python - Groupby + 计算特定项目(不是全部)，将结果放入新列

Question

我有这套:

df=pd.DataFrame({'user':[1,1,2,2,2,3,3,3,3,3,4,4],
                  'date':['1995-09-01','1995-09-02','1995-10-03','1995-10-04','1995-10-05','1995-11-07','1995-11-08','1995-11-09','1995-11-10','1995-11-15','1995-12-18','1995-12-20'],
                  'type':['a','a','b','a','c','a','b','a','b','b','a','b']})

这给了我:

user    date    type
 1  1995-09-01   a
 1  1995-09-02   a
 2  1995-10-03   b
 2  1995-10-04   a
 2  1995-10-05   c
 3  1995-11-07   a
 3  1995-11-08   b
 3  1995-11-09   a
 3  1995-11-10   b
 3  1995-11-15   b
 4  1995-12-18   a
 4  1995-12-20   b

我想创建一个新列，其中显示“类型”列上a值的计数，并按“用户”列分组

这是预期的结果:

user    date    type    cta_a
1   1995-09-01    a       2
1   1995-09-02    a       2
2   1995-10-03    b       1
2   1995-10-04    a       1
2   1995-10-05    c       1
3   1995-11-07    a       2
3   1995-11-08    b       2
3   1995-11-09    a       2
3   1995-11-10    b       2
3   1995-11-15    b       2
4   1995-12-18    a       1
4   1995-12-20    b       1

我尝试了以下方法，但没有成功。

df['ct_a'] = df.groupby('user')[df['type']== 'a'].transform('count')

Answer 1

屏蔽列type中的非a值，然后groupby和transform 使用计数:

df['ct_a'] = df['type'].mask(lambda x: x.ne('a'))\
                       .groupby(df['user']).transform('count')

---

    user        date type  ct_a
0      1  1995-09-01    a     2
1      1  1995-09-02    a     2
2      2  1995-10-03    b     1
3      2  1995-10-04    a     1
4      2  1995-10-05    c     1
5      3  1995-11-07    a     2
6      3  1995-11-08    b     2
7      3  1995-11-09    a     2
8      3  1995-11-10    b     2
9      3  1995-11-15    b     2
10     4  1995-12-18    a     1
11     4  1995-12-20    b     1