获取熊猫中多个ID的加权平均值

约瑟夫·科南

我有一个带有两个ID，一个计数和一个平均值的熊猫数据框。如何将两个ID分组并得到加权平均值，因此请使用以下数据集：

id1         id2       count   average
Person A    class 1   200     0.2
Person A    class 1   400     0.4
Person B    class 2   800     0.6
Person C    class 2   200     0.4
Person B    class 3   800     0.6
Person A    class 4   400     0.2
Person B    class 2   100     0.5

获取以下结果（以任何行顺序）：

id1         id2       count   average
Person A    class 1   600     0.33
Person B    class 2   900     0.59
Person C    class 2   200     0.4
Person B    class 3   800     0.6
Person A    class 4   400     0.2

以供参考：

pd.DataFrame({"id1" : ["Person A","Person A","Person B","Person C","Person B","Person A","Person B"],
              "id2" : ["class 1","class 1","class 2","class 2","class 3","class 4","class 2"],
              "count" : [200, 400, 800, 200, 800, 400, 100],
              "average" : [0.2, 0.4, 0.6, 0.4, 0.6, 0.2, 0.5]})

二凡

使用GroupBy.sum和GroupBy.apply：

df['average'] = df['count'].mul(df['average'])
grps = df.groupby(['id1', 'id2'], sort=False)
g1 = grps['count'].sum()
g2 = grps.apply(lambda x: x['average'].sum() / x['count'].sum())

dfn = pd.concat([g1, g2.rename('average').round(2)], axis=1).reset_index()

        id1      id2  count  average
0  Person A  class 1    600     0.33
1  Person B  class 2    900     0.59
2  Person C  class 2    200     0.40
3  Person B  class 3    800     0.60
4  Person A  class 4    400     0.20

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