python 实现分组求和与分组累加求和代码

我就废话不多说了,大家还是直接看代码吧!

# -*- 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)

运行:

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

补充知识:python项目篇-对符合条件的某个字段进行求和,聚合函数annotate(),aggregate()函数

对符合条件的某个字段求和

需求是,计算每日的收入和

1、

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'])

输出结果:total_income 132

2、

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'])

输出结果:total_income 572

第二种输出的是正确的数字

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