pandas根据另一列中的值创建一个列，该列选择作为条件

Question

I have the following 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

I want to create a column 0_amount_sum that indicates (boolean) if the amount sum is <= 0 or not for each id of a particular match_type , eg the following is the result df ;

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

for id=1 and match_type=exact , the amount sum is 30, so 0_amount_sum is False . The code is as follows,

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))

I am wondering if there is a better way/more efficient to do that, especially when the values of match_type is unknown, so the code can automatically enumerate all the possible values and then do the calculation accordingly.

Answer 1

I believe need groupby by 2 Series (columns) instead filtering:

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