[英]Applying function to Pandas dataframe with groupby ('Too many indexers' error)
I am trying to compute mean
and var
along axis=1
of dataframe using only first k columns (compute as .iloc[:,:-5]
),naively, I would run as: 我试图仅使用前k列(计算为
.iloc[:,:-5]
)沿数据帧的axis=1
计算mean
和var
,但我会这样运行:
df.groupby('id').agg([lambda x: x.iloc[:,:-5].mean(axis=1), lambda x: x.iloc[:,:-5].var(axis=1)])
but it throws the 'too many indexers' error. 但它会引发“索引器过多”错误。
EDIT 编辑
Some data: 一些数据:
0 1 2 3 4 5 6 7 8 9 Q1 Q2 Q3 Q4 id
0 3.0 3.0 4.0 4.0 3.0 3.0 3.0 3.0 3.0 3.0 12.0 0.83 80.0 1.000 11.0
1 3.0 3.0 4.0 4.0 4.0 3.0 3.0 3.0 3.0 3.0 14.0 1.60 80.0 1.000 11.0
2 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 5.0 13.0 1.40 75.0 1.000 11.0
3 3.0 3.0 4.0 4.0 4.0 3.0 3.0 3.0 3.0 3.0 12.0 0.50 80.0 0.848 11.0
4 3.0 4.0 4.0 4.0 7.0 7.0 5.0 4.0 4.0 2.0 13.0 1.74 70.0 0.883 11.0
13 3.0 3.0 2.0 2.0 2.0 2.0 3.0 2.0 3.0 3.0 12.0 3.67 45.0 1.000 14.0
14 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 13.0 3.67 48.0 0.848 14.0
15 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 12.0 1.67 70.0 0.848 14.0
16 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 NaN 2.0 12.0 3.33 60.0 0.848 14.0
17 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 12.0 3.33 60.0 0.848 14.0
25 4.0 4.0 6.0 5.0 NaN 6.0 4.0 3.0 NaN 4.0 11.0 3.36 85.0 0.796 17.0
26 4.0 5.0 4.0 7.0 6.0 5.0 4.0 6.0 7.0 5.0 8.0 4.76 50.0 0.725 17.0
27 4.0 4.0 3.0 4.0 5.0 4.0 5.0 3.0 3.0 5.0 9.0 3.33 50.0 0.725 17.0
28 3.0 4.0 4.0 3.0 4.0 4.0 NaN 3.0 NaN 3.0 10.0 3.12 75.0 0.725 17.0
29 3.0 3.0 2.0 NaN 2.0 1.0 NaN NaN 1.0 2.0 15.0 3.05 79.0 0.725 17.0
39 3.0 3.0 5.0 4.0 4.0 4.0 4.0 4.0 NaN 5.0 12.0 1.19 80.0 0.883 18.0
40 5.0 4.0 5.0 5.0 5.0 5.0 4.0 5.0 7.0 4.0 9.0 1.83 75.0 0.883 18.0
41 5.0 6.0 4.0 4.0 4.0 4.0 4.0 4.0 7.0 7.0 12.0 1.71 35.0 1.000 18.0
42 5.0 5.0 5.0 5.0 4.0 NaN 4.0 4.0 3.0 2.0 13.0 0.86 85.0 1.000 18.0
43 3.0 3.0 3.0 3.0 3.0 3.0 3.0 5.0 3.0 3.0 11.0 1.36 75.0 1.000 18.0
48 1
It seems you need first: 看来您首先需要:
df['m'] = df.iloc[:,:-5].mean(axis=1)
df['v'] = df.iloc[:,:-5].var(axis=1)
and then aggregate if necesary. 然后根据需要进行汇总。
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