使用 Pandas 计算面对面的统计数据答案

【问题标题】：Calculate head-to-head statistics using Pandas使用 Pandas 计算面对面的统计数据
【发布时间】：2020-06-15 12:15:13
【问题描述】：

我有一个如下所示的 DataFrame：

  home_team  away_team  score  home_goals  away_goals  winner
1  Arsenal    Chelsea    3-0        3          0       Arsenal
2  ManCity    Arsenal    1-1        1          1       draw
3  Chelsea    Arsenal    2-1        2          1       Chelsea
4  Arsenal    Chelsea    5-5        5          5       draw
5  Arsenal    ManCity    1-2        1          2       ManCity

我的问题是：我如何计算阿森纳与另一支球队的胜负/对战记录？

潜在的期望结果可能如下所示：

   team      opponent  games_played  wins  draws  losses  goals_scored  goals_conceded
1  Arsenal   Chelsea        3          1     1      1          9              7
2  Arsenal   ManCity        2          0     1      1          2              3

非常感谢任何帮助。请注意，数据框不是真实的（以防有任何英超联赛专家潜伏）！

【问题讨论】：

标签： python python-3.x pandas dataframe pandas-groupby

【解决方案1】：

检查此代码：

import pandas as pd

df_in = pd.read_csv('data.csv')
df_out = pd.DataFrame(columns = ['team', 'opponent', 'games_played', 'wins', 'draws', 'losses', 'goals_scored', 'goals_conceded'])

team = 'Arsenal'

for index, row in df_in.iterrows():
    if row['home_team'] == team:
        opponent = row['away_team']
        if row['home_goals'] > row['away_goals']:
            win = 1
            draw = 0
            loss = 0
        elif row['home_goals'] < row['away_goals']:
            win = 0
            draw = 0
            loss = 1
        else:
            win = 0
            draw = 1
            loss = 0
        goals_scored = row['home_goals']
        goals_conceded = row['away_goals']
    else:
        opponent = row['home_team']
        if row['home_goals'] > row['away_goals']:
            win = 0
            draw = 0
            loss = 1
        elif row['home_goals'] < row['away_goals']:
            win = 1
            draw = 0
            loss = 0
        else:
            win = 0
            draw = 1
            loss = 0
        goals_scored = row['away_goals']
        goals_conceded = row['home_goals']

    games_played = 1



    if opponent not in df_out['opponent'].unique():
        match = pd.DataFrame({'team': team,
                              'opponent': opponent,
                              'games_played': games_played,
                              'wins': win,
                              'draws': draw,
                              'losses': loss,
                              'goals_scored': goals_scored,
                              'goals_conceded': goals_conceded},
                             index = [0])
        df_out = pd.concat([df_out, match], ignore_index = True)
    else:
        df_out.loc[df_out['opponent'] == opponent, 'games_played'] += games_played
        df_out.loc[df_out['opponent'] == opponent, 'wins'] += win
        df_out.loc[df_out['opponent'] == opponent, 'draws'] += draw
        df_out.loc[df_out['opponent'] == opponent, 'losses'] += loss
        df_out.loc[df_out['opponent'] == opponent, 'goals_scored'] += goals_scored
        df_out.loc[df_out['opponent'] == opponent, 'goals_conceded'] += goals_conceded

此代码会将您的数据加载为df_in，并使用您想要的数据创建一个df_out。
输出：

      team opponent games_played wins draws losses goals_scored goals_conceded
0  Arsenal  Chelsea            3    1     1      1            9              7
1  Arsenal  ManCity            2    0     1      1            2              3

【讨论】：

太棒了——谢谢安德里亚。您是否知道是否有可能对此进行进一步细分（例如 home_games_played、away_games_played、games_played ....、home_losses、away_losses）？
我认为最好再提出一个问题

【解决方案2】：

首先，您需要复制数据并翻转主/客队以获得您想要的团队/对手风格的统计数据。

这是因为您需要计算每场比赛两次，一次是获胜者，一次是失败者。复制 df 并翻转字段，然后使用 df.concat 将数据帧放在一起。

现在你可以聚合

您应该汇总主客场和获胜者。计算这一步中的行数和目标。使用df.groupby(dimensions).agg(metrics)

现在您需要将索引重置回 df 以便您可以再次使用获胜者列。使用df.reset_index(inplace=True) 来完成此操作。

一旦你有了这个，你创建新的列win, loss, draw'你将获胜者与 home_team 列或静态字符串“draw”进行比较

您现在可以再次聚合 df 并总结赢/输/平局列。

【讨论】：