模拟数据

import pandas as pd
import numpy as np
employees = ["小明","小周","小孙","小王","小张"]   # 5位员工
time = ["上半年", "下半年"]


df=pd.DataFrame({
    "employees":np.random.choice(employees,10),  # 在员工中重复选择10次
    # 另一种写法
    #"employees":[employees[x] for x in np.random.randint(0,len(employees),10)],  
    "time":np.random.choice(time,10),
    "salary":np.random.randint(800,1000,10),  # 800-1000之间的薪资选择10个数值
    "score":np.random.randint(6,12,10)  # 6-11的分数选择10个
})

df

Python技巧分享之groupby基础用法详解

groupby+单个字段+单个聚合

求解每个人的总薪资金额:

total_salary = df.groupby("employees")["salary"].sum().reset_index()
total_salary

Python技巧分享之groupby基础用法详解

使用agg也能够实现上面的效果:

df.groupby("employees").agg({"salary":"sum"}).reset_index()

Python技巧分享之groupby基础用法详解

df.groupby("employees").agg({"salary":np.sum}).reset_index()

Python技巧分享之groupby基础用法详解

groupby+单个字段+多个聚合

求解每个人的总薪资金额和薪资的平均数

方法1:使用groupby+merge

mean_salary = df.groupby("employees")["salary"].mean().reset_index()
mean_salary

Python技巧分享之groupby基础用法详解

然后将上面的两个结果进行组合;在合并之前为了字段的名字更加的直观,我们重命名下:

total_salary.rename(columns={"employees":"total_salary"})
mean_salary.columns = ["employees","mean_salary"]
total_mean = total_salary.merge(mean_salary)
total_mean

Python技巧分享之groupby基础用法详解

方法2:使用groupby+agg

total_mean = df.groupby("employees")\
            .agg(total_salary=("salary", "sum"), 
                 mean_salary=("salary", "mean"))\
            .reset_index()
total_mean

Python技巧分享之groupby基础用法详解

groupby+多个字段+单个聚合

针对多个字段的同时聚合:

df.groupby(["employees","time"])["salary"].sum().reset_index()

Python技巧分享之groupby基础用法详解

# 使用agg来实现

df.groupby(["employees","time"]).agg({"salary":"sum"}).reset_index()

Python技巧分享之groupby基础用法详解

groupby+多个字段+多个聚合

使用的方法是:

agg(’新列名‘=(’原列名‘, ’统计函数/方法‘))

df.groupby(["employees","time"])\
            .agg(total_salary=("salary", "sum"), 
                mean_salary=("salary", "mean"),
                total_score=("score", "sum") 
                )\
            .reset_index()

Python技巧分享之groupby基础用法详解

原文地址:https://blog.csdn.net/qq_34160248/article/details/127522857

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