python 按组与前一行的差异

Question

我试图通过对“组”列进行分组来获取数据框中前一行的差异值，有几个类似的问题，但我无法正常工作。

         date group  value
0  2020-01-01     A    808
1  2020-01-01     B    331
2  2020-01-02     A    612
3  2020-01-02     B   1391
4  2020-01-03     A    234
5  2020-01-04     A    828
6  2020-01-04     B    820
6  2020-01-05     A   1075
8  2020-01-07     B    572
9  2020-01-10     B    736
10 2020-01-10     A   1436

df.sort_values(['group','date'], inplace=True)
df['diff'] = df['value'].diff()
print(df)

         date  value group   diff
1  2020-01-03    234     A    NaN
8  2020-01-01    331     B   97.0
2  2020-01-07    572     B  241.0
9  2020-01-02    612     A   40.0
5  2020-01-10    736     B  124.0
17 2020-01-01    808     A   72.0
14 2020-01-04    820     B   12.0
4  2020-01-04    828     A    8.0
18 2020-01-05   1075     A  247.0
7  2020-01-02   1391     B  316.0
10 2020-01-10   1436     A   45.0

这就是我需要的结果

         date group  value diff
0  2020-01-01     A    808 Na
2  2020-01-02     A    612 -196
4  2020-01-03     A    234 -378
5  2020-01-04     A    828 594
6  2020-01-05     A   1075 247
10 2020-01-10     A   1436 361
1  2020-01-01     B    331 Na
3  2020-01-02     B   1391 1060
6  2020-01-04     B    820 -571
8  2020-01-07     B    572 -248
9  2020-01-10     B    736 164

Answer 1

遍历每个组以创建计算列。从原始值列中减去该列以创建差异列。

df.sort_values(['group','date'], ascending=[True,True], inplace=True)
df['shift'] = df.groupby('group')['value'].shift()
df['diff'] = df['value'] - df['shift']
df = df[['date','group','value','diff']]

1
df
date    group   value   diff
0   2020-01-01  A   808 NaN
2   2020-01-02  A   612 -196.0
4   2020-01-03  A   234 -378.0
5   2020-01-04  A   828 594.0
6   2020-01-05  A   1075    247.0
10  2020-01-10  A   1436    361.0
1   2020-01-01  B   331 NaN
3   2020-01-02  B   1391    1060.0
6   2020-01-04  B   820 -571.0
8   2020-01-07  B   572 -248.0
9   2020-01-10  B   736 164.0