你需要在query/auto generated words（zip）对上使用列表解析，并使用set来进行有效的成员测试：

df['corrected keywords'] = ['|'.join(w for w in l if w in S)
                            for S, l in zip(df['query_text'].apply(lambda x: set(x.split())),
                                            df['auto generated keywords'].str.split('|'))]

输出量：

S.No                       query_text auto generated keywords corrected keywords
0     1                      I want ring            diamond|ring               ring
1     2              I want wedding band            band|wedding       band|wedding
2     3   I look for sapphire collection           ring|sapphire           sapphire
3     4           I want diamond earring         diamond|earring    diamond|earring
4     5  I am looking for stackable ring          ring|stackable     ring|stackable
5     6             I need gold bracelet           bracelet|gold      bracelet|gold
6     7             I look for gold ring               gold|ring          gold|ring
7     8             I need sapphire ring           ring|sapphire      ring|sapphire

我用了一个不那么优雅的方法来解决这个问题。
我首先清理了每行自动生成的关键字，|和中间的空格，并使用函数将它们转换为列表。

def stripdown(column):
    for index in range(1,len(column)+1):
        return_list = []
        raw_string = column[index]
        split_string = raw_string.split('|')
        for text in split_string:
            return_list.append(text.strip())
        column[index] = return_list

然后检查自动生成的关键字中的每个单词是否在query_text中。

for index in range(1,len(new_df['auto generated keywords'])+1):
    for item in new_df['auto generated keywords'][index]:
        if item not in new_df['query_text'][index]:
            new_df['auto generated keywords'][index].remove(item)
        else:
            continue

这是Result