Python提取包含单词的句子

Question

我试图从文本中提取包含指定单词的所有句子.

txt="I like to eat apple. Me too. Let's go buy some apples."
txt = "." + txt
re.findall(r"\."+".+"+"apple"+".+"+"\.", txt)

但它正在归还我:

[".I like to eat apple. Me too. Let's go buy some apples."]

代替 :

[".I like to eat apple., "Let's go buy some apples."]

有什么帮助吗？

Answer 1

不需要正则表达式:

>>> txt = "I like to eat apple. Me too. Let's go buy some apples."
>>> [sentence + '.' for sentence in txt.split('.') if 'apple' in sentence]
['I like to eat apple.', " Let's go buy some apples."]

Answer 2

In [3]: re.findall(r"([^.]*?apple[^.]*\.)",txt)                                                                                                                             
Out[4]: ['I like to eat apple.', " Let's go buy some apples."]

Answer 3

In [7]: import re

In [8]: txt=".I like to eat apple. Me too. Let's go buy some apples."

In [9]: re.findall(r'([^.]*apple[^.]*)', txt)
Out[9]: ['I like to eat apple', " Let's go buy some apples"]

但请注意,@ jamylak的split解决方案更快:

In [10]: %timeit re.findall(r'([^.]*apple[^.]*)', txt)
1000000 loops, best of 3: 1.96 us per loop

In [11]: %timeit [s+ '.' for s in txt.split('.') if 'apple' in s]
1000000 loops, best of 3: 819 ns per loop

对于较大的字符串,速度差异较小,但仍然很重要:

In [24]: txt = txt*10000

In [25]: %timeit re.findall(r'([^.]*apple[^.]*)', txt)
100 loops, best of 3: 8.49 ms per loop

In [26]: %timeit [s+'.' for s in txt.split('.') if 'apple' in s]
100 loops, best of 3: 6.35 ms per loop