blocks|key|232990|text|最简单的方法是自己过滤glob结果。下面是如何使用简单的循环理解来完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|232991|import+glob
res+=+[f+for+f+in+glob.glob("*.txt")+if+"abc"+in+f+or+"123"+in+f+or+"a1b"+in+f]
for+f+in+res:
++++print+f|code-block|syntax|javascript|232992|您也可以使用正则表达式而不使用glob|offset|length|style|CODE|232993|import+os
import+re
res+=+[f+for+f+in+os.listdir(path)+if+re.search(r'(abc%7C123%7Ca1b).*\.txt$',+f)]
for+f+in+res:
++++print+f|232994|(顺便说一句，将变量命名为list不是一个好主意，因为list是Python类型...)|232995|entityMap^0|0|0|F|4|0|0|D|4|R|4|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

The easiest way would be to filter the glob results yourself. Here is how to do it using a simple loop comprehension:

<pre><code>import glob
res = [f for f in glob.glob("*.txt") if "abc" in f or "123" in f or "a1b" in f]
for f in res:
 print f
</code></pre>

You could also use a regexp and no <code>glob</code>:

<pre><code>import os
import re
res = [f for f in os.listdir(path) if re.search(r'(abc|123|a1b).*\.txt$', f)]
for f in res:
 print f
</code></pre>

(By the way, naming a variable <code>list</code> is a bad idea since <code>list</code> is a Python type...)

blocks|key|233016|text|基于其他答案，这里有一个现成的方法来实现这一点。它不是最关键的性能，但它的工作方式如所描述的那样；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|233017|def+reglob(path,+exp,+invert=False):
++++"""glob.glob()+style+searching+which+uses+regex

++++:param+exp:+Regex+expression+for+filename
++++:param+invert:+Invert+match+to+non+matching+files
++++"""

++++m+=+re.compile(exp)

++++if+invert+is+False:
++++++++res+=+[f+for+f+in+os.listdir(path)+if+m.search(f)]
++++else:
++++++++res+=+[f+for+f+in+os.listdir(path)+if+not+m.search(f)]

++++res+=+map(lambda+x:+"%25s/%25s"+%25+(+path,+x,+),+res)
++++return+res|code-block|syntax|javascript|233018|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a ready to use way of doing this, based on the other answers. It's not the most performance critical, but it works as described;

<pre class="lang-py prettyprint-override"><code>def reglob(path, exp, invert=False):
 """glob.glob() style searching which uses regex

 :param exp: Regex expression for filename
 :param invert: Invert match to non matching files
 """

 m = re.compile(exp)

 if invert is False:
 res = [f for f in os.listdir(path) if m.search(f)]
 else:
 res = [f for f in os.listdir(path) if not m.search(f)]

 res = map(lambda x: "%s/%s" % ( path, x, ), res)
 return res
</code></pre>

blocks|key|450967|text|for+filename+in+glob.iglob(path_to_directory+%2B+"*.txt"):
++++if+filename.find("abc")+!=+-1+or+filename.find("123")+!=+-1+or+filename.find("a1b")+!=+-1:
++++++++print+filename|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|450968|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>for filename in glob.iglob(path_to_directory + "*.txt"):
 if filename.find("abc") != -1 or filename.find("123") != -1 or filename.find("a1b") != -1:
 print filename
</code></pre>

blocks|key|450998|text|令我惊讶的是，这里没有答案使用过滤器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|450999|import+os
import+re

def+glob_re(pattern,+strings):
++++return+filter(re.compile(pattern).match,+strings)

filenames+=+glob_re(r'.*(abc%7C123%7Ca1b).*\.txt',+os.listdir())|code-block|syntax|javascript|451000|它接受任何返回字符串的迭代器，包括列表、元组、字典(如果所有键都是字符串)等。如果您希望支持部分匹配，可以将.match更改为.search。请注意，这显然会返回一个生成器，因此，如果您希望使用结果而不对其进行迭代，您可以自己将结果转换为列表，或者使用list(...)包装返回语句。|offset|length|style|CODE|451001|entityMap^0|0|0|1I|6|1R|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

I'm surprised that no answers here used filter.

<pre><code>import os
import re

def glob_re(pattern, strings):
 return filter(re.compile(pattern).match, strings)

filenames = glob_re(r'.*(abc|123|a1b).*\.txt', os.listdir())
</code></pre>

This accepts any iterator that returns strings, including lists, tuples, dicts(if all keys are strings), etc. If you want to support partial matches, you could change <code>.match</code> to <code>.search</code>. Please note that this obviously returns a generator, so if you want to use the results without iterating over them, you could convert the result to a list yourself, or wrap the return statement with list(...).

<pre><code>import glob

list = glob.glob(r'*abc*.txt') + glob.glob(r'*123*.txt') + glob.glob(r'*a1b*.txt')

for i in list:
 print i
</code></pre>
This code works to list files in the current folder which have <code>'abc'</code>, <code>'123'</code> or <code>'a1b'</code> in their names.
How would I use one <code>glob</code> to perform this function?

Regular expression usage in glob.glob?

import globlist = glob.glob(r'*abc*.txt') + glob.glob(r'*123*.txt') + glob.glob(r'*a1b*.txt')for i in list:  print i此代码用于列出当前文件夹中名称中包含'abc'、'123'或'a1b'的文件。如何使用一...

问正则表达式在glob.glob中的使用？
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