python - 创建分组比例向量而不丢失行

Question

我很难准确地解释问题，但能够准确地 演示一下，我只想找到组内的比例。

使用以下数据:

import pandas as pd
df1 = pd.DataFrame(
    {
        "large": ["L1" for _ in range(8)],
        "small": ["S1" for i in range(4)] + ["S2" for _ in range(4)],
        "who": ["D", "E", "F", "G"] + ["D", "E", "F", "G"],
        "amount": [1, 3, 2, 0, 3, 10, 2, 1],
        "total": [22 for _ in range(8)],
    }
)

df2 = pd.DataFrame(
    {
        "large": ["L2" for _ in range(8)],
        "small": ["S3" for _ in range(4)] + ["S4" for _ in range(4)],
        "who": ["D", "E", "F", "G"] + ["D", "E", "F", "G"],
        "amount": [0, 8, 1, 1, 5, 3, 4, 1],
        "total": [23 for _ in range(8)],
    }
)

df = pd.concat([df1, df2]).reset_index(drop=True)

哪些输出:

In [82]: df
Out[82]:
   large small who  amount  total
0     L1    S1   D       1     22
1     L1    S1   E       3     22
2     L1    S1   F       2     22
3     L1    S1   G       0     22
4     L1    S2   D       3     22
5     L1    S2   E      10     22
6     L1    S2   F       2     22
7     L1    S2   G       1     22
8     L2    S3   D       0     23
9     L2    S3   E       8     23
10    L2    S3   F       1     23
11    L2    S3   G       1     23
12    L2    S4   D       5     23
13    L2    S4   E       3     23
14    L2    S4   F       4     23
15    L2    S4   G       1     23

我想计算(大范围内的金额)/(大范围内的总数) 对于每个 who，因此会有一些重复。

我可以按如下方式计算每个who的值

In [85]: df.groupby(['large','who']).agg('sum')
Out[85]:
           amount  total
large who
L1    D         4     44
      E        13     44
      F         4     44
      G         1     44
L2    D         5     46
      E        11     46
      F         5     46
      G         2     46

其中的金额列很有趣。

使用large_proportions来表示我将进行的计算 以下(我用分数表示，以便更清楚地了解发生了什么):

   large small who  amount  total  large_proportions
0     L1    S1   D       1     22                  4/22
1     L1    S1   E       3     22                  13/22
2     L1    S1   F       2     22                  4/22
3     L1    S1   G       0     22                  1/22
4     L1    S2   D       3     22                  4/22
5     L1    S2   E      10     22                  13/22
6     L1    S2   F       2     22                  4/22
7     L1    S2   G       1     22                  1/22
8     L2    S3   D       0     23                  5/23
9     L2    S3   E       8     23                  11/23
10    L2    S3   F       1     23                  5/23
11    L2    S3   G       1     23                  2/23
12    L2    S4   D       5     23                  5/23
13    L2    S4   E       3     23                  11/23
14    L2    S4   F       4     23                  5/23
15    L2    S4   G       1     23                  2/23

摘要

所以问题是，给定原始数据帧df，如何构造 最终输出包含 large_proportions

列

Answer 1

您可以在计算中使用变换，因此它仍保留原始尺寸:

df['large_proportions'] = df.groupby(['large','who'])['amount'].transform('sum') / df['total']

Out[32]: 
   large small who  amount  total  large_proportions
0     L1    S1   D       1     22           0.181818
1     L1    S1   E       3     22           0.590909
2     L1    S1   F       2     22           0.181818
3     L1    S1   G       0     22           0.045455
4     L1    S2   D       3     22           0.181818
5     L1    S2   E      10     22           0.590909
6     L1    S2   F       2     22           0.181818
7     L1    S2   G       1     22           0.045455
8     L2    S3   D       0     23           0.217391
9     L2    S3   E       8     23           0.478261
10    L2    S3   F       1     23           0.217391
11    L2    S3   G       1     23           0.086957
12    L2    S4   D       5     23           0.217391
13    L2    S4   E       3     23           0.478261
14    L2    S4   F       4     23           0.217391
15    L2    S4   G       1     23           0.086957

Transform 将聚合您的值并重复它们，以便您的结果具有与原始系列相同的长度，即使在 groupby 生效之后也是如此。