Python Dateframe列中匹配查找数据(list)添加新列的方法

数据：

      ID                   Found_IDs  
0  12345        [15443, 15533, 3433]   
1  15533  [2234, 16608, 12002, 7654]   
2   6789      [43322, 876544, 36789] 

待查数据：

bad_ids = [15533, 876544]

1、使用np.intersect1d实现

df['bad_id'] = df['Found_IDs'].apply(lambda x: np.intersect1d(x, bad_ids))
      ID                   Found_IDs    bad_id
0  12345        [15443, 15533, 3433]   [15533]
1  15533  [2234, 16608, 12002, 7654]        []
2   6789      [43322, 876544, 36789]  [876544]

2、使用set求交集

bad_ids_set = set(bad_ids)
df['Found_IDs'].apply(lambda x: list(set(x) & bad_ids_set))
      ID                   Found_IDs    bad_id
0  12345        [15443, 15533, 3433]   [15533]
1  15533  [2234, 16608, 12002, 7654]        []
2   6789      [43322, 876544, 36789]  [876544]

3、使用列表表达式实现

bad_ids = [15533, 876544]
df['bad_id'] = [any(c in l for c in bad_ids) for l  in df['Found_IDs']]
print (df)
      ID                   Found_IDs  bad_id
0  12345        [15443, 15533, 3433]    True
1  15533  [2234, 16608, 12002, 7654]   False
2   6789      [43322, 876544, 36789]    True

4、使用apply和np.any实现

df['bad_id'] = df['Found_IDs'].apply(lambda x: np.any([c in x for c in bad_ids]))

或者

df['bad_id'] = df['Found_IDs'].apply(lambda x: [*filter(lambda x: c in x, bad_ids)])

5、 使用pd.concat和merge实现

bad_ids = [15533, 876544, 36789, 11111]
df2 = pd.concat(
    [
        df,
        pd.merge(
            df["Found_IDs"].explode().reset_index(),
            pd.Series(bad_ids, name="bad_ids"),
            left_on="Found_IDs",
            right_on="bad_ids",
            how="inner",
        )
        .groupby("index")
        .agg(bad_ids=("bad_ids", list)),
    ],
    axis=1,
).fillna(False)
print(df2)

      ID                   Found_IDs          bad_ids
0  12345        [15443, 15533, 3433]          [15533]
1  15533  [2234, 16608, 12002, 7654]            False
2   6789      [43322, 876544, 36789]  [876544, 36789]