blocks|key|38494|text|只需使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|38495|if+file_name.endswith(tuple(extensions)):|code-block|syntax|javascript|38496|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|$]]

Just use:

<pre><code>if file_name.endswith(tuple(extensions)):
</code></pre>

blocks|key|38513|text|虽然不是很广为人知，但str.endswith也接受元组。你不需要循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|38514|>>>+'test.mp3'.endswith(('.mp3',+'.avi'))
True|code-block|syntax|javascript|38515|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/stdtypes.html#str.endswith^0|B|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Though not widely known, <a href="http://docs.python.org/2/library/stdtypes.html#str.endswith" rel="noreferrer">str.endswith</a> also accepts a tuple. You don't need to loop.

<pre><code>&gt;&gt;&gt; 'test.mp3'.endswith(('.mp3', '.avi'))
True
</code></pre>

blocks|key|38525|text|从文件中提取一个扩展名，看看它是否在扩展名集中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|38526|>>>+import+os
>>>+extensions+=+set(['.mp3','.avi'])
>>>+file_name+=+'test.mp3'
>>>+extension+=+os.path.splitext(file_name)[1]
>>>+extension+in+extensions
True|code-block|syntax|javascript|38527|使用集合，因为在集合中查找的时间复杂度是O(1)+(docs)。|offset|length|38528|entityMap|0|LINK|mutability|MUTABLE|url|http://wiki.python.org/moin/TimeComplexity^0|0|0|Q|4|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Take an extension from the file and see if it is in the set of extensions:

<pre><code>&gt;&gt;&gt; import os
&gt;&gt;&gt; extensions = set(['.mp3','.avi'])
&gt;&gt;&gt; file_name = 'test.mp3'
&gt;&gt;&gt; extension = os.path.splitext(file_name)[1]
&gt;&gt;&gt; extension in extensions
True
</code></pre>

Using a set because time complexity for lookups in sets is O(1) (<a href="http://wiki.python.org/moin/TimeComplexity" rel="noreferrer">docs</a>).

blocks|key|255223|text|我有这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255224|def+has_extension(filename,+extension):

++++ext+=+"."+%2B+extension
++++if+filename.endswith(ext):
++++++++return+True
++++else:
++++++++return+False|code-block|syntax|javascript|255225|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|$]]

I have this:

<pre><code>def has_extension(filename, extension):

 ext = "." + extension
 if filename.endswith(ext):
 return True
 else:
 return False
</code></pre>

blocks|key|38552|text|有两种方法:正则表达式和字符串(str)方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|38553|字符串方法通常更快(~2倍)。|38554|import+re,+timeit
p+=+re.compile('.*(.mp3%7C.avi)$',+re.IGNORECASE)
file_name+=+'test.mp3'
print(bool(t.match(file_name))
%25timeit+bool(t.match(file_name)|code-block|syntax|javascript|38555|每环792+ns±1.83+ns+(平均值±标准dev.共7次运行，每次1000000次循环)|38556|file_name+=+'test.mp3'
extensions+=+('.mp3','.avi')
print(file_name.lower().endswith(extensions))
%25timeit+file_name.lower().endswith(extensions)|38557|每环274+ns±4.22+ns+(平均值±标准dev.共7次运行，每次1000000次循环)|38558|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

There is two ways: regular expressions and string (str) methods. 

String methods are usually faster ( ~2x ).

<pre><code>import re, timeit
p = re.compile('.*(.mp3|.avi)$', re.IGNORECASE)
file_name = 'test.mp3'
print(bool(t.match(file_name))
%timeit bool(t.match(file_name)
</code></pre>

792 ns ± 1.83 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

<pre><code>file_name = 'test.mp3'
extensions = ('.mp3','.avi')
print(file_name.lower().endswith(extensions))
%timeit file_name.lower().endswith(extensions)
</code></pre>

274 ns ± 4.22 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

blocks|key|255261|text|我只是在寻找其他东西的时候偶然发现了这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255262|我建议使用os包中的方法。这是因为您可以使其更通用，以补偿任何奇怪的情况。|offset|length|style|CODE|255263|你可以这样做：|255264|import+os

the_file+=+'aaaa/bbbb/ccc.ddd'

extensions_list+=+['ddd',+'eee',+'fff']

if+os.path.splitext(the_file)[-1]+in+extensions_list:
++++#+Do+your+thing.|code-block|syntax|javascript|255265|entityMap^0|0|5|2|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I just came across this, while looking for something else. 

I would recommend to go with the methods in the <code>os</code> package. This is because you can make it more general, compensating for any weird case. 

You can do something like: 

<pre><code>import os

the_file = 'aaaa/bbbb/ccc.ddd'

extensions_list = ['ddd', 'eee', 'fff']

if os.path.splitext(the_file)[-1] in extensions_list:
 # Do your thing.
</code></pre>

blocks|key|38574|text|另一种可能是使用IN语句：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|38575|extensions+=+['.mp3','.avi']
file_name++=+'test.mp3'
if+"."+in+file_name+and+file_name[file_name.rindex("."):]+in+extensions:
++++print(True)|code-block|syntax|javascript|38576|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|$]]

Another possibility could be to make use of IN statement:

<pre><code>extensions = ['.mp3','.avi']
file_name = 'test.mp3'
if "." in file_name and file_name[file_name.rindex("."):] in extensions:
 print(True)
</code></pre>

blocks|key|255290|text|返回匹配字符串列表的另一种方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|255291|sample+=+"alexis+has+the+control"
matched_strings+=+filter(sample.endswith,+["trol",+"ol",+"troll"])
print+matched_strings
['trol',+'ol']|code-block|syntax|javascript|255292|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|$]]

another way which can return the list of matching strings is 

<pre><code>sample = "alexis has the control"
matched_strings = filter(sample.endswith, ["trol", "ol", "troll"])
print matched_strings
['trol', 'ol']
</code></pre>

What is the pythonic way of writing the following code?

<pre><code>extensions = ['.mp3','.avi']
file_name = 'test.mp3'

for extension in extensions:
 if file_name.endswith(extension):
 #do stuff
</code></pre>

I have a vague memory that the explicit declaration of the <code>for</code> loop can be avoided and be written in the <code>if</code> condition. Is this true?

Check if string ends with one of the strings from a list

编写以下代码的pythonic方式是什么？extensions = ['.mp3','.avi']file_name = 'test.mp3'for extension in extensions:    if file_name.endswith(extension):        #do stuff我有一个模糊的...

问检查字符串是否以列表中的一个字符串结尾
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