函数在Pandas数据中的应用(“太多索引器”错误)

Question

我试图仅使用第一个k列(计算为mean作为.iloc[:,:-5])沿dataframe的axis=1计算axis=1和var，天真地说，我将运行如下：

df.groupby('id').agg([lambda x: x.iloc[:,:-5].mean(axis=1), lambda x: x.iloc[:,:-5].var(axis=1)])

但它会抛出“太多索引器”的错误。

编辑

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

Answer 1

看来你首先需要：

df['m'] = df.iloc[:,:-5].mean(axis=1)
df['v'] = df.iloc[:,:-5].var(axis=1)

如果有必要的话。