我更喜欢使用glob模块，因为它可以进行模式匹配和扩展。

import glob
print(glob.glob("/home/adam/*"))

它可以直观地进行模式匹配

import glob
# All files and directories ending with .txt and that don't begin with a dot:
print(glob.glob("/home/adam/*.txt")) 
# All files and directories ending with .txt with depth of 2 folders, ignoring names beginning with a dot:
print(glob.glob("/home/adam/*/*.txt")) 

它将返回一个包含查询文件和目录的列表：

['/home/adam/file1.txt', '/home/adam/file2.txt', .... ]

注意，glob忽略以点开头的文件和目录。，因为这些被认为是隐藏的文件和目录，除非模式类似于.*。

使用glob.escape转义不应该是模式的字符串：

print(glob.glob(glob.escape(directory_name) + "/*.txt"))

从目录及其所有子目录获取完整文件路径

import os

def get_filepaths(directory):
    """
    This function will generate the file names in a directory 
    tree by walking the tree either top-down or bottom-up. For each 
    directory in the tree rooted at directory top (including top itself), 
    it yields a 3-tuple (dirpath, dirnames, filenames).
    """
    file_paths = []  # List which will store all of the full filepaths.

    # Walk the tree.
    for root, directories, files in os.walk(directory):
        for filename in files:
            # Join the two strings in order to form the full filepath.
            filepath = os.path.join(root, filename)
            file_paths.append(filepath)  # Add it to the list.

    return file_paths  # Self-explanatory.

# Run the above function and store its results in a variable.   
full_file_paths = get_filepaths("/Users/johnny/Desktop/TEST")

我在上述函数中提供的路径包含3个文件，其中两个位于根目录中，另一个位于名为“subfolder”的子文件夹中打印将打印列表的完整文件路径：['/Users/johnny/Desktop/TEST/file1.txt'，'/Users/johnny/Desctop/TEST-file2.txt'，'/Users/johnny/Desktop/STEST/SUBFOLDER/file3.dat']

如果愿意，您可以打开并阅读内容，或者只关注扩展名为“.dat”的文件，如下面的代码所示：

for f in full_file_paths:
  if f.endswith(".dat"):
    print f

/用户/johnny/Desktop/TEST/SUBFOLDER/file3.dat

def list_files(path):
    # returns a list of names (with extension, without full path) of all files 
    # in folder path
    files = []
    for name in os.listdir(path):
        if os.path.isfile(os.path.join(path, name)):
            files.append(name)
    return files 

使用发电机

import os
def get_files(search_path):
     for (dirpath, _, filenames) in os.walk(search_path):
         for filename in filenames:
             yield os.path.join(dirpath, filename)
list_files = get_files('.')
for filename in list_files:
    print(filename)

import os
os.listdir("somedirectory")

将返回“somedirectory”中所有文件和目录的列表。

listdir（）返回目录中的所有内容——包括文件和目录。

os.path的isfile（）只能用于列出文件：

from os import listdir
from os.path import isfile, join
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]

或者，os.walk（）为它访问的每个目录生成两个列表——一个用于文件，一个用于目录。如果您只想要顶级目录，则可以在第一次生成时中断：

from os import walk

f = []
for (dirpath, dirnames, filenames) in walk(mypath):
    f.extend(filenames)
    break

或更短：

from os import walk

filenames = next(walk(mypath), (None, None, []))[2]  # [] if no file