python如何使用groupby对数据进行分类并计算其他列的平均值

Question

我有一个数据框，如下所示。我想使用“part1”列作为将数据分类为 3 个部分（每个部分具有相同数量的数据集）的基准，并计算每个组的 part2 均值的均值。如row0和row1为groupB，均值为(0.67+(-0.03))/2。import pandas as pddf = pd.DataFrame({    "date":["20130101","20130101","20130103","20130103","20130105","20130105"],    "part1":[0.5,0.7,1.3,1.5,0.1,0.3],    "part2":[0.67,-0.03,1.95,-3.25,-0.3,0.6]})    date    part1   part2   output0   20130101    0.5 0.67    0.321   20130101    0.7 -0.03   0.322   20130103    1.3 1.95    -0.653   20130103    1.5 -3.25   -0.654   20130105    0.1 -0.3    0.155   20130105    0.3 0.6 0.15

Answer 1

如果你想计算每天的平均值，你可以使用groupby如下：

import pandas as pd

df = pd.DataFrame({

    "date":["20130101","20130101","20130103","20130103","20130105","20130105"],

    "part1":[0.5,0.7,1.3,1.5,0.1,0.3],

    "part2":[0.67,-0.03,1.95,-3.25,-0.3,0.6]

})

df.groupby("date").mean().reset_index()

结果：

       date  part1  part2

0  20130101    0.6   0.32

1  20130103    1.4  -0.65

2  20130105    0.2   0.15

Answer 2

您可以by为熊猫groupby方法的参数传递一个函数。

from functools import partial

import pandas as pd

df = pd.DataFrame({

    "date":["20130101","20130101","20130103","20130103","20130105","20130105"],

    "part1":[0.5,0.7,1.3,1.5,0.1,0.3],

    "part2":[0.67,-0.03,1.95,-3.25,-0.3,0.6]

})

def grouper(df, val):

    foo = df.iloc[val]['part1']

    if foo > 0.0 and foo < 0.4:

         return 0

    elif foo > 0.3 and foo < 1.0:

         return 1

    elif foo > 1.0:

         return 2

grouped = df['part2'].groupby(by=partial(grouper, df)).mean()

这导致

1    0.15

2    0.32

3   -0.65

Name: part2, dtype: float64