如何使用glob.glob模块搜索子文件夹?

本文介绍了如何使用glob.glob模块搜索子文件夹?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我想在一个文件夹中打开一系列子文件夹，找到一些文本文件并打印一些文本文件行.我正在使用这个:

I want to open a series of subfolders in a folder and find some text files and print some lines of the text files. I am using this:

configfiles = glob.glob('C:/Users/sam/Desktop/file1/*.txt')

但是，这也无法访问子文件夹.有谁知道我也可以使用相同的命令来访问子文件夹?

But this cannot access the subfolders as well. Does anyone know how I can use the same command to access subfolders as well?

推荐答案

在Python 3.5及更高版本中，使用新的递归**/功能:

In Python 3.5 and newer use the new recursive **/ functionality:

configfiles = glob.glob('C:/Users/sam/Desktop/file1/**/*.txt', recursive=True)

设置recursive时，**后跟路径分隔符将匹配0个或多个子目录.

When recursive is set, ** followed by a path separator matches 0 or more subdirectories.

在早期的Python版本中，glob.glob()无法递归列出子目录中的文件.

In earlier Python versions, glob.glob() cannot list files in subdirectories recursively.

在这种情况下，我将使用 os.walk() 与 fnmatch.filter() 组合:

In that case I'd use os.walk() combined with fnmatch.filter() instead:

import os import fnmatch path = 'C:/Users/sam/Desktop/file1' configfiles = [os.path.join(dirpath, f) for dirpath, dirnames, files in os.walk(path) for f in fnmatch.filter(files, '*.txt')]

这将递归遍历您的目录，并将所有绝对路径名返回到匹配的.txt文件.在这种特定情况下，fnmatch.filter()可能会过大，您也可以使用.endswith()测试:

This'll walk your directories recursively and return all absolute pathnames to matching .txt files. In this specific case the fnmatch.filter() may be overkill, you could also use a .endswith() test:

import os path = 'C:/Users/sam/Desktop/file1' configfiles = [os.path.join(dirpath, f) for dirpath, dirnames, files in os.walk(path) for f in files if f.endswith('.txt')]