如何使用glob()递归查找文件? [英] How to use glob() to find files recursively?

问题描述

这就是我所拥有的:

glob(os.path.join('src','*.c'))

但是我想搜索src的子文件夹.这样的事情会起作用:

but I want to search the subfolders of src. Something like this would work:

glob(os.path.join('src','*.c'))
glob(os.path.join('src','*','*.c'))
glob(os.path.join('src','*','*','*.c'))
glob(os.path.join('src','*','*','*','*.c'))

但这显然是有限且笨拙的.

But this is obviously limited and clunky.

推荐答案

Python 3.5 +

由于您使用的是新的python，因此应使用 pathlib 模块中的pathlib.Path.rglob . >

Since you're on a new python, you should use pathlib.Path.rglob from the the pathlib module.

from pathlib import Path

for path in Path('src').rglob('*.c'):
    print(path.name)

如果您不想使用pathlib，只需使用 ，但不要忘记传递recursive关键字参数.

If you don't want to use pathlib, just use glob.glob, but don't forget to pass in the recursive keyword parameter.

如果匹配的文件以点(.)开头；例如当前目录中的文件或基于Unix的系统上的隐藏文件，请使用 c4> 下面的解决方案.

For cases where matching files beginning with a dot (.); like files in the current directory or hidden files on Unix based system, use the os.walk solution below.

Python的旧版本

对于较旧的Python版本，请使用 os.walk 来递归遍历目录，然后 fnmatch.filter 来匹配一个简单的目录表达式:

For older Python versions, use os.walk to recursively walk a directory and fnmatch.filter to match against a simple expression:

import fnmatch
import os

matches = []
for root, dirnames, filenames in os.walk('src'):
    for filename in fnmatch.filter(filenames, '*.c'):
        matches.append(os.path.join(root, filename))

这篇关于如何使用glob()递归查找文件?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！