map带过滤功能，我认为这是一个有效的解决方案

df['fruits'].map(lambda s: [x for x in s if x in fruit_list])

0             [apple]
1         [pineapple]
2    [orange, tomato]
Name: fruits, dtype: object

您可以使用apply并定义function：

def in_fruits(row, lst):
    ret_lst = []
    for f in row:
        if f in lst:
            ret_lst.append(f)
    return ret_lst

interest_fruits = ['apple','orange','pineapple','banana','tomato']
df1 = pd.DataFrame({'Fruits':[['apple', 'mango'], ['mango', 'cabbage', 'pineapple', 'carrot'], ['egg', 'sandwich', 'orange', 'carrot', 'tomato']]})
df1.Fruits.apply(lambda x: in_fruits(x, interest_fruits))

它返回以下内容：

0             [apple]
1         [pineapple]
2    [orange, tomato]
Name: Fruits, dtype: object

使用set.intersection获取常见水果

df.fruits.apply(lambda x: list(set(x) & set(fruit_list)))

输出为

0             [apple]
1         [pineapple]
2    [orange, tomato]
Name: fruits, dtype: object