在Python列表中查找并替换字符串值

Question

我有这个清单:

words = ['how', 'much', 'is[br]', 'the', 'fish[br]', 'no', 'really']

我想要的是取代[br]一些类似的奇妙价值<br />,从而得到一个新的清单:

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

Answer 1

words = [w.replace('[br]', '<br />') for w in words]

这称为列表理解.

Answer 2

除列表理解外,您还可以尝试地图

>>> map(lambda x: str.replace(x, "[br]", "<br/>"), words)
['how', 'much', 'is<br/>', 'the', 'fish<br/>', 'no', 'really']

Answer 3

您可以使用,例如:

words = [word.replace('[br]','<br />') for word in words]

Answer 4

如果你想知道不同方法的表现,这里有一些时间:

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

正如你可以看到的那样简单的模式,接受的列表理解是最快的,但请看下面的内容:

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)中可以(更快)地更快.这真的取决于你的问题和reg-exp的最短部分.

Answer 5

for 循环的示例（我更喜欢列表推导式）。

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<br/>', 'the', 'fish<br/>', 'no', 'really']