基于多列 pandas 的分组依据和总和的运行总计

Question

我正在使用以下数据集，但在根据团队 ID 计算总分时遇到了麻烦。一支球队可以是主场也可以是客场，我希望计算他们的总得分。

我已经成功创建了基于 home_id 和away_id 的运行总计以及基于 home/away id 的运行平均值，但我很难根据这两列进行计算

例如，如果在第 1 场比赛中主队得分 1，然后在第 2 场比赛中他们是客队并得分 3，我想创建一个列，表示到目前为止他们在数据集中总共得分 4

到目前为止我的代码是：

import pandas as pd

game_data = pd.read_csv('game_data.csv')

game_data['home_avg_home_games'] = game_data.groupby('home_id')['home_score'].transform(lambda x: x.rolling(165, min_periods = 0).mean())
game_data['home_avg_against_home_games'] = game_data.groupby('home_id')['away_score'].transform(lambda x: x.rolling(165, min_periods = 0).mean())

game_data['away_avg_away_games'] = game_data.groupby('away_id')['away_score'].transform(lambda x: x.rolling(165, min_periods = 0).mean())
game_data['away_avg_against_away_games'] = game_data.groupby('away_id')['home_score'].transform(lambda x: x.rolling(165, min_periods = 0).mean())

game_data['scored_home_total'] = game_data.groupby('home_id')['home_score'].cumsum()
game_data['scored_away_total'] = game_data.groupby('away_id')['away_score'].cumsum()

游戏_id	离开_id	home_id	客场得分	首页_分数	home_avg_home_games	主场平均对抗主场比赛	away_avg_away_games	away_avg_against_away_games
446877	138	134	1	4	4	1	1	4
446911	141	139	5	3	3	5	5	3
446873	121	118	3	4	4	3	3	4
446875	137	158	12	3	3	12	12	3
446872	142	110	2	3	3	2	2	3
446876	136	140	2	3	3	2	2	3
446874	143	113	2	6	6	2	2	6
446879	120	144	4	3	3	4	4	3
446871	119	135	15	0	0	15	15	0
446878	141	139	5	3	3	5	5	3
446869	115	109	10	5	5	10	10	5
446889	112	108	9	0	0	9	9	0
446868	145	133	4	3	3	4	4	3
446870	117	147	5	3	3	5	5	3
446867	111	114	6	2	2	6	6	2
446896	121	118	2	0	2	2.5	2.5	2
446910	138	134	5	6	5	3	3	5
446887	141	139	2	3	3	4	4	3
446883	116	146	8	7	7	8	8	7
446886	136	140	10	2	2.5	6	6	2.5
446885	137	158	2	1	2	7	7	2
446882	115	109	6	11	8	8	8	8
446880	112	108	6	1	0.5	7.5	7.5	0.5
446881	145	133	5	4	3.5	4.5	4.5	3.5
446884	119	135	3	0	0	9	9	0
446901	141	139	3	5	3.5	3.75	3.75	3.5
446898	137	158	3	4	2.666666667	5.666666667	5.666666667	2.666666667
446899	136	140	9	5	3.333333333	7	7	3.333333333
446891	115	109	4	3	6.333333333	6.666666667	6.666666667	6.333333333

我想要的输出是：

游戏_id	离开_id	home_id	客场得分	首页_分数	home_avg_for_home_games	主场平均对抗主场比赛	away_avg_away_games	away_avg_against_away_games	home_total_score	客场总分
446877	1	2	1	4	4	1	1	4	4	1
446911	2	3	5	3	3	5	5	3	3	5
446873	1	3	3	4	3.5	4	2	4	7	4

Answer 1

我创建了名为

total_score

的新列，总结了每个团队在

home_score

和

away_score

上的得分 将 pandas 导入为 pd

data = {
    'game_id': [446877, 446911, 446873, 446875, 446872, 446876, 446874, 446879, 446871, 446878, 446869, 446889, 446868, 446870, 446867, 446896, 446910, 446887, 446883, 446886, 446885, 446882, 446880, 446881, 446884, 446901, 446898, 446899, 446891],
    'away_id': [138, 141, 121, 137, 142, 136, 143, 120, 119, 141, 115, 112, 145, 117, 111, 121, 138, 141, 116, 136, 137, 115, 112, 145, 119, 141, 137, 136, 115],
    'home_id': [134, 139, 118, 158, 110, 140, 113, 144, 135, 139, 109, 108, 133, 147, 114, 118, 134, 139, 146, 140, 158, 109, 108, 133, 135, 139, 158, 140, 109],
    'away_score': [1, 5, 3, 12, 2, 2, 2, 4, 15, 5, 10, 9, 4, 5, 6, 2, 5, 3, 8, 10, 2, 6, 6, 5, 3, 3, 3, 9, 4],
    'home_score': [4, 3, 4, 3, 3, 3, 6, 3, 0, 3, 5, 0, 3, 3, 2, 0, 6, 3, 7, 2, 1, 11, 1, 4, 0, 5, 4, 5, 3] }

df = pd.DataFrame(data)

df['total_score'] = df.groupby('home_id')['home_score'].cumsum().fillna(0) + df.groupby('away_id')['away_score'].cumsum().fillna(0)

print(df)

最终输出是这样的：

    game_id  away_id  home_id  away_score  home_score  total_score
0    446877      138      134           1           4            5
1    446911      141      139           5           3            8
2    446873      121      118           3           4            7
3    446875      137      158          12           3           15
4    446872      142      110           2           3            5
5    446876      136      140           2           3            5
6    446874      143      113           2           6            8
7    446879      120      144           4           3            7
8    446871      119      135          15           0           15
9    446878      141      139           5           3           16
10   446869      115      109          10           5           15
11   446889      112      108           9           0            9
12   446868      145      133           4           3            7
13   446870      117      147           5           3            8
14   446867      111      114           6           2            8
15   446896      121      118           2           0            9
16   446910      138      134           5           6           16
17   446887      141      139           3           3           22
18   446883      116      146           8           7           15
19   446886      136      140          10           2           17
20   446885      137      158           2           1           18
21   446882      115      109           6          11           32
22   446880      112      108           6           1           16
23   446881      145      133           5           4           16
24   446884      119      135           3           0           18
25   446901      141      139           3           5           30
26   446898      137      158           3           4           25
27   446899      136      140           9           5           31
28   446891      115      109           4           3           39