查找包含特定单词的所有句子

Question

我有一个由句子组成的字符串，并且想要查找包含至少一个特定关键字的所有句子，即keyword1 1 或keyword2 ：

import re

s = "This is a sentence which contains keyword1. And keyword2 is inside this sentence. "

pattern = re.compile(r"([A-Z][^\.!?].*(keyword1)|(keyword2).*[\.!?])\s")
for match in pattern.findall(s):
    print(match)

输出：

('This is a sentence which contains keyword1', 'keyword1', '')
('keyword2 is inside this sentence. ', '', 'keyword2')

预期输出：

('This is a sentence which contains keyword1', 'keyword1', '')
('And keyword2 is inside this sentence. ', '', 'keyword2')

如您所见，第二个匹配项不包含第一组中的整个句子。 我在这里想念什么？

Answer 1

您可以使用否定字符类来不匹配. ! 和? 并将关键字放在同一组中以防止结果中出现空字符串。

然后 re.findall 返回捕获组值，即整个匹配的第 1 组，以及其中一个关键字的第 2、3 组等。

([A-Z][^.!?]*(?:(keyword1)|(keyword2))[^.!?]*[.!?])\s

解释

• (捕获组 1
  • [AZ][^.!?]*匹配大写字符 AZ 和可选的任何字符，除了.!?
  • (?:(keyword1)|(keyword2))捕获自己组中的关键字之一
  • [^.!?]*[.!?]匹配除.!?之外的任何字符然后匹配.!?之一
• )关闭第 1 组
• \s匹配一个空白字符

请参阅正则表达式演示和Python 演示。

例子

import re

s = "This is a sentence which contains keyword1. And keyword2 is inside this sentence. "

pattern = re.compile(r"([A-Z][^.!?]*(?:(keyword1)|(keyword2))[^.!?]*[.!?])\s")
for match in pattern.findall(s):
    print(match)

输出

('This is a sentence which contains keyword1.', 'keyword1', '')
('And keyword2 is inside this sentence.', '', 'keyword2')

Answer 2

您可以尝试以下正则表达式：

[.?!]*\s*(.*(keyword1)[^.?!]*[.?!]|.*(keyword2)[^.?!]*[.?!])

代码：

import re

s = "This is a sentence which contains keyword1. And keyword2 is inside this sentence. "

pattern = re.compile(r"[.?!]*\s*(.*(keyword1)[^.?!]*[.?!]|.*(keyword2)[^.?!]*[.?!])")
for match in pattern.findall(s):
    print(match)

输出：

('This is a sentence which contains keyword1.', 'keyword1', '')
('And keyword2 is inside this sentence.', '', 'keyword2')