使用groupby时的问题，如何计算groupby后的增量

pd.DataFrame({'date': {0: Timestamp('2021-08-01 00:00:00'),
1: Timestamp('2022-08-01 00:00:00'),
2: Timestamp('2021-08-01 00:00:00'),
3: Timestamp('2021-08-01 00:00:00'),
4: Timestamp('2022-08-01 00:00:00'),
5: Timestamp('2022-08-01 00:00:00')},
'customer_nr': {0: 2, 1: 3, 2: 2, 3: 3, 4: 2, 5: 2},
'product_nr': {0: 3, 1: 2, 2: 2, 3: 1, 4: 2, 5: 1},
'age': {0: 32.0, 1: 32.0, 2: 32.0, 3: 32.0, 4: 32.0, 5: 37.0},
'gender': {0: 'M', 1: 'M', 2: 'M', 3: 'M', 4: 'M', 5: 'M'},
'age_group': {0: '25-34',
1: '25-34',
2: '25-34',
3: '25-34',
4: '25-34',
5: '35-44'}} )

t = df.groupby(['date','product_nr','age_group']).age.count().unstack()

在product_nr上应用百分比更改分组：(如果您有多个日期，此代码也适用(

output = pd.DataFrame()
for group,df in t.groupby('product_nr'):
temp = ((df/df.shift(1))-1)*100
output = pd.concat([output,temp])
output.reset_index(inplace=True)

输出：

age_group       date  product_nr  25-34  35-44
0         2021-08-01           1    NaN    NaN
1         2022-08-01           1    NaN    NaN
2         2021-08-01           2    NaN    NaN
3         2022-08-01           2  100.0    NaN
4         2021-08-01           3    NaN    NaN

获取所需日期的输出：

output[output['date'] == '2022-08-01']

最终输出：

age_group       date  product_nr  25-34  35-44
1         2022-08-01           1    NaN    NaN
3         2022-08-01           2  100.0    NaN

df_group = df.groupby(['date','product_nr','age_group']).age.count().unstack().fillna(0)

df_2022 = df_group.loc["2022-08-01"]
df_2021 = df_group.loc["2021-08-01"]

(df_2022 - df_2021).divide(df_2021)