如何在Python中列出目录中的所有文件并将其添加到列表中?


当前回答

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

其他回答

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)

dircache是“自2.6版以来已弃用:Python 3.0中已删除dircache模块。”

import dircache
list = dircache.listdir(pathname)
i = 0
check = len(list[0])
temp = []
count = len(list)
while count != 0:
  if len(list[i]) != check:
     temp.append(list[i-1])
     check = len(list[i])
  else:
    i = i + 1
    count = count - 1

print temp

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

从3.4版开始,有内置的迭代器,比os.listdir()高效得多:

pathlib:3.4版新增。

>>> import pathlib
>>> [p for p in pathlib.Path('.').iterdir() if p.is_file()]

根据PEP428,pathlib库的目的是提供一个简单的类层次结构来处理文件系统路径和用户对它们进行的常见操作。

os.scandir():3.5版新增。

>>> import os
>>> [entry for entry in os.scandir('.') if entry.is_file()]

注意,os.walk()使用os.scandir()而不是3.5版的os.listdir(),根据PEP471,它的速度提高了2-20倍。

让我也推荐阅读ShadowRanger在下面的评论。