Python 3.x 熊猫基于另一列中选定为条件的值创建列

我有以下

df

id    match_type    amount    negative_amount
1     exact         10        False
1     exact         20        False
1     name          30        False
1     name          40        False
1     amount        15        True
1     amount        15        True 
2     exact         0         False
2     exact         0         False

---

我想创建一个列

0\u amount\u sum

，该列指示（布尔值）如果

amount

sum是我认为需要

groupby

by 2

系列

（列）而不是筛选：

df['0_amount_sum_'] = ((df.amount * np.where(df.negative_amount, -1, 1))
                           .groupby([df['id'], df['match_type']])
                           .transform('sum')
                           .le(0))

   id match_type  amount  negative_amount  0_amount_sum_
0   1      exact      10            False          False
1   1      exact      20            False          False
2   1       name      30            False          False
3   1       name      40            False          False
4   1     amount      15             True           True
5   1     amount      15             True           True
6   2      exact       0            False           True
7   2      exact       0            False           True

---

@ScottBoston因为

金额

对于

id=1

和

match_type=amount

，总和为-30，即

df = df.loc[df.match_type=='exact']

df['0_amount_sum_'] = (df.assign(
    amount_n=df.amount * np.where(df.negative_amount, -1, 1)).groupby(
    'id')['amount_n'].transform(lambda x: sum(x) <= 0))

df = df.loc[df.match_type=='name']

df['0_amount_sum_'] = (df.assign(
    amount_n=df.amount * np.where(df.negative_amount, -1, 1)).groupby(
    'id')['amount_n'].transform(lambda x: sum(x) <= 0))

df = df.loc[df.match_type=='amount']

df['0_amount_sum_'] = (df.assign(
    amount_n=df.amount * np.where(df.negative_amount, -1, 1)).groupby(
    'id')['amount_n'].transform(lambda x: sum(x) <= 0))

df['0_amount_sum_'] = ((df.amount * np.where(df.negative_amount, -1, 1))
                           .groupby([df['id'], df['match_type']])
                           .transform('sum')
                           .le(0))

   id match_type  amount  negative_amount  0_amount_sum_
0   1      exact      10            False          False
1   1      exact      20            False          False
2   1       name      30            False          False
3   1       name      40            False          False
4   1     amount      15             True           True
5   1     amount      15             True           True
6   2      exact       0            False           True
7   2      exact       0            False           True