我想从文本中提取包含指定单词的所有句子。包含单词的Python提取语句
txt="I like to eat apple. Me too. Let's go buy some apples."
txt = "." + txt
re.findall(r"\."+".+"+"apple"+".+"+"\.", txt)
，但它返回我：的
[".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."]
任何帮助吗？
In [3]: re.findall(r"([^.]*?apple[^.]*\.)",txt)
Out[4]: ['I like to eat apple.', " Let's go buy some apples."]
您可以使用str.split，
>>> txt="I like to eat apple. Me too. Let's go buy some apples."
>>> txt.split('. ')
['I like to eat apple', 'Me too', "Let's go buy some apples."]
>>> [ t for t in txt.split('. ') if 'apple' in t]
['I like to eat apple', "Let's go buy some apples."]
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"]
但需要注意的是@ jamylak的split为基础的解决方案是更快：
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
速度差异较小，但仍然显著，对于较大字符串：
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
无需正则表达式：
>>> 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."]
谢谢jamylak – user2187202
@ user2187202你可以接受我的答案，如果你想要或接受正则表达式的解决方案，如果这实际上是你所需要的，因为你确实把它标记为正则表达式问题，我不确定这是否是必要的或不 – jamylak
r"\."+".+"+"apple"+".+"+"\."
这条线是一个有点古怪;为什么连接这么多单独的字符串？你可以使用r'.. + apple。+。'。
无论如何，你的正则表达式的问题是它的贪婪。默认情况下，x+将尽可能多地匹配x。所以你的.+将尽可能匹配尽可能多的字符（任何字符）;包括点和apple s。
你想使用的是一个非贪婪的表达式;您通常可以通过在末尾添加?来完成此操作：.+?。
这会让你得到以下结果：
['.I like to eat apple. Me too.']
正如你所看到的你不再同时获得苹果的句子，但仍是Me too.。这是因为您仍然匹配apple之后的.，因此无法捕捉下面的句子。
一个工作正则表达式将是这样：r'\.[^.]*?apple[^.]*?\.'
在这里，你不看就在任何字符，但只有那些不是字符圆点自己。我们也允许不匹配任何字符（因为在第一句中apple之后没有非点字符）。使用表达式的结果是：
['.I like to eat apple.', ". Let's go buy some apples."]
显然，有问题的样品是extract sentence containing substring而不是
extract sentence containing word。如何通过python解决extract sentence containing word问题如下：
一句话可以在句子的开头。不限于问题的例子，我将提供一个例句检索词的一般功能：
def searchWordinSentence(word,sentence):
pattern = re.compile(' '+word+' |^'+word+' | '+word+' $')
if re.search(pattern,sentence):
return True
仅限于问题的例子，我们可以解决，如：
txt="I like to eat apple. Me too. Let's go buy some apples."
word = "apple"
print [ t for t in txt.split('. ') if searchWordofSentence(word,t)]
相应的输出是：
['I like to eat apple']
+1不错的答案！如果你创建一个'txt = txt * 10000'，那么'％timeit'结果会更清晰 – Kent
谢谢Kent。我为更大的字符串添加了'％timeit'基准。 – unutbu