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.")