试试这个:

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

你可以使用

os.listdir(path)

参考和更多的操作系统函数看这里:

Python 2文档:https://docs.python.org/2/library/os.html#os.listdir Python 3文档:https://docs.python.org/3/library/os.html#os.listdir

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')

如果你需要通配符，也有一个模块。例如:

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

将返回如下内容:

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

请在这里查看文档。

这是另一种选择。

os.scandir(path='.')

它返回os的迭代器。对应于path指定目录中的条目(以及文件属性信息)的DirEntry对象。

例子:

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文档

下面的代码将列出目录和目录下的文件

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)