大熊猫中的独立群

Question

我需要帮助熊猫群：

我有以下df：

A     B          C          D
04547 2022-07-04 2022-07-04 1000000
04547 2022-07-11 2022-07-11 1000000
04547 2022-08-08 2022-08-08 1000000
04547 2022-10-11 2022-10-11 0100000
04547 2022-10-18 2022-10-18 0100000
04547 2022-10-24 2022-10-24 1000000
04547 2022-11-01 2022-11-01 0100000
04547 2022-11-08 2022-11-08 0100000
04548 2022-10-11 2022-10-11 0100000
04548 2022-10-18 2022-10-18 0100000
04548 2022-10-24 2022-10-24 1000000
04548 2022-11-01 2022-11-01 0100000
04548 2022-11-08 2022-11-08 0100000

我需要的输出应该是：

    A           B           C        D
04547  2022-07-04  2022-08-08  1000000
04547  2022-10-11  2022-10-18  0100000
04547  2022-10-24  2022-10-24  1000000
04547  2022-11-01  2022-11-08  0100000
04548  2022-10-11  2022-10-18  0100000
04548  2022-10-24  2022-10-24  1000000
04548  2022-11-01  2022-11-08  0100000

但有：

a = {'A':'first','B':'first','C':'last','D':'first'}
df = df.groupby(['A','D']).agg(a)

A     B          C          D
4547 2022-10-11 2022-11-08  0100000
4547 2022-07-04 2022-10-24  1000000
4548 2022-10-11 2022-11-08  0100000
4548 2022-10-24 2022-10-24  1000000

因为当D列中的新序列开始为每一列A分离时，我不得不中断分组

Answer 1

您可以使用df['D'].ne(df['D'].shift().cumsum()作为第D列的石斑鱼，以确保按连续值进行分组(shift允许您访问上一行值)：

a = {'A':'first','B':'first','C':'last','D':'first'}
out = (df.groupby(['A', df['D'].ne(df['D'].shift()).cumsum()],
                  as_index=False)
         .agg(a)
      )

产出：

       A           B           C        D
0  04547  2022-07-04  2022-08-08  1000000
1  04547  2022-10-11  2022-10-18  0100000
2  04547  2022-10-24  2022-10-24  1000000
3  04547  2022-11-01  2022-11-08  0100000
4  04548  2022-10-11  2022-10-18  0100000
5  04548  2022-10-24  2022-10-24  1000000
6  04548  2022-11-01  2022-11-08  0100000

Answer 2

如果需要按A,D列的连续值分组，则使用DataFrame.shifted值comapre表示DataFrame.ne不相等，DataFrame.any表示Series，然后按Series.cumsum添加累积和

g = df[['A','D']].ne(df[['A','D']].shift()).any(1).cumsum()
a = {'A':'first','B':'first','C':'last','D':'first'}
df = df.groupby(g).agg(a)
print (df)
       A           B           C        D
1  04547  2022-07-04  2022-08-08  1000000
2  04547  2022-10-11  2022-10-18  0100000
3  04547  2022-10-24  2022-10-24  1000000
4  04547  2022-11-01  2022-11-08  0100000
5  04548  2022-10-11  2022-10-18  0100000
6  04548  2022-10-24  2022-10-24  1000000
7  04548  2022-11-01  2022-11-08  0100000