我想计算足球队的积分。我有每场比赛的积分，我得到了本垒打或客场积分的总和。我不知道如何获得每支球队的总积分（主场+客场积分）
这是我到目前为止所拥有的：
df = pd.DataFrame([
["Gothenburg", "Malmo", 2018, 1, 1],
["Malmo","Gothenburg", 2018, 1, 1],
["Malmo", "Gothenburg", 2018, 0, 3],
["Gothenburg", "Malmo", 2018, 1, 1],
["Gothenburg", "Malmo" ,2018, 0, 3],
["Gothenburg", "Malmo", 2018, 1, 1],
["Gothenburg", "Malmo", 2018, 0, 3],
["Malmo", "Gothenburg", 2018, 0, 3],
["Gothenburg", "Malmo", 2018, 1, 1],
["Malmo", "Gothenburg", 2018, 0, 3],
[ "Malmo","Gothenburg", 2018, 1, 1],
[ "Malmo", "Gothenburg",2018, 0, 3],
])
df.columns = ['H_team', 'A_team', "Year", 'H_points', 'A_points']
# Cumulaive sum for home/ away team with shift 1 row
df["H_cumsum"] = df.groupby(['H_team', "Year"])['H_points'].transform(
lambda x: x.cumsum().shift())
df["A_cumsum"] = df.groupby(['A_team', "Year"])['A_points'].transform(
lambda x: x.cumsum().shift())
print(df)
H_team A_team Year H_points A_points H_cumsum A_cumsum
0 Gothenburg Malmo 2018 1 1 NaN NaN
1 Malmo Gothenburg 2018 1 1 NaN NaN
2 Malmo Gothenburg 2018 0 3 1.0 1.0
3 Gothenburg Malmo 2018 1 1 1.0 1.0
4 Gothenburg Malmo 2018 0 3 2.0 2.0
5 Gothenburg Malmo 2018 1 1 2.0 5.0
6 Gothenburg Malmo 2018 0 3 3.0 6.0
7 Malmo Gothenburg 2018 0 3 1.0 4.0
8 Gothenburg Malmo 2018 1 1 3.0 9.0
9 Malmo Gothenburg 2018 0 3 1.0 7.0
10 Malmo Gothenburg 2018 1 1 1.0 10.0
11 Malmo Gothenburg 2018 0 3 2.0 11.0
此表为我提供了每支球队的累计出发点和出站点的信息，移动了1行。但是我需要主场和客场比赛的总得分。 H_cumsum和A_cumsum应该添加主场和客场比赛的先前得分。
所需的输出：
row 0: Malmo = NaN, Gothenburg = NaN
row 1: Gothenburg = 1, Malmo = 1
row 2: Malmo = 1 + 1 = 2, Gothenburg = 1 + 1 = 2
row 3: Gothenburg = 1 + 1 + 3 = 5, Malmo = 1 + 1 + 0 = 2
row 4: Gothenburg = 1 + 1 + 3 + 1 = 6, Malmo = 1 + 1 + 0 + 1 = 3
And so on...
最后11行应为：
H_cumsum (team Malmo) = 12 H_cumsum (team Gothenburg) = 15
答案 0 :(得分：0)
这对我来说似乎还可以。这有点长。
df.columns = ['H_team', 'A_team', "Year", 'H_points', 'A_points']
# H_team cumsum() for science.
df['H_cumsum'] = df[['H_team', 'H_points']].groupby(['H_team']).cumsum()
# A_team cumsum() for more science.
df['A_cumsum'] = df[['A_team', 'A_points']].groupby(['A_team']).cumsum()
# Creating a column for the sum of the two, or total points scored by either side.
df['T_sum'] = df['H_points'] + df['A_points']
# Creating the cumsum() column for T_sum
df['T_cumsum'] = df['T_sum'].cumsum()
print(df)
答案 1 :(得分：0)
我找到了一种使用堆栈的解决方案，但这不是一个好方法：
df = pd.DataFrame([
["Gothenburg", "Malmo", 2018, 1, 1],
["Malmo","Gothenburg", 2018, 1, 1],
["Malmo", "Gothenburg", 2018, 0, 3],
["Gothenburg", "Malmo", 2018, 1, 1],
["Gothenburg", "Malmo" ,2018, 0, 3],
["Gothenburg", "Malmo", 2018, 1, 1],
["Gothenburg", "Malmo", 2018, 0, 3],
["Malmo", "Gothenburg", 2018, 0, 3],
["Gothenburg", "Malmo", 2018, 1, 1],
["Malmo", "Gothenburg", 2018, 0, 3],
[ "Malmo","Gothenburg", 2018, 1, 1],
[ "Malmo", "Gothenburg",2018, 0, 3],
])
df.columns = [['Team', 'Team', "Year", 'Points', 'Points'],
['Home', 'Away', 'Year', 'Home', 'Away']]
d1 = df.stack()
total = d1.groupby('Team').Points.apply(lambda x: x.shift().cumsum())
df = d1.assign(Total=total).unstack()
print(df)
Points Team Year Total
Away Home Year Away Home Year Away Home Year Away Home Year
0 1.0 1.0 NaN Malmo Gothenburg NaN NaN NaN 2018.0 NaN NaN NaN
1 1.0 1.0 NaN Gothenburg Malmo NaN NaN NaN 2018.0 1.0 1.0 NaN
2 3.0 0.0 NaN Gothenburg Malmo NaN NaN NaN 2018.0 2.0 2.0 NaN
3 1.0 1.0 NaN Malmo Gothenburg NaN NaN NaN 2018.0 2.0 5.0 NaN
4 3.0 0.0 NaN Malmo Gothenburg NaN NaN NaN 2018.0 3.0 6.0 NaN
5 1.0 1.0 NaN Malmo Gothenburg NaN NaN NaN 2018.0 6.0 6.0 NaN
6 3.0 0.0 NaN Malmo Gothenburg NaN NaN NaN 2018.0 7.0 7.0 NaN
7 3.0 0.0 NaN Gothenburg Malmo NaN NaN NaN 2018.0 7.0 10.0 NaN
8 1.0 1.0 NaN Malmo Gothenburg NaN NaN NaN 2018.0 10.0 10.0 NaN
9 3.0 0.0 NaN Gothenburg Malmo NaN NaN NaN 2018.0 11.0 11.0 NaN
10 1.0 1.0 NaN Gothenburg Malmo NaN NaN NaN 2018.0 14.0 11.0 NaN
11 3.0 0.0 NaN Gothenburg Malmo NaN NaN NaN 2018.0 15.0 12.0 NaN
总计/离开和总计/家庭下的点是正确的。但是，该表变得很难用所有多余的列进行概览。 （在此示例中，我没有为每行显示另外10列，所以这真是一团糟。）
所需的输出是：
H_team A_team Year H_points A_points H_cumsum A_cumsum
0 Gothenburg Malmo 2018 1 1 NaN NaN
1 Malmo Gothenburg 2018 1 1 1.0 1.0
2 Malmo Gothenburg 2018 0 3 2.0 2.0
3 Gothenburg Malmo 2018 1 1 5.0 2.0
4 Gothenburg Malmo 2018 0 3 6.0 3.0
5 Gothenburg Malmo 2018 1 1 6.0 6.0
6 Gothenburg Malmo 2018 0 3 7.0 7.0
7 Malmo Gothenburg 2018 0 3 10.0 7.0
8 Gothenburg Malmo 2018 1 1 10.0 10.0
9 Malmo Gothenburg 2018 0 3 11.0 11.0
10 Malmo Gothenburg 2018 1 1 11.0 14.0
11 Malmo Gothenburg 2018 0 3 12.0 15.0