如何为 Python 中的每个分组应用用户定义的函数

How can I apply a user defined function for each grouping in Python

我有一个数据框 df1 如下:

Country|Month|Revenue
-------|-----|-------
US     |Jan  |100
US     |Feb  |200
US     |Mar  |300
Canada |Jan  |200
Canada |Feb  |400
Canada |Mar  |500

我想按如下方式应用用户定义函数:

df3=df1.groupby(['Country'])['Revenue'].my_cool_func()
def my_cool_func():
    b = max(Revenue)-Min(Revenue)
    c=b/2
    return c

我对 df3 的最终输出应该是:

Country|my_cool_func_rev
-------|----------------
US     |100
Canada |150

如何使用用户定义函数获得上述输出?

您可以使用 GroupBy.apply and in function working with Series, so is possible use Series.max and Series.min:

def my_cool_func(x):
    #print (x)
    return (x.max() - x.min()) / 2

df3=df1.groupby(['Country'])['Revenue'].apply(my_cool_func).reset_index()
print (df3)
  Country  Revenue
0  Canada    150.0
1      US    100.0

或者:

df3=df1.groupby(['Country'])['Revenue'].apply(lambda x:(x.max() - x.min()) / 2).reset_index()
print (df3)
  Country  Revenue
0  Canada    150.0
1      US    100.0

编辑:使用 Series.std:

def my_cool_func(x):
    b = x.std()
    c=b/2
    return c

df3=df1.groupby(['Country'])['Revenue'].apply(my_cool_func).reset_index()
print (df3)
  Country    Revenue
0  Canada  76.376262
1      US  50.000000

如果要聚合多个列,您可以尝试的另一件事是 groupby + agg:

def my_cool_func(x):
    return (x.max() - x.min()) / 2

你可以直接:

df.groupby("Country")
  .agg({
   "column1": "sum",
   "Revenue": my_cool_func,
   "columnOther": ...
  })