blocks|key|3614461|text|由于Python中的{}不受glob()的支持，所以您可能想要的是|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3614462|import+os
import+re

...

match_dir+=+re.compile('(faint%7Cbright.*)/(science%7Ccalib)(/chip)?')
for+dirpath,+dirnames,+filenames+in+os.walk("/your/top/dir")
++++if+match_dir.search(dirpath):
++++++++do_whatever_with_files(dirpath,+files)
++++++++#+OR
++++++++do_whatever_with_subdirs(dirpath,+dirnames)|code-block|syntax|javascript|3614463|entityMap^0|A|2|E|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Since <code>{}</code> aren't supported by <code>glob()</code> in Python, what you probably want is something like
<pre><code>import os
import re

...

match_dir = re.compile('(faint|bright.*)/(science|calib)(/chip)?')
for dirpath, dirnames, filenames in os.walk(&quot;/your/top/dir&quot;)
 if match_dir.search(dirpath):
 do_whatever_with_files(dirpath, files)
 # OR
 do_whatever_with_subdirs(dirpath, dirnames)
</code></pre>

blocks|key|3071486|text|将球状和支撑展开相结合。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3071487|pip+install+braceexpand|code-block|syntax|javascript|3071488|示例：|3071489|from+glob+import+glob
from+braceexpand+import+braceexpand

def+braced_glob(path):
++++l+=+[]
++++for+x+in+braceexpand(path):
++++++++l.extend(glob(x))
++++++++++++
++++return+l|3071490|>>>+braced_glob('/usr/bin/{x,z}*k')++
['/usr/bin/xclock',+'/usr/bin/zipcloak']|3071491|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Combining globbing with brace expansion.
<pre><code>pip install braceexpand
</code></pre>
Sample:
<pre class="lang-py prettyprint-override"><code>from glob import glob
from braceexpand import braceexpand

def braced_glob(path):
 l = []
 for x in braceexpand(path):
 l.extend(glob(x))
 
 return l
</code></pre>
<pre><code>&gt;&gt;&gt; braced_glob('/usr/bin/{x,z}*k') 
['/usr/bin/xclock', '/usr/bin/zipcloak']
</code></pre>

blocks|key|3614454|text|{..}被称为支撑展开，它是在球化发生之前应用的一个单独的步骤。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3614455|它不是globs的一部分，也不受python函数的支持。|3614456|entityMap^0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

<code>{..}</code> is known as brace expansion, and is a separate step applied before globbing takes place. 

It's not part of globs, and not supported by the python glob function.

blocks|key|589071|text|作为that+other+guy+pointed+out，Python不直接支持大括号扩展。但是，由于在评估通配符之前就已经完成了支撑扩展，所以您可以自己做，例如，|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|589072|result+=+glob('{faint,bright*}/{science,calib}/chip?/')|code-block|syntax|javascript|589073|变成了|589074|result+=+[
++++f+
++++for+b+in+['faint',+'bright*']+
++++for+s+in+['science',+'calib']+
++++for+f+in+glob('{b}/{s}/chip?/'.format(b=b,+s=s))
]|589075|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/22996810/3830997^0|2|E|H|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

As <code>that other guy</code> <a href="https://stackoverflow.com/a/22996810/3830997">pointed out</a>, Python doesn't support brace expansion directly. But since brace expansion is done before the wildcards are evaluated, you could do that yourself, e.g., 

<pre><code>result = glob('{faint,bright*}/{science,calib}/chip?/')
</code></pre>

becomes

<pre><code>result = [
 f 
 for b in ['faint', 'bright*'] 
 for s in ['science', 'calib'] 
 for f in glob('{b}/{s}/chip?/'.format(b=b, s=s))
]
</code></pre>

blocks|key|588597|text|正如其他答案所述，支撑展开是glob的预处理步骤:您展开所有的大括号，然后对每个结果运行glob。(大括号展开将一个字符串转换为字符串列表。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|588598|奥威尔亲信推荐braceexpand库。在我看来，这个问题太小，无法证明依赖是合理的(尽管这是标准库中的一个常见问题，最好打包在glob模块中)。|offset|length|style|CODE|588599|因此，这里有一种方法可以用几行代码来完成。|588600|import+itertools
import+re

def+expand_braces(text,+seen=None):
++++if+seen+is+None:
++++++++seen+=+set()

++++spans+=+[m.span()+for+m+in+re.finditer("\{[%5E\{\}]*\}",+text)][::-1]
++++alts+=+[text[start+%2B+1+:+stop+-+1].split(",")+for+start,+stop+in+spans]

++++if+len(spans)+==+0:
++++++++if+text+not+in+seen:
++++++++++++yield+text
++++++++seen.add(text)

++++else:
++++++++for+combo+in+itertools.product(*alts):
++++++++++++replaced+=+list(text)
++++++++++++for+(start,+stop),+replacement+in+zip(spans,+combo):
++++++++++++++++replaced[start:stop]+=+replacement

++++++++++++yield+from+expand_braces("".join(replaced),+seen)

###+testing

text_to_expand+=+"{{pine,}apples,oranges}+are+{tasty,disgusting}+to+m{}e+}{"

for+result+in+expand_braces(text_to_expand):
++++print(result)|code-block|syntax|javascript|588601|版画|588602|pineapples+are+tasty+to+me+}{
oranges+are+tasty+to+me+}{
apples+are+tasty+to+me+}{
pineapples+are+disgusting+to+me+}{
oranges+are+disgusting+to+me+}{
apples+are+disgusting+to+me+}{|588603|这里发生的是：|588604|嵌套括号可以产生非唯一的结果，因此我们使用seen只产生尚未见过的结果。|ordered-list-item|588605|spans是text中所有内部平衡括号的起始和停止索引。这个顺序被[::-1]切片颠倒，使得索引从最高到最低(稍后将相关)。|588606|alts的每个元素都是以逗号分隔的选项的对应列表。|588607|如果没有匹配(+text不包含平衡括号)，则生成text本身，确保它在seen中是唯一的。|588608|否则，使用itertools.product迭代逗号分隔的选项的笛卡儿积。|588609|将括号内的文本替换为当前的备选案文。因为我们要就地替换数据，所以它必须是一个可变序列(list，而不是str)，而且我们必须首先替换最高的索引。如果我们首先替换最低的索引，那么后面的索引就会与spans中的索引不同。这就是为什么我们在第一次创建spans时将其反转的原因。|588610|text可能在花括号中有花括号。正则表达式只找到不包含任何其他花括号的平衡花括号，但嵌套的花括号是合法的。因此，我们需要递归，直到没有嵌套的花括号(+len(spans)+==+0大小写)。使用Python生成器的递归使用yield+from重新生成递归调用的每个结果。|588611|588612|588613|在输出中，{{pine,}apples,oranges}首先扩展为{pineapples,oranges}和{apples,oranges}，然后每个扩展。如果我们没有请求使用oranges的唯一结果，那么seen结果将出现两次。|588614|空括号(如m{}e中的括号)扩展为零，因此这只是me。|588615|不平衡的括号，如}{，被保留为-原样。|588616|如果需要高性能的大型数据集，这不是一种可以使用的算法，但它是合理大小的数据的通用解决方案。|588617|entityMap^0|0|7|B|0|0|0|0|0|0|L|4|0|0|5|6|4|X|6|0|0|4|0|8|4|O|4|Z|4|0|5|H|0|17|4|1F|3|2O|5|3E|5|0|0|4|23|F|33|A|0|0|0|5|N|X|K|1I|G|2G|7|2V|4|0|5|4|O|2|0|8|2|0|0^^$0|@$1|2|3|4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1M|8|@$D|1N|E|1O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1P|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|1Q|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1R|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|1S|8|@]|9|@]|A|$M|N]]|$1|S|3|T|5|6|7|1T|8|@]|9|@]|A|$]]|$1|U|3|V|5|W|7|1U|8|@$D|1V|E|1W|F|G]]|9|@]|A|$]]|$1|X|3|Y|5|W|7|1X|8|@$D|1Y|E|1Z|F|G]|$D|20|E|21|F|G]|$D|22|E|23|F|G]]|9|@]|A|$]]|$1|Z|3|10|5|W|7|24|8|@$D|25|E|26|F|G]]|9|@]|A|$]]|$1|11|3|12|5|W|7|27|8|@$D|28|E|29|F|G]|$D|2A|E|2B|F|G]|$D|2C|E|2D|F|G]]|9|@]|A|$]]|$1|13|3|14|5|W|7|2E|8|@$D|2F|E|2G|F|G]]|9|@]|A|$]]|$1|15|3|16|5|W|7|2H|8|@$D|2I|E|2J|F|G]|$D|2K|E|2L|F|G]|$D|2M|E|2N|F|G]|$D|2O|E|2P|F|G]]|9|@]|A|$]]|$1|17|3|18|5|W|7|2Q|8|@$D|2R|E|2S|F|G]|$D|2T|E|2U|F|G]|$D|2V|E|2W|F|G]]|9|@]|A|$]]|$1|19|3|-4|5|6|7|2X|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|2Y|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|6|7|2Z|8|@$D|30|E|31|F|G]|$D|32|E|33|F|G]|$D|34|E|35|F|G]|$D|36|E|37|F|G]|$D|38|E|39|F|G]]|9|@]|A|$]]|$1|1D|3|1E|5|6|7|3A|8|@$D|3B|E|3C|F|G]|$D|3D|E|3E|F|G]]|9|@]|A|$]]|$1|1F|3|1G|5|6|7|3F|8|@$D|3G|E|3H|F|G]]|9|@]|A|$]]|$1|1H|3|1I|5|6|7|3I|8|@]|9|@]|A|$]]|$1|1J|3|-4|5|6|7|3J|8|@]|9|@]|A|$]]]|1K|$]]

As stated in other answers, brace-expansion is a pre-processing step for glob: you expand all the braces, then run glob on each of the results. (Brace-expansion turns one string into a list of strings.)
Orwellophile recommends the <code>braceexpand</code> library. This feels to me like too small of a problem to justify a dependency (though it's a common problem that ought to be in the standard library, ideally packaged in the glob module).
So here's a way to do it with a few lines of code.
<pre class="lang-py prettyprint-override"><code>import itertools
import re

def expand_braces(text, seen=None):
 if seen is None:
 seen = set()

 spans = [m.span() for m in re.finditer(&quot;\{[^\{\}]*\}&quot;, text)][::-1]
 alts = [text[start + 1 : stop - 1].split(&quot;,&quot;) for start, stop in spans]

 if len(spans) == 0:
 if text not in seen:
 yield text
 seen.add(text)

 else:
 for combo in itertools.product(*alts):
 replaced = list(text)
 for (start, stop), replacement in zip(spans, combo):
 replaced[start:stop] = replacement

 yield from expand_braces(&quot;&quot;.join(replaced), seen)

### testing

text_to_expand = &quot;{{pine,}apples,oranges} are {tasty,disgusting} to m{}e }{&quot;

for result in expand_braces(text_to_expand):
 print(result)
</code></pre>
prints
<pre><code>pineapples are tasty to me }{
oranges are tasty to me }{
apples are tasty to me }{
pineapples are disgusting to me }{
oranges are disgusting to me }{
apples are disgusting to me }{
</code></pre>
What's happening here is:
<ol>
<li>Nested brackets can produce non-unique results, so we use <code>seen</code> to only yield results that haven't yet been seen.</li>
<li><code>spans</code> is the starting and stopping indexes of all innermost, balanced brackets in the <code>text</code>. The order is reversed by the <code>[::-1]</code> slice, such that indexes go from highest to lowest (will be relevant later).</li>
<li>Each element of <code>alts</code> is the corresponding list of comma-delimited alternatives.</li>
<li>If there aren't any matches (the <code>text</code> does not contain balanced brackets), yield the <code>text</code> itself, ensuring that it is unique with <code>seen</code>.</li>
<li>Otherwise, use <code>itertools.product</code> to iterate over the Cartesian product of comma-delimited alternatives.</li>
<li>Replace the curly-bracketed text with the current alternative. Since we're replacing data in-place, it has to be a mutable sequence (<code>list</code>, rather than <code>str</code>), and we have to replace the highest indexes first. If we replaced the lowest indexes first, the later indexes would have changed from what they were in the <code>spans</code>. This is why we reversed <code>spans</code> when it was first created.</li>
<li>The <code>text</code> might have curly brackets within curly brackets. The regular expression only found balanced curly brackets that do not contain any other curly brackets, but nested curly brackets are legal. Therefore, we need to recurse until there are no nested curly brackets (the <code>len(spans) == 0</code> case). Recursion with Python generators uses <code>yield from</code> to re-yield each result from the recursive call.</li>
</ol>
In the output, <code>{{pine,}apples,oranges}</code> is first expanded to <code>{pineapples,oranges}</code> and <code>{apples,oranges}</code>, and then each of these is expanded. The <code>oranges</code> result would appear twice if we didn't request unique results with <code>seen</code>.
Empty brackets like the ones in <code>m{}e</code> expand to nothing, so this is just <code>me</code>.
Unbalanced brackets, like <code>}{</code>, are left as-is.
This is not an algorithm to use if high performance for large datasets is required, but it's a general solution for reasonably sized data.

I have python 2.7 and am trying to issue:

<pre><code>glob('{faint,bright*}/{science,calib}/chip?/')
</code></pre>

I obtain no matches, however from the shell <code>echo {faint,bright*}/{science,calib}/chip?</code> gives:

<pre><code>faint/science/chip1 faint/science/chip2 faint/calib/chip1 faint/calib/chip2 bright1/science/chip1 bright1/science/chip2 bright1w/science/chip1 bright1w/science/chip2 bright2/science/chip1 bright2/science/chip2 bright2w/science/chip1 bright2w/science/chip2 bright1/calib/chip1 bright1/calib/chip2 bright1w/calib/chip1 bright1w/calib/chip2 bright2/calib/chip1 bright2/calib/chip2 bright2w/calib/chip1 bright2w/calib/chip2
</code></pre>

What is wrong with my expression?

Brace expansion in python glob

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有python 2.7，并试图发布：glob('{faint,bright*}/{science,calib}/chip?/')我没有获得匹配，但是从shell echo {faint,bright*}/{science,calib}/chip?给出：faint/science/chip1 faint/science/chip2 faint/calib/chip1 faint/calib/chi

问巨蟒球体中的支撑膨胀
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