如何将列表形式的输出拆分为单独的行？
最后一列派生自定义函数GROUPBY A、B、C.应用GROUPBY函数后，最后一列没有列标题
A B C
202311 X Y [147.01, 140.01, 133.01, 126.01, 119.01, 112.0...
202312 X Y [175, 168, 161, 154, 147, 140, 133, 126, 119, ...
202313 X Y [175, 168, 161, 154, 147, 140, 133, 126, 119, ...
202314 X Y [175, 168, 161, 154, 147, 140, 133, 126, 119, ...
202315 X Y [175, 168, 161, 154, 147, 140, 133, 126, 119, ...
202316 X Y [95.09, 88.09, 81.09, 74.09, 67.09, 60.09, 53....
202317 X Y [60.1, 53.1, 46.1, 39.1, 32.1, 25.1, 18.1, 11....
202318 X Y [57.02, 50.02, 43.02, 36.02, 29.02, 22.02, 15....
202319 X Y [43.31, 36.31, 29.31, 22.31, 15.31, 8.31, 1.31...
202320 X Y [66.84, 59.84, 52.84, 45.84, 38.84, 31.84, 24....
202321 X Y [175, 168, 161, 154, 147, 140, 133, 126, 119, ...
202322 X Y [155.18, 148.18, 141.18, 134.18, 127.18, 120.1...
202323 X Y [170.63, 163.63, 150.18, 143.18, 136.18, 129.1...
预期结果:
A B C D E
202311 X Y 202311 147.01
202311 X Y 202312 140.01
202311 X Y 202313 133.01
202311 X Y 202314 126.01
202311 X Y 202315 119.01
202311 X Y 202316 112.0
202311 X Y 202317 ..
202311 X Y 202318 ..
202312 X Y 202312 175
202312 X Y 202313 168
202312 X Y 202314 161
202312 X Y 202315 154
202312 X Y 202316 147
列D是现有列.
谢谢!
已try
df['E'] = df.groupby(['A', 'B', 'C']).apply(lambda d: customfunction(d['X'], d['Y']).str.split(',')个
错误消息:
incompatible index of inserted column with frame index个
