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)

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)]

对于Python 2

#!/bin/python2

import os

def scan_dir(path):
    print map(os.path.abspath, os.listdir(pwd))

对于Python 3

对于filter和map，你需要用list()来包装它们

#!/bin/python3

import os

def scan_dir(path):
    print(list(map(os.path.abspath, os.listdir(pwd))))

现在的建议是用生成器表达式或列表推导式替换map和filter的使用:

#!/bin/python

import os

def scan_dir(path):
    print([os.path.abspath(f) for f in os.listdir(path)])

试试这个:

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