Python列表目录、子目录和文件

I'm trying to make a script to list all directory, subdirectory, and files in a given directory.
I tried this:

我正在尝试编写一个脚本，列出给定目录中的所有目录、子目录和文件。我试着这样的:

import sys,os

root = "/home/patate/directory/"
path = os.path.join(root, "targetdirectory")

for r,d,f in os.walk(path):
    for file in f:
        print os.path.join(root,file)

Unfortunatly it doesn't work properly.
I get all the files, but not their complete paths.

不幸的是它不能正常工作。我得到所有的文件，但不是它们的完整路径。

For example if the dir struct would be:

例如，如果目录结构是:

/home/patate/directory/targetdirectory/123/456/789/file.txt

It would print:

它会打印:

/home/patate/directory/targetdirectory/file.txt

What I need is the first result. Any help would be greatly appreciated! Thanks.

我需要的是第一个结果。非常感谢您的帮助!谢谢。

6 个解决方案

#1

Use os.path.join to concatenate the directory and file name:

使用os.path。连接到连接目录和文件名:

for path, subdirs, files in os.walk(root):
    for name in files:
        print os.path.join(path, name)

Note the usage of path and not root in the concatenation, since using root would be incorrect.

注意在连接中使用路径而不是根，因为使用根将是不正确的。

---

In Python 3.4, the pathlib module was added for easier path manipulations. So the equivalent to os.path.join would be:

在Python 3.4中，为更容易的路径操作添加了pathlib模块。也就是os。path。加入是:

pathlib.PurePath(path, name)

The advantage of pathlib is that you can use a variety of useful methods on paths. If you use the concrete Path variant you can also do actual OS calls through them, like chanding into a directory, deleting the path, opening the file it points to and much more.

pathlib的优点是可以在路径上使用各种有用的方法。如果您使用具体的路径变体，您还可以通过它们进行实际的操作系统调用，比如对目录进行搜索、删除路径、打开它指向的文件等等。

#2

Just in case... Getting all files in the directory and subdirectories matching some pattern (*.py for example):

以防……在目录和子目录中获取与某些模式匹配的所有文件(*。py为例):

import os
from fnmatch import fnmatch

root = '/some/directory'
pattern = "*.py"

for path, subdirs, files in os.walk(root):
    for name in files:
        if fnmatch(name, pattern):
            print os.path.join(path, name)

#3

Here is a one-liner:

import os

[val for sublist in [[os.path.join(i[0], j) for j in i[2]] for i in os.walk('./')] for val in sublist]
# Meta comment to ease selecting text

The outer most val for sublist in ... loop flattens the list to be one dimensional. The j loop collects a list of every file basename and joins it to the current path. Finally, the i loop iterates over all directories and sub directories.

子列表中最外层的val。loop将列表变平为一维。j循环收集每个文件basename的列表并将其连接到当前路径。最后，i循环遍历所有目录和子目录。

This example uses the hard-coded path ./ in the os.walk(...) call, you can supplement any path string you like.

本例使用硬编码路径。/在os.walk(…)调用中，您可以补充任何您喜欢的路径字符串。

Note: os.path.expanduser and/or os.path.expandvars can be used for paths strings like ~/

注意:os.path。expanduser和/或os.path。扩展字符可以用于像~/这样的路径字符串

Extending this example:

Its easy to add in file basename tests and directoryname tests.

它很容易添加到文件basename测试和directoryname测试中。

For Example, testing for *.jpg files:

例如，测试*.jpg文件:

... for j in i[2] if j.endswith('.jpg')] ...

Additionally, excluding the .git directory:

此外，不包括.git目录:

... for i in os.walk('./') if '.git' not in i[0].split('/')]

#4

You should use 'r' in your join instead of 'root'

你应该在连接中使用“r”而不是“root”

#5

You can take a look at this sample I made. It uses the os.path.walk function which is deprecated beware.Uses a list to store all the filepaths

你可以看看我做的这个样品。它使用os.path。不赞成的步行功能请注意。使用列表存储所有文件路径

root = "Your root directory"
ex = ".txt"
where_to = "Wherever you wanna write your file to"
def fileWalker(ext,dirname,names):
    '''
    checks files in names'''
    pat = "*" + ext[0]
    for f in names:
        if fnmatch.fnmatch(f,pat):
            ext[1].append(os.path.join(dirname,f))


def writeTo(fList):

    with open(where_to,"w") as f:
        for di_r in fList:
            f.write(di_r + "\n")






if __name__ == '__main__':
    li = []
    os.path.walk(root,fileWalker,[ex,li])

    writeTo(li)

#6

A bit simpler one-liner:

一点简单的一行程序:

import os
from itertools import product, chain

chain.from_iterable([["\\".join(w) for w in product([i[0]], i[2])] for i in os.walk(dir)])