熊猫根据按列分组的指标计算差异

小精灵

df['New']=df.b.loc[df.k==1]# get all value b when k equal to 1
df.New=df.groupby('a').New.apply(lambda x : x.ffill().shift()) # fillna by froward method , then we need shift.
df.b-df['New']# yield 
Out[260]: 
0     NaN
1     NaN
2     1.0
3     2.0
4     3.0
5     1.0
6     NaN
7     NaN
8     1.0
9     2.0
10    1.0
dtype: float64

在后面创建行数据的分区 k == 1 一直到下一个 K=1 对每组使用cumsum和shift a

parts = df.groupby('a').k.apply(lambda x: x.shift().cumsum())

按分组 df.a 和; parts 计算两者之间的差额 b 和; b.min() 在每组中

vals = df.groupby([df.a, parts]).b.apply(lambda x: x-x.min()+1)

当part==0时将值设置为空,并将其分配回数据帧

df['diff_b'] = np.select([parts!=0], [vals], np.nan)

输出:

    a  b  k  diff_b
0   0  0  0     NaN
1   0  1  1     NaN
2   0  2  0     1.0
3   0  3  0     2.0
4   0  4  1     3.0
5   0  5  0     1.0
6   1  0  0     NaN
7   1  1  1     NaN
8   1  2  0     1.0
9   1  3  1     2.0
10  1  4  0     1.0