1

我有一个 DataFrame 如下：

df = pd.DataFrame()
df['Team1']   = ['A','B','C','D','E','F','A','B','C','D','E','F']
df['Score1']  = [1,2,3,1,2,4,1,2,3,1,2,4]
df['Team2']   = ['U','V','W','X','Y','Z','U','V','W','X','Y','Z']
df['Score2']  = [2,1,2,2,3,3,2,1,2,2,3,3]
df['Match']   = df['Team1']  + ' Vs '+ df['Team2']
df['Match_no']= [1,2,3,4,5,6,1,2,3,4,5,6]
df['model']  = ['ELO','ELO','ELO','ELO','ELO','ELO','xG','xG','xG','xG','xG','xG']
winner = df.Score1>df.Score2
df['winner']  = np.where(winner,df['Team1'],df['Team2'])

我想做的是为下一阶段的锦标赛创建另一个日期框架。在下一阶段，我们将为每个模型（ELO 和 xG）进行 3 场比赛。我想按模型分组。这些比赛按模型分组，第 1 场和第 1 场比赛的获胜者，第 3 场比赛的获胜者对第 4 场比赛的获胜者等将进行比赛（即 U 对 B，C 对 X，Y 对 F）。那么谁能告诉我如何提取这些团队？

我预期的新数据框如下：

df1 =pd.DataFrame()
df1['Team1']   = ['U','C','Y','U','C','Y']
df1['Team2']   = ['B','X','F','B','X','F']

df1['Match']   = df1['Team1']  + ' Vs '+ df1['Team2']
df1['Match_no']= [1,2,3,1,2,3]
df1['model']  = ['ELO','ELO','ELO','xG','xG','xG']

我该如何设置？

4

2 回答 2

1

您可以使用GroupBy.cumcount每组计数：

df1 = pd.DataFrame()
df1['Team1'] = df.loc[::2, 'winner'].values
df1['Team2'] = df.loc[1::2, 'winner'].values
df1['Match'] = df1['Team1']  + ' Vs '+ df1['Team2']
model = df.loc[::2, 'model'].values
df1['Match_no'] = df1.groupby(model).cumcount() + 1
df1['model'] = model
print (df1)
  Team1 Team2   Match  Match_no model
0     U     B  U Vs B         1   ELO
1     C     X  C Vs X         2   ELO
2     Y     F  Y Vs F         3   ELO
3     U     B  U Vs B         1    xG
4     C     X  C Vs X         2    xG
5     Y     F  Y Vs F         3    xG
于 2018-06-29T07:07:57.630 回答
1

我会尽力给你一个答案，虽然我很难理解你所说的“奇数比赛号码和偶数比赛号码获胜”是什么意思。

如果这意味着匹配 1 和 2 的获胜者配对，那么 3 和 4 等等，你可以做一些简单的事情

df1['Team1'] = df.loc[::2, 'winner']
df1['Team2'] = df.loc[1::2, 'winner']

假设您的数据按呈现方式排序。你可以通过

df[df['model'] == 'ELO'].sort_values('Match_no')

如果我没听错的话，似乎不需要 pandas-groupby 等。

于 2018-06-29T06:23:36.200 回答