如何迭代输出某文件夹下的所有文件的路径?(os.walk用法)

查看目录结构: tree
查看文件结构: os.walk
查看os.walk用法: help(os.walk)

For each directory in the directory tree rooted at top (including top itself, but excluding '.' and '..'), yields a 3-tuple dirpath, dirnames, filenames
dirpath is a string, the path to the directory. dirnames is a list of the names of the subdirectories in dirpath (excluding '.' and '..').
filenames is a list of the names of the non-directory files in dirpath. Note that the names in the lists are just names, with no path components.
To get a full path (which begins with top) to a file or directory in dirpath, do os.path.join(dirpath, name).

根据以上的内容,发现可以迭代输出文件下的所有文件的路径地址:

import os
for root,dirs,files in os.walk("/home/cbb/桌面/"):
    for dir in dirs:
        for file in files:
            print(os.path.join(root,dir,file))
原文地址:https://www.cnblogs.com/everfight/p/7977326.html