过滤os.walk（）目录和文件

Question

我正在寻找一种包含/排除文件模式并从os.walk()呼叫中排除目录的方法。

这是我现在正在做的事情：

import fnmatch
import os

includes = ['*.doc', '*.odt']
excludes = ['/home/paulo-freitas/Documents']

def _filter(paths):
    for path in paths:
        if os.path.isdir(path) and not path in excludes:
            yield path

        for pattern in (includes + excludes):
            if not os.path.isdir(path) and fnmatch.fnmatch(path, pattern):
                yield path

for root, dirs, files in os.walk('/home/paulo-freitas'):
    dirs[:] = _filter(map(lambda d: os.path.join(root, d), dirs))
    files[:] = _filter(map(lambda f: os.path.join(root, f), files))

    for filename in files:
        filename = os.path.join(root, filename)

        print(filename)

有一个更好的方法吗？怎么样？

Answer 1

此解决方案用于fnmatch.translate将glob模式转换为正则表达式（假定include仅用于文件）：

import fnmatch
import os
import os.path
import re

includes = ['*.doc', '*.odt'] # for files only
excludes = ['/home/paulo-freitas/Documents'] # for dirs and files

# transform glob patterns to regular expressions
includes = r'|'.join([fnmatch.translate(x) for x in includes])
excludes = r'|'.join([fnmatch.translate(x) for x in excludes]) or r'$.'

for root, dirs, files in os.walk('/home/paulo-freitas'):

    # exclude dirs
    dirs[:] = [os.path.join(root, d) for d in dirs]
    dirs[:] = [d for d in dirs if not re.match(excludes, d)]

    # exclude/include files
    files = [os.path.join(root, f) for f in files]
    files = [f for f in files if not re.match(excludes, f)]
    files = [f for f in files if re.match(includes, f)]

    for fname in files:
        print fname