我如何在Python中获得给定目录中的所有文件(和目录)的列表?


当前回答

这是一种遍历目录树中每个文件和目录的方法:

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')

其他回答

我写了一个很长的版本,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:])

和我一起工作的是萨利赫在本页其他地方的回答的修改版本。

代码如下:

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

下面是一行python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)]

这段代码列出给定目录名中所有文件和目录的完整路径。

下面是我经常使用的一个辅助函数:

import os

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

试试这个:

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