import os

for filename in os.listdir("C:\\temp"):
    print  filename

import os

for filename in os.listdir("C:\\temp"):
    print  filename

只递归列出文件的漂亮的一行代码。我在setup.py package_data指令中使用了这个:

import os

[os.path.join(x[0],y) for x in os.walk('<some_directory>') for y in x[2]]

我知道这不是问题的答案，但可能会派上用场

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

简单的方法:

list_output_files = [os.getcwd()+"\\"+f for f in os.listdir(os.getcwd())]

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)]