df = pd.DataFrame([
    {"meta": ['Home', 'grocery', 'cake']},
    {"meta": ['Home', 'grocery', 'Biscuit', 'Oreo']}
    ])

terms = ['cake', 'biscuit']

df["match_value"] = df["meta"].apply(lambda row: list(set(map(str.lower, row)).intersection(set(terms))))

你可以使用正则表达式模式在分解列表后提取子字符串：

import re

pattern = re.compile(f"({'|'.join(terms)})", re.IGNORECASE)

df['match value'] = (df['meta'].explode().str.extractall(pattern)[0]
                               .groupby(level=0).agg(list)

df['result'] = df['match value'].str.len().astype(bool)

输出：

>>> df
                             meta   match value  result
0  [Home, grocery, cake, biscuit]        [cake]    True
1  [Home, grocery, Biscuit, Oreo]     [Biscuit]    True

我们使用带有矢量化字符串匹配的explode

exploded = df['meta'].explode().str.lower()
df['match'] = exploded[exploded.isin(terms)].groupby(level=0).agg(list)
df['result'] = df['match'].str.len() > 0

为获得最佳结果，请确保terms中的术语也是小写的。
让我们尝试另一组术语，看看这是如何工作的

terms = ['home', 'biscuit']

exploded = df['meta'].explode().str.lower()
df['match'] = exploded[exploded.isin(terms)].groupby(level=0).agg(list)
df['result'] = df['match'].str.len() > 0
df
    
                             meta            match  result
0           [Home, grocery, cake]           [home]    True
1  [Home, grocery, Biscuit, Oreo]  [home, biscuit]    True