Question

我有一个这样的数据集，其中每一行代表gameID指定的特定匹配中的一个。

  gameID          Won/Lost   Home   Away  metric2 metric3 metric4   team1 team2 team3 team4
2017020001         1          1      0      10      10      10      1     0     0      0
2017020001         0          0      1      10      10      10      0     1     0      0

我要做的是创建一个函数，该函数将具有相同gameID的行合并起来。正如您在下面的数据示例中看到的那样，这两行代表一个比赛，该比赛分为一个主队（row_1）和一个客队（row_2）。我希望这两行只能坐在一行上。

Won/Lost  h_metric2 h_metric3 h_metric4 a_metric2 a_metric3 a_metric4 h_team1 h_team2 h_team3 h_team4 a_team1 a_team2 a_team3 a_team4
1            10       10         10        10         10        10      1       0        0      0         0      1        0      0

如何获得此结果？

编辑：我造成了太多混乱，发布了我的代码，因此您可以更好地了解我要解决的问题。

import numpy as np
import pandas as pd
import requests
import json
from sklearn import preprocessing
from sklearn.preprocessing import OneHotEncoder

results = []
for game_id in range(2017020001, 2017020010, 1):
    url = 'https://statsapi.web.nhl.com/api/v1/game/{}/boxscore'.format(game_id)
r = requests.get(url)
game_data = r.json()

for homeaway in ['home','away']:

    game_dict = game_data.get('teams').get(homeaway).get('teamStats').get('teamSkaterStats')
    game_dict['team'] = game_data.get('teams').get(homeaway).get('team').get('name')
    game_dict['homeaway'] = homeaway
    game_dict['game_id'] = game_id
    results.append(game_dict)

df = pd.DataFrame(results)

df['Won/Lost'] = df.groupby('game_id')['goals'].apply(lambda g: (g == g.max()).map({True: 1, False: 0}))

df["faceOffWinPercentage"] = df["faceOffWinPercentage"].astype('float')
df["powerPlayPercentage"] = df["powerPlayPercentage"].astype('float')
df["team"] = df["team"].astype('category')
df = pd.get_dummies(df, columns=['homeaway'])
df = pd.get_dummies(df, columns=['team'])

Answer 1

我只是假设，您正在使用面包和黄油： numpy，pandas＆co？

如果是这样，我还假设您的表当前存储在名为'df'的pandas.DataFrame-instance中：

将您的df分为两个df，然后将它们加入：

df_team1 = df[df['Won/Lost']==1]
df_team2 = df[df['Won/Lost']==0]
final_df = df_team1.join(df_team2, lsuffix='_team1', rsuffix='_team2', on='gameID')

您当然可以编辑它，以更好地满足您的目的。例如，根据“居家/离开”列等创建df。

BR 本：]

Answer 2

这是基于以下假设：每个gameID恰好有两行，并且您要根据该ID进行分组。（它也假设我理解这个问题。）

改进的解决方案

给出一个数据帧df，例如

       gameID  Won/Lost  Home  Away  metric2  metric3  metric4  team1  team2  team3  team4
0  2017020001         1     1     0       10       10       10      1      0      0      0
1  2017020001         0     0     1       10       10       10      0      1      0      0
2  2017020002         1     1     0       10       10       10      1      0      0      0
3  2017020002         0     0     1       10       10       10      0      1      0      0

您可以像这样使用pd.merge（以及一些数据处理）：

>>> is_home = df['Home'] == 1                                                                                                                                                                                                                   
>>> home = df[is_home].drop(['Home', 'Away'], axis=1).add_prefix('h_').rename(columns={'h_gameID':'gameID'})                                                                                                                                    
>>> away = df[~is_home].drop(['Won/Lost', 'Home', 'Away'], axis=1).add_prefix('a_').rename(columns={'a_gameID':'gameID'})                                                                                                                       
>>> pd.merge(home, away, on='gameID')                                                                                                                                                                                                           
       gameID  h_Won/Lost  h_metric2  h_metric3  h_metric4  h_team1  h_team2  h_team3  h_team4  a_metric2  a_metric3  a_metric4  a_team1  a_team2  a_team3  a_team4
0  2017020001           1         10         10         10        1        0        0        0         10         10         10        0        1        0        0
1  2017020002           1         10         10         10        1        0        0        0         10         10         10        0        1        0        0

（我保留了Won/Lost的前缀，因为它表明这是主队的统计数据。此外，如果有人知道如何更优雅地添加前缀而不必重命名gameID请发表评论。）

原始尝试

您可以在分组后应用以下功能

def munge(group): 
     is_home = group.Home == 1 
     wonlost = group.loc[is_home, 'Won/Lost'].reset_index(drop=True) 
     group = group.loc[:, 'metric2':] 
     home = group[is_home].add_prefix('h_').reset_index(drop=True) 
     away = group[~is_home].add_prefix('a_').reset_index(drop=True) 
     return pd.concat([wonlost, home, away], axis=1)

...像这样

>>> df.groupby('gameID').apply(munge).reset_index(level=1, drop=True)                                                                                                                                                                           
            Won/Lost  h_metric2  h_metric3  h_metric4  h_team1  h_team2  h_team3  h_team4  a_metric2  a_metric3  a_metric4  a_team1  a_team2  a_team3  a_team4
gameID                                                                                                                                                        
2017020001         1         10         10         10        1        0        0        0         10         10         10        0        1        0        0
2017020002         1         10         10         10        1        0        0        0         10         10         10        0        1        0        0

如何根据列值合并数据框中的行？

2 个答案: