合并2个不同长度的独立列表，无需索引

Question

这 2 个不同的列表有 2 个不同的长度。所以我想根据分数合并它们。

我的示例输出如下所示

list1 = [
    "ld Liked", "ld Disliked", "ld Liked", "ld Liked", "ld Disliked",
    "ld Liked", "ld Disliked", "ld Liked", "ld Disliked", "ld Disliked",
    "ld Liked", "ld Disliked", "ld Liked"
]

list2 = [
    "score   10", "score   10", "score   10", "score   8.0", "score   10",
    "score   6.0", "score   10", "score   8.0"
]


example of the output must be as follows after match
 
list[0] = liked and disliked : 10
list[1] = liked : 10
list[2] = liked and disliked : 10
list[3] = liked and disliked : 8.0
list[4] = liked and disliked : 10
list[5] = disliked : 6.0
list[6] = liked and disliked : 10
list[7] = liked : 8.0

我尝试使用 zip 来迭代它们，并使用一些 if 语句来检查它是否匹配，但它不起作用

我输出的目的是将评论与评分进行匹配，因为每个“评论”只有一个分数，但可能包含喜欢和不喜欢的字符串。

因此，如我的示例所示，评分将检查评论是否有“喜欢”或“不喜欢”，并将其视为单个实体。

每个评论都会以“喜欢”开始，除非没有“喜欢”，否则根据我的示例输出，只会有“不喜欢”

Answer 1

假设连续的喜欢和不喜欢应该被组合在一起贪婪地，你可以这样实现：

result = []
i = 0
likes = [like.split()[-1].lower() for like in list1]
for score in list2:
    end = i + 1 + (likes[i:i+2] == ["liked", "disliked"])
    result.append(f"{' and '.join(likes[i:end])} : {score.split()[-1]}")
    i = end

print(result)

这假设列表的大小符合此逻辑的预期。您可能需要添加代码来验证情况是否确实如此（错误检查）。