我有一张由2名球员互相竞争的桌子:
date plA plB ptsA ptsB
0 01/01/2013 Jeff Tom 78 72
1 15/01/2013 Jeff Tom 52 67
2 01/02/2013 Tom Jeff 91 93
3 15/02/2013 Jeff Tom 83 87
4 01/03/2013 Tom Jeff 65 76
我想应用展开式平均值,以便每个玩家的ptsA
和ptsB
计入(并且不会留给)净结果。最终输出应该更清楚:
date plA plB ptsA ptsB meanA meanB
0 01/01/2013 Jeff Tom 78 72 78 72 # init mean
1 15/01/2013 Jeff Tom 52 67 65 69.5
2 01/02/2013 Tom Jeff 91 93 74.3 76.6 # Tom: (72+67+91)/3, Jeff: (78+52+93)/3
3 15/02/2013 Jeff Tom 83 87 76.5 79.25 # Jeff: (78+52+93+83)/4, Tom: (72+67+91+87)/4
4 01/03/2013 Tom Jeff 65 76 76.4 76.4 # Tom: (72+67+91+87+65)/5, Jeff: (78+52+93+83+76)/5
现在,我开始按plA
对数据进行分组,如下所示:
by_A = players.sort(columns='date').groupby('plA')
players['meanA'] = by_A['ptsA'].apply(pd.expanding_mean)
players['meanB'] = by_A['ptsB'].apply(pd.expanding_mean)
显然,我需要做同样的事情,groupby('plB')
然后我正在绘制一个空白如何正确加入这两个结果。
也许熊猫提供了内置功能或者您有解决方案吗?
@EDIT Saullo Castro的解决方案,数据略有不同
date studentA studentB scoreA scoreB meanJeff meanTom meanMaggie
0 2013-01-01 Jeff Tom 78 72 78.000000 72.000000 0.000000
1 2013-01-15 Jeff Maggie 52 67 65.000000 36.000000 33.500000
2 2013-02-01 Tom Jeff 91 93 74.333333 54.333333 22.333333
3 2013-02-15 Jeff Tom 83 87 76.500000 62.500000 16.750000
4 2013-03-01 Tom Jeff 65 76 76.400000 63.000000 13.400000
Maggie的意思应该一直保持67
。
答案 0 :(得分:1)
(请参阅下面的固定解决方案)
一种方法是首先找出所有玩家的名字:
names = pd.concat((df.plA, df.plB)).unique()
然后创建一个新列,每个玩家的扩展均值:
for name in names:
df['mean'+name] = pd.expanding_mean(df.ptsA*(df.plA==name) + df.ptsB*(df.plB==name))
导致:
date plA plB ptsA ptsB meanJeff meanTom
0 2013-01-01 00:00:00 Jeff Tom 78 72 78.000000 72.000000
1 15/01/2013 Jeff Tom 52 67 65.000000 69.500000
2 2013-01-02 00:00:00 Tom Jeff 91 93 74.333333 76.666667
3 15/02/2013 Jeff Tom 83 87 76.500000 79.250000
4 2013-01-03 00:00:00 Tom Jeff 65 76 76.400000 76.400000
编辑:修复解决方案:
对于两个以上的名称,您可以使用以下方法构建扩展均值的公式:
df = pd.read_excel('stack.xlsx', 'tabelle1')
names = pd.concat((df.plA, df.plB)).unique()
for name in names:
nA = df.plA==name
nB = df.plB==name
df['mean'+name] = np.cumsum(df.ptsA*nA + df.ptsB*nB)/np.maximum(1.,
np.cumsum(1.*np.logical_or(nA, nB)))
导致:
date plA plB ptsA ptsB meanJeff meanTom meanMaggie
0 2013-01-01 00:00:00 Jeff Tom 78 72 78.000000 72.000000 0
1 2013-01-15 00:00:00 Jeff Maggie 52 67 65.000000 72.000000 67
2 2013-02-01 00:00:00 Tom Jeff 91 93 74.333333 81.500000 67
3 2013-02-15 00:00:00 Jeff Tom 83 87 76.500000 83.333333 67
4 2013-03-01 00:00:00 Tom Jeff 65 76 76.400000 78.750000 67