports.py
./apps/polls/admin.py

解决方案3使用“ pathlib”

# lookup in current dir
path_ = pathlib.Path('.')
tuple(path_.glob(pattern))

# lookup recursive
tuple(path_.rglob(pattern))

笔记:

  1. 在Python 3.4上测试
  2. 仅在Python 3.4中添加了模块“ pathlib”
  3. Python 3.5添加了glob.glob https://docs.python.org/3.5/library/glob.html#glob.glob递归查找的功能 。由于我的机器安装了Python 3.4,因此尚未进行测试。

Preliminary code

import glob
import fnmatch
import pathlib
import os

pattern = '*.py'
path = '.'

Solution 1 – use “glob”

# lookup in current dir
glob.glob(pattern)

In [2]: glob.glob(pattern)
Out[2]: ['wsgi.py', 'manage.py', 'tasks.py']

Solution 2 – use “os” + “fnmatch”

Variant 2.1 – Lookup in current dir

# lookup in current dir
fnmatch.filter(os.listdir(path), pattern)

In [3]: fnmatch.filter(os.listdir(path), pattern)
Out[3]: ['wsgi.py', 'manage.py', 'tasks.py']

Variant 2.2 – Lookup recursive

# lookup recursive
for dirpath, dirnames, filenames in os.walk(path):

    if not filenames:
        continue

    pythonic_files = fnmatch.filter(filenames, pattern)
    if pythonic_files:
        for file in pythonic_files:
            print('{}/{}'.format(dirpath, file))

Result

./wsgi.py
./manage.py
./tasks.py
./temp/temp.py
./apps/diaries/urls.py
./apps/diaries/signals.py
./apps/diaries/actions.py
./apps/diaries/querysets.py
./apps/library/tests/test_forms.py
./apps/library/migrations/0001_initial.py
./apps/polls/views.py
./apps/polls/formsets.py
./apps/polls/reports.py
./apps/polls/admin.py

Solution 3 – use “pathlib”

# lookup in current dir
path_ = pathlib.Path('.')
tuple(path_.glob(pattern))

# lookup recursive
tuple(path_.rglob(pattern))

Notes:

  1. Tested on the Python 3.4
  2. The module “pathlib” was added only in the Python 3.4
  3. The Python 3.5 added a feature for recursive lookup with glob.glob https://docs.python.org/3.5/library/glob.html#glob.glob. Since my machine is installed with Python 3.4, I have not tested that.

回答 6

使用os.walk递归列出您的文件

import os
root = "/home"
pattern = "145992"
alist_filter = ['jpg','bmp','png','gif'] 
path=os.path.join(root,"mydir_to_scan")
for r,d,f in os.walk(path):
    for file in f:
        if file[-3:] in alist_filter and pattern in file:
            print os.path.join(root,file)

use os.walk to recursively list your files

import os
root = "/home"
pattern = "145992"
alist_filter = ['jpg','bmp','png','gif'] 
path=os.path.join(root,"mydir_to_scan")
for r,d,f in os.walk(path):
    for file in f:
        if file[-3:] in alist_filter and pattern in file:
            print os.path.join(root,file)

回答 7

import os

dir="/path/to/dir"
[x[0]+"/"+f for x in os.walk(dir) for f in x[2] if f.endswith(".jpg")]

这将为您提供jpg文件及其完整路径的列表。您可以替换x[0]+"/"+ff的只是文件名。您也可以f.endswith(".jpg")用所需的任何字符串条件替换。

import os

dir="/path/to/dir"
[x[0]+"/"+f for x in os.walk(dir) for f in x[2] if f.endswith(".jpg")]

This will give you a list of jpg files with their full path. You can replace x[0]+"/"+f with f for just filenames. You can also replace f.endswith(".jpg") with whatever string condition you wish.


回答 8

您可能还需要更高级的方法(我已经实现并打包为findtools):

from findtools.find_files import (find_files, Match)


# Recursively find all *.txt files in **/home/**
txt_files_pattern = Match(filetype='f', name='*.txt')
found_files = find_files(path='/home', match=txt_files_pattern)

for found_file in found_files:
    print found_file

可以安装

pip install findtools

you might also like a more high-level approach (I have implemented and packaged as findtools):

from findtools.find_files import (find_files, Match)


# Recursively find all *.txt files in **/home/**
txt_files_pattern = Match(filetype='f', name='*.txt')
found_files = find_files(path='/home', match=txt_files_pattern)

for found_file in found_files:
    print found_file

can be installed with

pip install findtools

回答 9

“ path / to / images”中带有“ jpg”和“ png”扩展名的文件名:

import os
accepted_extensions = ["jpg", "png"]
filenames = [fn for fn in os.listdir("path/to/images") if fn.split(".")[-1] in accepted_extensions]

Filenames with “jpg” and “png” extensions in “path/to/images”:

import os
accepted_extensions = ["jpg", "png"]
filenames = [fn for fn in os.listdir("path/to/images") if fn.split(".")[-1] in accepted_extensions]

回答 10

您可以使用Python标准库3.4及更高版本中提供的pathlib

from pathlib import Path

files = [f for f in Path.cwd().iterdir() if f.match("145592*.jpg")]

You can use pathlib that is available in Python standard library 3.4 and above.

from pathlib import Path

files = [f for f in Path.cwd().iterdir() if f.match("145592*.jpg")]

回答 11

您可以定义模式并进行检查。在这里,我采用了开始和结束模式,并在文件名中查找它们。FILES包含目录中所有文件的列表。

import os
PATTERN_START = "145592"
PATTERN_END = ".jpg"
CURRENT_DIR = os.path.dirname(os.path.realpath(__file__))
for r,d,FILES in os.walk(CURRENT_DIR):
    for FILE in FILES:
        if PATTERN_START in FILE and PATTERN_END in FILE:
            print FILE

You can define pattern and check for it. Here I have taken both start and end pattern and looking for them in the filename. FILES contains the list of all the files in a directory.

import os
PATTERN_START = "145592"
PATTERN_END = ".jpg"
CURRENT_DIR = os.path.dirname(os.path.realpath(__file__))
for r,d,FILES in os.walk(CURRENT_DIR):
    for FILE in FILES:
        if PATTERN_START in FILE.startwith(PATTERN_START) and PATTERN_END in FILE.endswith(PATTERN_END):
            print FILE

回答 12

str.split()怎么样?没什么可导入的。

import os

image_names = [f for f in os.listdir(path) if len(f.split('.jpg')) == 2]

How about str.split()? Nothing to import.

import os

image_names = [f for f in os.listdir(path) if len(f.split('.jpg')) == 2]

回答 13

您可以使用subprocess.check_ouput()作为

import subprocess

list_files = subprocess.check_output("ls 145992*.jpg", shell=True) 

当然,引号之间的字符串可以是您要在shell中执行并存储输出的任何内容。

You can use subprocess.check_ouput() as

import subprocess

list_files = subprocess.check_output("ls 145992*.jpg", shell=True) 

Of course, the string between quotes can be anything you want to execute in the shell, and store the output.