如何在Python中将三句文档字符串拆分为含标点的句子列表？

修复你的现有代码 #

你的代码存在两处问题：未初始化存储结果的sentencelist列表，以及找到句子结尾标点后没有重置sentence变量，导致后续句子会拼接前面的内容。修复后的代码如下：

sentencelist = []
sentence = ''
docstring = 'The rain in #Spain in 2019, rained "mainly" on the plain.\
There is a nice function to split a string into a list based on a given \
delimiter! Why do I try to do too much?'

for character in docstring:
    sentence += character
    if character in ('.', '?', '!'):
        sentencelist.append(sentence.strip())  # 可选：去除句首可能的多余空格
        sentence = ''  # 重置变量，准备收集下一句

print(sentencelist)

执行后输出：

['The rain in #Spain in 2019, rained "mainly" on the plain.', 'There is a nice function to split a string into a list based on a given delimiter!', 'Why do I try to do too much?']

更高效的正则表达式方案 #

如果不想用循环遍历字符，Python的re模块可以更简洁地实现需求：通过正向后顾断言拆分字符串，同时保留句子结尾的标点符号。

import re

docstring = 'The rain in #Spain in 2019, rained "mainly" on the plain.\
There is a nice function to split a string into a list based on a given \
delimiter! Why do I try to do too much?'

# 匹配.?!的位置进行拆分，同时保留标点；过滤空内容并去除多余空格
sentencelist = [s.strip() for s in re.split(r'(?<=[.!?])', docstring) if s.strip()]

print(sentencelist)

这个方案更适合处理较长的文本，代码可读性和效率都更高。

内容的提问来源于stack exchange，提问作者Beginner