在分组中各选择一个

Question

我正在尝试创建所有可能的球员配对组合，根据A、B、C或D型障碍分配到4人高尔夫球队

我尝试过各种itertools方法，比如组合和排列，但都没有找到正确的方法。你知道吗

from itertools import combinations, product, permutations
g = player_df.groupby(by = 'hcp_ABCD')
teams_listoflists = [group[1].index for group in g]
teams_combo_ndx = [player for player in permutations(teams_listoflists, 4)]

这是我的熊猫桌：

        handicap      name hcp_ABCD
0         24   Player1        D
1         21   Player2        D
2          8   Player3        B
3         14   Player4        C
4         20   Player5        D
5         13   Player6        C
6         -1   Player7        A
7          5   Player8        A
8          8   Player9        B
9          6  Player10        B
10        20  Player11        D
11        15  Player12        C
12         0  Player13        A
13        12  Player14        C
14         0  Player15        A
15        10  Player16        B

我想输出的所有组合（没有重复）的球员组合（队），这样每个队有一个类型a，B，C，和D的每一个。此输出可以是一个类似于上面按“选项”分组的表

编辑： 为了清楚起见，我添加了这个输出示例。你知道吗

                       A Player     B Player     C Player   D Player
    option 1  team1    Player7      Player3      Player4    Player1
              team2    Player8      Player9      Player6    Player2
              team3    Player13     Player10     Player12   Player5
              team4    Player15     Player16     Player14   Player11

    option 2  team1    Player7      Player16     Player4    Player1
              team2    Player8      Player3      Player6    Player2
              team3    Player13     Player9      Player12   Player5
              team4    Player15     Player10     Player14   Player11

    ...


                       A Player     B Player     C Player   D Player
    option n  team1    Player7      Player3      Player4    Player11
              team2    Player8      Player9      Player6    Player1
              team3    Player13     Player10     Player12   Player2
              team4    Player15     Player16     Player14   Player5

上面的观点是，我试图找到一个生成器，它循环遍历每个残障组中所有球员的组合，这样就可以清楚地看到球队的选择组合。你知道吗

编辑#2 我已确定此代码生成所有潜在团队组合的组合：

g = df.groupby(by = 'hcp_ABCD')
combinations = [list(group[1].index) for group in g]

这将创建一个列表，其中a个玩家在列表[0]中，B个玩家在列表[1]中，等等

这会得到所有可能的团队组合的索引器：

from itertools import product
options = [option for option in product(*combinations)]

但是，如何将这些分配到“选项”（见上面的例子）中并确保没有重复是我一直坚持的。你知道吗

编辑#3更简单的版本（思考此问题的方式）是使用以下集合：

A = ['A1', 'A2', 'A3', 'A4']
B = ['B1', 'B2', 'B3', 'B4']
C = ['C1', 'C2', 'C3', 'C4']
D=  ['D1', 'D2', 'D3', 'D4']

这基本上与groupby在上面所做的相同（按hcp\u ABCD分组），但将每个人命名为“A Player”、“B Player”等

可能的团队组合：

team_combinations = [team for team in product(A, B, C, D)]

然后，下一个技巧是将这些分配到4个团队的组合中，没有重复的玩家。你知道吗

Answer 1

我在评论中提出了一个建议。下面是一个实现：

import pandas as pd
from functools import reduce

data = [
    (24,'Player1','D'),
    (21,'Player2','D'),
    (8,'Player3','B'),
    (8,'Player4','B'),
    (14,'Player5','C'),
    (13,'Player6','C'),
    (-1,'Player7','A'),
    (5,'Player8','A')
]
df = pd.DataFrame(
    data,
    columns=['handicap', 'name', 'hcp_ABCD']
)

dfs = [
    grp_df.drop(columns="hcp_ABCD")
          .rename(columns={"name": f"player_{hndcp}",
                           "handicap": f"handicap_{hndcp}"})
    for hndcp, grp_df in df.assign(key=1)
                           .groupby("hcp_ABCD")
]
result = reduce(
    lambda left, right: left.merge(right, how="outer", on="key"),
    dfs
).drop(columns="key")
print(result)

输出：

    handicap_A player_A  handicap_B player_B  handicap_C player_C  handicap_D player_D
0           -1  Player7           8  Player3          14  Player5          24  Player1
1           -1  Player7           8  Player3          14  Player5          21  Player2
2           -1  Player7           8  Player3          13  Player6          24  Player1
3           -1  Player7           8  Player3          13  Player6          21  Player2
4           -1  Player7           8  Player4          14  Player5          24  Player1
5           -1  Player7           8  Player4          14  Player5          21  Player2
6           -1  Player7           8  Player4          13  Player6          24  Player1
7           -1  Player7           8  Player4          13  Player6          21  Player2
8            5  Player8           8  Player3          14  Player5          24  Player1
9            5  Player8           8  Player3          14  Player5          21  Player2
10           5  Player8           8  Player3          13  Player6          24  Player1
11           5  Player8           8  Player3          13  Player6          21  Player2
12           5  Player8           8  Player4          14  Player5          24  Player1
13           5  Player8           8  Player4          14  Player5          21  Player2
14           5  Player8           8  Player4          13  Player6          24  Player1
15           5  Player8           8  Player4          13  Player6          21  Player2