blocks|key|1211899|text|您可以使用，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1211900|words+=+[word.replace('[br]','<br+/>')+for+word+in+words]|code-block|syntax|javascript|1211901|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|$]]

You can use, for example:

<pre><code>words = [word.replace('[br]','&lt;br /&gt;') for word in words]
</code></pre>

blocks|key|997664|text|words+=+[w.replace('[br]',+'<br+/>')+for+w+in+words]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|997665|这些被称为List+Comprehensions。|unstyled|offset|length|997666|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/datastructures.html#list-comprehensions^0|0|5|J|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>words = [w.replace('[br]', '&lt;br /&gt;') for w in words]
</code></pre>

These are called <a href="http://docs.python.org/tutorial/datastructures.html#list-comprehensions" rel="noreferrer">List Comprehensions</a>.

blocks|key|997693|text|如果你想知道不同方法的性能，这里有一些计时：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|997694|In+[1]:+words+=+[str(i)+for+i+in+range(10000)]

In+[2]:+%25timeit+replaced+=+[w.replace('1',+'<1>')+for+w+in+words]
100+loops,+best+of+3:+2.98+ms+per+loop

In+[3]:+%25timeit+replaced+=+map(lambda+x:+str.replace(x,+'1',+'<1>'),+words)
100+loops,+best+of+3:+5.09+ms+per+loop

In+[4]:+%25timeit+replaced+=+map(lambda+x:+x.replace('1',+'<1>'),+words)
100+loops,+best+of+3:+4.39+ms+per+loop

In+[5]:+import+re

In+[6]:+r+=+re.compile('1')

In+[7]:+%25timeit+replaced+=+[r.sub('<1>',+w)+for+w+in+words]
100+loops,+best+of+3:+6.15+ms+per+loop|code-block|syntax|javascript|997695|正如您所看到的，对于这样的简单模式，可接受的列表理解是最快的，但请看以下内容：|997696|In+[8]:+%25timeit+replaced+=+[w.replace('1',+'<1>').replace('324',+'<324>').replace('567',+'<567>')+for+w+in+words]
100+loops,+best+of+3:+8.25+ms+per+loop

In+[9]:+r+=+re.compile('(1%7C324%7C567)')

In+[10]:+%25timeit+replaced+=+[r.sub('<\1>',+w)+for+w+in+words]
100+loops,+best+of+3:+7.87+ms+per+loop|997697|这表明对于更复杂的替换，预编译的reg-exp+(如在9-10中)可能(快得多)。这真的取决于你的问题和reg-exp中最短的部分。|offset|length|style|CODE|997698|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

In case you're wondering about the performance of the different approaches, here are some timings:

<pre><code>In [1]: words = [str(i) for i in range(10000)]

In [2]: %timeit replaced = [w.replace('1', '&lt;1&gt;') for w in words]
100 loops, best of 3: 2.98 ms per loop

In [3]: %timeit replaced = map(lambda x: str.replace(x, '1', '&lt;1&gt;'), words)
100 loops, best of 3: 5.09 ms per loop

In [4]: %timeit replaced = map(lambda x: x.replace('1', '&lt;1&gt;'), words)
100 loops, best of 3: 4.39 ms per loop

In [5]: import re

In [6]: r = re.compile('1')

In [7]: %timeit replaced = [r.sub('&lt;1&gt;', w) for w in words]
100 loops, best of 3: 6.15 ms per loop
</code></pre>

as you can see for such simple patterns the accepted list comprehension is the fastest, but look at the following:

<pre><code>In [8]: %timeit replaced = [w.replace('1', '&lt;1&gt;').replace('324', '&lt;324&gt;').replace('567', '&lt;567&gt;') for w in words]
100 loops, best of 3: 8.25 ms per loop

In [9]: r = re.compile('(1|324|567)')

In [10]: %timeit replaced = [r.sub('&lt;\1&gt;', w) for w in words]
100 loops, best of 3: 7.87 ms per loop
</code></pre>

This shows that for more complicated substitutions a pre-compiled reg-exp (as in <code>9-10</code>) can be (much) faster. It really depends on your problem and the shortest part of the reg-exp.

blocks|key|997727|text|一个使用for循环的例子(我更喜欢列表理解)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|997728|a,+b+=+'[br]',+'<br+/>'
for+i,+v+in+enumerate(words):
++++if+a+in+v:
++++++++words[i]+=+v.replace(a,+b)
print(words)
#+['how',+'much',+'is ',+'the',+'fish ',+'no',+'really']|code-block|syntax|javascript|997729|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|$]]

An example with for loop (I prefer List Comprehensions).

<pre><code>a, b = '[br]', '&lt;br /&gt;'
for i, v in enumerate(words):
 if a in v:
 words[i] = v.replace(a, b)
print(words)
# ['how', 'much', 'is&lt;br/&gt;', 'the', 'fish&lt;br/&gt;', 'no', 'really']
</code></pre>

I got this list:

<pre><code>words = ['how', 'much', 'is[br]', 'the', 'fish[br]', 'no', 'really']
</code></pre>

What I would like is to replace <code>[br]</code> with some fantastic value similar to <code>&amp;lt;br /&amp;gt;</code> and thus getting a new list:

<pre><code>words = ['how', 'much', 'is&lt;br /&gt;', 'the', 'fish&lt;br /&gt;', 'no', 'really']
</code></pre>

Find and replace string values in list

我有一份清单：words = ['how', 'much', 'is[br]', 'the', 'fish[br]', 'no', 'really']我想要的是用一些类似于&lt;br /&gt;的奇妙的值来替换[br]，从而得到一个新的列表：words = ['how', 'much', 'is ', 't...

问查找和替换列表中的字符串值
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