python difference with previous row by group

Question

i am trying to take diff value from previous row in a dataframe by grouping column "group", there are several similar questions but i can't get this working.

         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

This is the result that i need

         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

Shifts through each group to create a calculated column. Subtract that column from the original value column to create the difference column.

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

Answer 2

You can group use diff()

df = df.sort_values('date')
df['diff'] = df.groupby(['group'])['value'].diff()

gives

          date group  value    diff
0   2020-01-01     A    808     NaN
1   2020-01-01     B    331     NaN
2   2020-01-02     A    612  -196.0
3   2020-01-02     B   1391  1060.0
4   2020-01-03     A    234  -378.0
5   2020-01-04     A    828   594.0
6   2020-01-04     B    820  -571.0
7   2020-01-05     A   1075   247.0
8   2020-01-07     B    572  -248.0
10  2020-01-10     A   1436   361.0
9   2020-01-10     B    736   164.0

If you want the dataset ordered as you have it you can add group to the sort but its not necessary for the operation and can be done before or after you get the differences.

df.sort_values(['group','date'])

          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
7   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