python 实现分组求和与分组累加求和代码

# -*- encoding=utf-8 -*-
import pandas as pd
data=['abc','abc','abc','asc','ase','ase','ase']
num=[1,2,2,1,2,1,2]
df1=pd.DataFrame({'name':data,'num':num})
print(df1)

df1['mmm']=df1['num']
df2=df1.groupby(['name', 'num'], as_index=False).count()
print(df2)
df2.sort_values(['name', 'num'], ascending=[1, 1], inplace=True)
print(df2)
df2['sum']=df2.groupby(['name'])['mmm'].cumsum()
print(df2)
kk=df2.groupby(['name'],as_index=False)['num'].sum()
print(kk)
df3 = pd.merge(df2, kk, on='name', how='left',)
print(df3)
df3['ratio']=df3['sum']/df3['num_y']
df3.columns = ['name', 'num', 'mmm', 'sum','numsum','ratio']
print(df3)
df4=df3.groupby(['mmm'],as_index=False)['ratio'].mean()

print(df4)

 name num
0 abc  1
1 abc  2
2 abc  2
3 asc  1
4 ase  2
5 ase  1
6 ase  2
 name num mmm
0 abc  1  1
1 abc  2  2
2 asc  1  1
3 ase  1  1
4 ase  2  2
 name num mmm
0 abc  1  1
1 abc  2  2
2 asc  1  1
3 ase  1  1
4 ase  2  2
 name num mmm sum
0 abc  1  1  1
1 abc  2  2  3
2 asc  1  1  1
3 ase  1  1  1
4 ase  2  2  3
 name num
0 abc  3
1 asc  1
2 ase  3
 name num_x mmm sum num_y
0 abc   1  1  1   3
1 abc   2  2  3   3
2 asc   1  1  1   1
3 ase   1  1  1   3
4 ase   2  2  3   3
 name num mmm sum numsum   ratio
0 abc  1  1  1    3 0.333333
1 abc  2  2  3    3 1.000000
2 asc  1  1  1    1 1.000000
3 ase  1  1  1    3 0.333333
4 ase  2  2  3    3 1.000000
  mmm   ratio
0  1 0.555556
1  2 1.000000

Process finished with exit code 0

 new_dayincome = request.POST.get("dayincome_time", None)

    # total_income = models.bathAccount.objects.filter(dayBath=new_dayincome).aggregate(nums=Sum('priceBath'))
    total_income = models.bathAccount.objects.values('priceBath').annotate(nums=Sum('priceBath')).filter(dayBath=new_dayincome)
    print("total_income",total_income[0]['nums'])

from django.db.models import Sum,Count
new_dayincome = request.POST.get("dayincome_time", None)

    total_income = models.bathAccount.objects.filter(dayBath=new_dayincome).aggregate(nums=Sum('priceBath'))
    print("total_income",total_income['nums'])