Question

我正在建立一个曲棍球比赛的数据集，需要根据'Game_id'和'Goals'列确定一支球队是赢还是输。每个游戏都有自己的ID，并跨越两行，因此，在2000行中存储了1000个游戏。

我的数据框如下：

Team  Home/Away  Goals  Game_id
CAL   Home       7      2017020001
PHY   Away       4      2017020001
CAP   Home       7      2017020002
WILD  Away       4      2017020002

我需要一个基于特定'Won/Lost'的目标的健身列'Game_id'。我正在努力创建一个为我做到这一点的循环。

我正在寻找的结果是：

Team  Home/Away  Goals  Game_id      Won/Lost
CAL   Home       7      2017020001   Won
PHY   Away       4      2017020001   Lost
CAP   Home       7      2017020002   Won
WILD  Away       4      2017020002   Lost

Answer 1

给出

>>> df                                                                                                                  
   Team Home/Away  Goals     Game_id
0   CAL      Home      7  2017020001
1   PHY      Away      4  2017020001
2   CAP      Home      7  2017020002
3  WILD      Away      4  2017020002
4  WILD      Away      1  2017020003
5   CAP      Home      1  2017020003

我要编写以下函数：

def win_loss_draw(group): 
     group = group == group.max() 
     if group.all(): 
         group[:] = 'Draw' 
     else: 
         group = group.map({True: 'Won', False: 'Lost'}) 
     return group

...并像这样应用它：

>>> df['Won/Lost'] = df.groupby('Game_id')['Goals'].apply(win_loss_draw)                                                
>>> df                                                                                                                  
   Team Home/Away  Goals     Game_id Won/Lost
0   CAL      Home      7  2017020001      Won
1   PHY      Away      4  2017020001     Lost
2   CAP      Home      7  2017020002      Won
3  WILD      Away      4  2017020002     Lost
4  WILD      Away      1  2017020003     Draw
5   CAP      Home      1  2017020003     Draw

考虑到冰球比赛只能在常规时间内以平局结束，因此我不考虑平局，但是我的数据随着时间的推移而变化，所以只有输赢

在这种情况下，发出就足够了

df['Won/Lost'] = df.groupby('Game_id')['Goals'].apply(lambda g: (g == g.max()).map({True: 'Won', False: 'Lost'}))

（这是版本1）

〜编辑〜

性能改进！

版本2：

is_winner = df.groupby('Game_id')['Goals'].transform('max') == df['Goals']
df['Won/Lost'] = is_winner.map({True: 'Won', False: 'Lost'})

版本3：

is_winner = df.groupby('Game_id')['Goals'].transform('max') == df['Goals']                                          
df['Won/Lost'] = np.where(is_winner.values, 'Won', 'Lost')

时间：

# Setup
>>> df = pd.concat([df]*1000, ignore_index=True)                                                                        
>>> df['Game_id'] = np.arange(len(df)//2).repeat(2)                                                                     
>>>                                                                                                                     
>>> df                                                                                                                  
      Team Home/Away  Goals  Game_id
0      CAL      Home      7        0
1      PHY      Away      4        0
2      CAP      Home      7        1
3     WILD      Away      4        1
4      CAL      Home      7        2
...    ...       ...    ...      ...
3995  WILD      Away      4     1997
3996   CAL      Home      7     1998
3997   PHY      Away      4     1998
3998   CAP      Home      7     1999
3999  WILD      Away      4     1999

# Timings (i5-6200U CPU @ 2.30GHz, only relative times are important though)
>>> %timeit df.groupby('Game_id')['Goals'].apply(lambda g: (g == g.max()).map({True: 'Won', False: 'Lost'})) # Version 1            
1.73 s ± 13.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
>>> %timeit (df.groupby('Game_id')['Goals'].transform('max') == df['Goals']).map({True: 'Won', False: 'Lost'}) # Version 2          
2.38 ms ± 37.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
>>> %timeit np.where((df.groupby('Game_id')['Goals'].transform('max') == df['Goals']).values, 'Won', 'Lost') # Version 3            
1.53 ms ± 6.19 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

根据熊猫得分栏确定游戏的获胜者

1 个答案: