Python实现：匹配列表子串并返回元素与子串的元组列表

可以用嵌套的列表推导式实现需求，同时遍历ls2的每个元素和ls1的每个子串，检查子串是否为当前ls2元素的子串，满足条件就生成对应的元组：

ls1 = ['apple','banana','pear']
ls2 = ['strawberry is not here',
       'blueberry is over there',
       'the pear tree is not ready yet',
       'we have lots of pear trees',
       'apples are yummy']

result = [(s2, s1) for s2 in ls2 for s1 in ls1 if s1 in s2]
print(result)

运行代码后，输出结果与预期一致：

[('the pear tree is not ready yet', 'pear'), ('we have lots of pear trees', 'pear'), ('apples are yummy', 'apple')]

逻辑说明 #

• 外层遍历ls2中的每个元素s2，内层遍历ls1中的每个子串s1
• 通过if s1 in s2判断s1是否是s2的子串
• 满足条件时，将s2和s1组成元组加入结果列表

如果存在单个s2匹配多个s1的情况（比如s2为"apple and pear"），这段代码会生成对应的多个元组，完全符合子串匹配的逻辑。

内容的提问来源于stack exchange，提问作者frank