Python 将groupby单独规范化

我想知道一个球员在总比赛中输赢的百分比。分析的列将为：

game | player | games
 aXa |  Jose  | has won
 aXb |  John  | has won
 aXb |  John  | has won
 uXu |  Adam  | lost
 bXb |  John  | lost
 oXo |  John  | lost
 pXp |  Jose  | has won

输出如下所示：

player | games | wins | losses
  John |   4   |  50% | 50%
  Jose |   2   | 100% | 0%
  Adam |   1   |   0% | 100%

用于计数器和与表连接的

：

在分组数据帧上使用
应用
的替代解决方案：

df_out = df.groupby('player')   \
           .apply(lambda x: pd.Series({'games': len(x.game),
                                       'wins': 100*sum(x.games == 'has won')/ len(x.game),
                                       'losses': 100*sum(x.games == 'lost')/ len(x.game)})) \
           .reset_index()

print(df_out)
   player  games   wins  losses
0  Adam      1.0    0.0   100.0
1  John      4.0   50.0    50.0
2  Jose      2.0  100.0     0.0

耸人听闻！我试着把输赢分开，然后恢复正常，但你这样做看起来很棒。非常感谢。我不明白normalize=0，你能解释一下吗？
df_out = df.groupby('player')   \
           .apply(lambda x: pd.Series({'games': len(x.game),
                                       'wins': 100*sum(x.games == 'has won')/ len(x.game),
                                       'losses': 100*sum(x.games == 'lost')/ len(x.game)})) \
           .reset_index()

print(df_out)
   player  games   wins  losses
0  Adam      1.0    0.0   100.0
1  John      4.0   50.0    50.0
2  Jose      2.0  100.0     0.0