Pandas Standard Deviation返回NaN

Question

我在Python 2.7中有以下Pandas Dataframe.

码:

import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.rand(10,6),columns=list('ABCDEF'))
df.insert(0,'Category',['A','C','D','D','B','E','F','F','G','H'])
print df.groupby('Category').std()

这是df:

Category         A         B         C         D         E         F
       A  0.500200  0.791039  0.498083  0.360320  0.965992  0.537068
       C  0.295330  0.638823  0.133570  0.272600  0.647285  0.737942
       D  0.912966  0.051288  0.055766  0.906490  0.078384  0.928538
       D  0.416582  0.441684  0.605967  0.516580  0.458814  0.823692
       B  0.714371  0.636975  0.153347  0.936872  0.000649  0.692558
       E  0.639271  0.486151  0.860172  0.870838  0.831571  0.404813
       F  0.375279  0.555228  0.020599  0.120947  0.896505  0.424233
       F  0.952112  0.299520  0.150623  0.341139  0.186734  0.807519
       G  0.384157  0.858391  0.278563  0.677627  0.998458  0.829019
       H  0.109465  0.085861  0.440557  0.925500  0.767791  0.626924

我希望执行a GROUP_BY然后计算平均值和标准差.标准偏差有时在分组超过1行后计算 - 这意味着除以N-1将有时会给出0将要打印的除法NaN.

以下是上述代码的输出:

OUTPUT:

                A         B         C         D         E         F
Category                                                            
A              NaN       NaN       NaN       NaN       NaN       NaN
B              NaN       NaN       NaN       NaN       NaN       NaN
C              NaN       NaN       NaN       NaN       NaN       NaN
D         0.350996  0.276052  0.389051  0.275708  0.269004  0.074137
E              NaN       NaN       NaN       NaN       NaN       NaN
F         0.407882  0.180813  0.091941  0.155699  0.501884  0.271025
G              NaN       NaN       NaN       NaN       NaN       NaN
H              NaN       NaN       NaN       NaN       NaN       NaN

对于我执行GROUP_BY超过1行的情况,有没有办法跳过标准偏差并只返回值本身.例如,我希望得到这个:

期望的输出

                 A         B         C         D         E         F
Category                                                            
A         0.500200  0.791039  0.498083  0.360320  0.965992  0.537068
B         0.714371  0.636975  0.153347  0.936872  0.000649  0.692558
C         0.295330  0.638823  0.133570  0.272600  0.647285  0.737942
D         0.350996  0.276052  0.389051  0.275708  0.269004  0.074137
E         0.639271  0.486151  0.860172  0.870838  0.831571  0.404813
F         0.407882  0.180813  0.091941  0.155699  0.501884  0.271025
G         0.384157  0.858391  0.278563  0.677627  0.998458  0.829019
H         0.109465  0.085861  0.440557  0.925500  0.767791  0.626924

是不是可以用熊猫做到这一点？

编辑:要在上面创建精确的Pandas Dataframe,选择它,复制到剪贴板,然后使用:

import pandas as pd
df = pd.read_clipboard(index_col='Category')
print df
print df.groupby('Category').std()

Answer 1

不完全是问题中的问题,但是如果你想避免使用NaN数值,用0 自由度计算人口标准差(即),除以:std(ddof=0)N

>>> print(df.groupby('Category').std(ddof=0))
                 A         B         C         D         E         F
Category                                                            
A         0.000000  0.000000  0.000000  0.000000  0.000000  0.000000
B         0.000000  0.000000  0.000000  0.000000  0.000000  0.000000
C         0.000000  0.000000  0.000000  0.000000  0.000000  0.000000
D         0.248192  0.195198  0.275101  0.194955  0.190215  0.052423
E         0.000000  0.000000  0.000000  0.000000  0.000000  0.000000
F         0.288417  0.127854  0.065012  0.110096  0.354885  0.191643
G         0.000000  0.000000  0.000000  0.000000  0.000000  0.000000
H         0.000000  0.000000  0.000000  0.000000  0.000000  0.000000

零表示没有方差,因为组中只有一个值或所有相同的值.

(注意,对于默认ddof为numpy.var是零,因此比熊猫的1默认不同).

Answer 2

您可以fillna替换缺失的值 -DataFrame使用每个组的最后一个值传入 a 。

In [86]: (df.groupby('Category').std()
    ...:    .fillna(df.groupby('Category').last()))

Out[86]: 
                 A         B         C         D         E         F
Category                                                            
A         0.500200  0.791039  0.498083  0.360320  0.965992  0.537068
B         0.714371  0.636975  0.153347  0.936872  0.000649  0.692558
C         0.295330  0.638823  0.133570  0.272600  0.647285  0.737942
D         0.350996  0.276052  0.389051  0.275708  0.269005  0.074137
E         0.639271  0.486151  0.860172  0.870838  0.831571  0.404813
F         0.407883  0.180813  0.091941  0.155699  0.501884  0.271024
G         0.384157  0.858391  0.278563  0.677627  0.998458  0.829019
H         0.109465  0.085861  0.440557  0.925500  0.767791  0.626924