如果列表 2l2中的子字符串与l1下面的元素匹配，我正在尝试遍历两个列表并连接元素。你能帮我解决这个问题吗？
l1=['Rule1','Rule2','Rule3']
l2=['This is Rule1 for boys', 'That is Rule2', 'That is Rule1','This is not Rule', 'This is Rule3','Which is Rule3 for girls']
预期输出：
l2=['This is Rule1 for boys|Rule1', 'That is Rule2|Rule2', 'That is Rule1|Rule1','This is not Rule', 'This is Rule3|Rule3','Which is Rule3 for girls|Rule3']
这应该对你有用。
这不是最优雅的方法，但算法很容易理解。
def fn(l1, l2):
l3 = []
for item in l2:
for substring in l1:
if substring in item:
l3.append(f"{item}|{substring}")
break
else: l3.append(item)
return l3
l1=['Rule1','Rule2','Rule3']
l2=['This is Rule1 for boys', 'That is Rule2', 'That is Rule1','This is not Rule', 'This is Rule3','Which is Rule3 for girls']
print(fn(l1, l2))
# OUTPUTS:
# ['This is Rule1 for boys|Rule1', 'That is Rule2|Rule2', 'That is Rule1|Rule1', 'This is not Rule', 'This is Rule3|Rule3', 'Which is Rule3 for girls|Rule3']
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