Python 带regex的os.walk

我想得到一个适用于我的正则表达式的文件列表。我想我应该使用os.walk，但是我如何将它与regex一起使用呢

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谢谢。

如果您的正则表达式可以翻译成shell表达式，例如

foo/*.txt

，那么您可以使用

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我不知道stdlib中有什么东西实现了这一点，但编写代码并不困难：

import os, os.path

def iter_matching(dirpath, regexp):
    """Generator yielding all files under `dirpath` whose absolute path
       matches the regular expression `regexp`.
       Usage:

           >>> for filename in iter_matching('/', r'/home.*\.bak'):
           ....    # do something
    """
    for dir_, dirnames, filenames in os.walk(dirpath):
        for filename in filenames:
            abspath = os.path.join(dir_, filename)
            if regexp.match(abspath):
                yield abspath

或更一般的：

import os, os.path

def filter_filenames(dirpath, predicate):
    """Usage:

           >>> for filename in filter_filenames('/', re.compile(r'/home.*\.bak').match):
           ....    # do something
    """
    for dir_, dirnames, filenames in os.walk(dirpath):
        for filename in filenames:
            abspath = os.path.join(dir_, filename)
            if predicate(abspath):
                yield abspath

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请发布一些您尝试使用的代码。请发布您建议的解决方案，包括您收到的任何错误消息。这不是

www.do-my-job-for-me.com

import os, os.path

def filter_filenames(dirpath, predicate):
    """Usage:

           >>> for filename in filter_filenames('/', re.compile(r'/home.*\.bak').match):
           ....    # do something
    """
    for dir_, dirnames, filenames in os.walk(dirpath):
        for filename in filenames:
            abspath = os.path.join(dir_, filename)
            if predicate(abspath):
                yield abspath