使用python将字符串拆分成句子

Question

我有以下字符串：

string = 'This is one sentence  ${w_{1},..,w_{i}}$. This is another sentence. '

现在，我想把它分成两句话。

但是，当我这样做时：

string.split('.')

我明白了：

['This is one sentence  ${w_{1},',
 '',
 ',w_{i}}$',
 ' This is another sentence',
 ' ']

任何人都知道如何改进它，以免检测到“。”在$ $ 内？

另外，你会怎么做：

string2 = 'This is one sentence  ${w_{1},..,w_{i}}$! This is another sentence. Is this a sentence? Maybe !  '

编辑 1：

期望的输出是：

对于字符串 1：

['This is one sentence  ${w_{1},..,w_{i}}$','This is another sentence']

对于字符串 2：

['This is one sentence  ${w_{1},..,w_{i}}$','This is another sentence', 'Is this a sentence', 'Maybe !  ']

Answer 1

对于更一般的情况，您可以像这样使用re.split：

import re

mystr = 'This is one sentence  ${w_{1},..,w_{i}}$. This is another sentence. '

re.split("[.!?]\s{1,}", mystr)
# ['This is one sentence  ${w_{1},..,w_{i}}$', 'This is another sentence', '']

str2 = 'This is one sentence  ${w_{1},..,w_{i}}$! This is another sentence. Is this a sentence? Maybe !  '

re.split("[.!?]\s{1,}", str2)
['This is one sentence  ${w_{1},..,w_{i}}$', 'This is another sentence', 'Is this a sentence', 'Maybe ', '']

括号中的字符是您选择的标点符号，并且您在\s{1,} 末尾添加至少一个空格以忽略其他没有空格的.。这也将处理您的感叹号案例

这是恢复标点符号的（有点老套）方法

punct = re.findall("[.!?]\s{1,}", str2)
['! ', '. ', '? ', '!  ']

sent = [x+y for x,y in zip(re.split("[.!?]\s{1,}", str2), punct)]
sent
['This is one sentence  ${w_{1},..,w_{i}}$! ', 'This is another sentence. ', 'Is this a sentence? ', 'Maybe !  ']

Answer 2

您可以将re.findall 与交替模式一起使用。为确保句子以非空格开头和结尾，请在开头使用正向lookahead 模式，在结尾使用正向lookbehind 模式：

re.findall(r'((?=[^.!?\s])(?:$.*?\$|[^.!?])*(?<=[^.!?\s]))\s*[.!?]', string)

这会返回，对于第一个字符串：

['This is one sentence  ${w_{1},..,w_{i}}$', 'This is another sentence']

对于第二个字符串：

['This is one sentence  ${w_{1},', ',w_{i}}$', 'This is another sentence', 'Is this a sentence', 'Maybe']

Answer 3

使用'. ' （在 . 之后有一个空格）因为它只存在于句子结束时，而不是句子中间。

string = 'This is one sentence  ${w_{1},..,w_{i}}$. This is another sentence. '

string.split('. ')

返回：

['这是一个句子 ${w_{1},..,w_{i}}$', '这是另一个句子', '']