python pandas groupby去重条件列并且求和原标数据都在没有删除的函数记录

import pandas as pd
data = {"number":[1,1,np.nan,np.nan,2,2,1,2,2],
        "letter":['a','b',np.nan,np.nan,'c','d','b','c','d'],
        "label":[1,1,9,9,2,2,1,2,2],"label2":[33,12,9,9,2,2,1,2,2]}
dataset1 = pd.DataFrame(data) #初始化DataFrame 得到数据集dataset1
print(dataset1)
t_optarr=['letter','number']
sumobj={'label':sum,'label2':sum}
##去重 合并求和 并且保留原列表的函数
def rechong_sum_hwj(odata,optarr,sumobj):
    duplicate_row = odata.duplicated(subset=optarr,keep=False)
    duplicate_data = odata.loc[duplicate_row,:]
    duplicate_data_sum = duplicate_data.groupby(by=optarr).agg(sumobj).reset_index(drop=True)
    duplicate_data_one = duplicate_data.drop_duplicates(subset=optarr, keep="first").reset_index(drop=True)
    no_duplicate=odata.drop_duplicates(subset=optarr,keep=False)
    for i in sumobj:
        duplicate_data_one[i] = duplicate_data_sum[i]
    result2 = pd.concat([no_duplicate, duplicate_data_one])
    return result2
##去重 合并求和 并且保留原列表的函数
print(rechong_sum_hwj(dataset,t_optarr,sumobj))

结果是这样：基本实现了又保留原列，又求和合并，

  number letter  label  label2
0      1      a      1      33
1      1      b      1      12
2    xxx    xxx      9       9
3    xxx    xxx      9       9
4      2      c      2       2
5      2      d      2       2
6      1      b      1       1
7      2      c      2       2
8      2      d      2       2
  number letter  label  label2
0      1      a      1      33
0      1      b      2      13
1    xxx    xxx      4       4
2      2      c      4       4
3      2      d     18      18

Process finished with exit code 0