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<br />
的奇妙的值来替换[br]
,从而得到一个新的列表:
words = ['how', 'much', 'is<br />', 'the', 'fish<br />', 'no', 'really']
发布于 2010-06-29 06:47:36
words = [w.replace('[br]', '<br />') for w in words]
这些被称为List Comprehensions。
发布于 2010-06-29 06:47:06
您可以使用,例如:
words = [word.replace('[br]','<br />') for word in words]
发布于 2015-05-26 22:38:38
如果你想知道不同方法的性能,这里有一些计时:
In [1]: words = [str(i) for i in range(10000)]
In [2]: %timeit replaced = [w.replace('1', '<1>') 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', '<1>'), words)
100 loops, best of 3: 5.09 ms per loop
In [4]: %timeit 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]: %timeit replaced = [r.sub('<1>', w) for w in words]
100 loops, best of 3: 6.15 ms per loop
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:
In [8]: %timeit 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|324|567)')
In [10]: %timeit replaced = [r.sub('<\1>', w) for w in words]
100 loops, best of 3: 7.87 ms per loop
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。
https://stackoverflow.com/questions/3136689
复制相似问题