Question

我有以下df，

id    a_id    b_id
1     25      50
1     25      50
2     26      51
2     26      51
3     25      52
3     28      52
3     28      52

我有以下代码将a_id和b_id分配给-1，具体取决于id中每个df值的每一行数。 }};如果a_id或b_id值中的每一个与id的特定值具有完全相同的行/子df，那么a_id和b_id的行得-1;

cluster_ids = df.loc[df['id'] > -1]['id'].unique()

types = ['a_id', 'b_id']

for cluster_id in cluster_ids:
    rows = df.loc[df['id'] == cluster_id]

    for type in types:
        ids = rows[type].values

        match_rows = df.loc[df[type] == ids[0]]

        if match_rows.equals(rows):
           df.loc[match_rows.index, type] = -1

所以结果df看起来像，

id    a_id    b_id
1     25      -1
1     25      -1
2     -1      -1
2     -1      -1
3     25      -1
3     28      -1
3     28      -1

我想知道是否有更有效的方法来实现它。

Answer 1

one_value_for_each_id = df.groupby('id').transform(lambda x: len(set(x)) == 1)

 a_id  b_id
0   True  True
1   True  True
2   True  True
3   True  True
4  False  True
5  False  True
6  False  True

one_id_for_each_value = pd.DataFrame({
    col: df.groupby(col).id.transform(lambda x: len(set(x)) == 1)
    for col in ['a_id', 'b_id']
})

   a_id  b_id
0  False  True
1  False  True
2   True  True
3   True  True
4  False  True
5   True  True
6   True  True

one_to_one_relationship = one_id_for_each_value & one_value_for_each_id

# Set all values that satisfy the one-to-one relationship to `-1`
df.loc[one_to_one_relationship.a_id, 'a_id'] = -1
df.loc[one_to_one_relationship.b_id, 'b_id'] = -1

a_id  b_id
0    25    -1
1    25    -1
2    -1    -1
3    -1    -1
4    25    -1
5    28    -1
6    28    -1

pandas assign columns values取决于df

1 个答案: