使用python将列表逐点拆分为一个句子

我有一个句子列表：

['hello', 'I would like to thank you', 'I would like to thank you. By the way']

当我找到“”时，我需要将每个句子分成一个列表

例如，在上面的示例中，预期结果是：

['hello', 'I would like to thank you', 'I would like to thank you'. 'By the way']

我尝试用python编写以下代码：

def split_pint(result):
    for i in result:
        i = re.split(r". ", i)
    return result

但判决没有分开

有什么想法吗

---

感谢您使用简单的迭代和str.split

Ex:

data = ['hello', 'I would like to thank you', 'I would like to thank you. By the way']

def split_pint(data):
    result = []
    for elem in data:
        result.extend(elem.split(". "))        
    return result

print(split_pint(data))

['hello', 'I would like to thank you', 'I would like to thank you', 'By the way']

输出：

data = ['hello', 'I would like to thank you', 'I would like to thank you. By the way']

def split_pint(data):
    result = []
    for elem in data:
        result.extend(elem.split(". "))        
    return result

print(split_pint(data))

['hello', 'I would like to thank you', 'I would like to thank you', 'By the way']

---

如您所见，这不是修改列表的方法：

l = [0, 0]
for x in l:
    x = 1
print(l)
# [0, 0]

无论如何，如果要使用

re.split

，则需要转义

字符：

import re

l = ['hello', 'I would like to thank you', 'I would like to thank you. By the way']
def split_pint(result):
    res = []
    for i in result:
        res += re.split("\. ", i)
    return res


print(split_pint(l))
['hello', 'I would like to thank you', 'I would like to thank you', 'By the way']

---

另一个选项，但只有一个线性和函数式编程方式：

>>> from functools import reduce
>>> a = ['hello', 'I would like to thank you', 'I would like to thank you. By the way']
>>> reduce(lambda i, j: i + j, map(lambda s: s.split('. '), a))
['hello', 'I would like to thank you', 'I would like to thank you', 'By the way']

首先，

map

从每个字符串生成一个列表，其次，

reduce

只是将所有列表串联起来