这些方法不依赖于Group列的值，只依赖于有两个有序组。

out = (df.set_index(['date', 'bucket', 'Group'])['purchase']
         .groupby(['date', 'bucket'])
         .diff(-1)
         .dropna()
         .droplevel(-1)
         .reset_index(name='difference'))
print(out)

# OR 

out = (df.groupby(['date', 'bucket'])
         .apply(lambda x: x.groupby('Group')['purchase']
                           .sum()
                           .diff(-1)
                           .dropna()))
out.columns = ['difference']
out = out.reset_index()
print(out)

输出：

date  bucket  difference
0  2020-01-01        1         1.0
1  2020-01-01        2         0.0
2  2020-02-01        1         1.0

1-Mozway的班轮：

(df.set_index(['date', 'bucket', 'Group'])['purchase']
   .agg(lambda x: x.xs('A', level='Group').sub(x.xs('B', level='Group')))
   .reset_index(name='difference'))

输出：

date  bucket  difference
0  2020-01-01       1           1
1  2020-01-01       2           0
2  2020-02-01       1           1

Use <code>DataFrame.pivot_table</code> with <code>DataFrame.sub</code>:
df1 = df.pivot_table(index=['date','bucket'], columns='purchase', 
values='value')
   .sub(df1['B'], axis=0)
   .rename(columns={'A':'difference'})
   .reset_index()
print (df1)
  purchase       date  bucket  difference
  0 2020-01-01 2020-01-01       1           1
  1 2020-02-01 2020-02-01       1           1

def function1(dd:pd.DataFrame):
    return dd.assign(difference=dd.purchase.diff(-1)).iloc[0]

df1.groupby(["date",'bucket'],as_index=False).apply(function1).drop("purchase",axis=1)

出局

date    bucket  Group   difference
0   2020-01-01  1   A   1.0
1   2020-01-01  2   A   0.0
2   2020-02-01  1   A   1.0

当需要选择和对齐时，通常更容易将列设置为索引。这里使用xs来选择组：

s = df.set_index(['date', 'bucket', 'Group'])['purchase'].rename('difference')

(s.xs('A', level='Group')-s.xs('B', level='Group')).reset_index()

使用DataFrame作为中间层，然后重命名的变量（用于处理多列）：

df2 = df.set_index(['date', 'bucket', 'Group'])

(df2.xs('A', level='Group')-df2.xs('B', level='Group')
 ).reset_index().rename(columns={'purchase': 'difference'})

输出：

date  bucket  difference
0  2020-01-01       1           1
1  2020-01-01       2           0
2  2020-02-01       1           1

您可以在计算diff之前重新调整 Dataframe ：

out = (df.set_index(['date', 'bucket', 'group'])['purchase']
         .unstack('group').diff(-1, axis=1)['A']
         .rename('difference').reset_index())
print(out)

# Output
         date  bucket  difference
0  2020-01-01       1           1
1  2020-01-01       2           0
2  2020-02-01       1           1

或者使用pivot：

out = (df.pivot(['date', 'bucket'], 'group', 'purchase')
         .diff(-1, axis=1)['A'].rename('difference').reset_index())