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我在Python 2.7中有以下Pandas Dataframe。熊猫标准偏差返回NaN
CODE:
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()
这里是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
我期待执行GROUP_BY
,然后计算平均值和标准偏差。标准偏差是有时分组后计算1行 - 这意味着除以N-1
将有时给予除以0
这将打印NaN
。
这里是上面的代码的输出:
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
对于我在哪里执行GROUP_BY
超过1行的情况下,有一个方法来跳过标准偏差只是返回值本身。例如,我希望得到这样的:
所需的输出
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
是否有可能与大熊猫做到这一点?
编辑: 要创建精确的熊猫据帧以上,选择它,复制到剪贴板,然后使用此:
import pandas as pd
df = pd.read_clipboard()
print df
print df.groupby('Category').std()
您是如何计算您所需的输出中的类别D和F的值(例如,类别D的A列为0.403709)? – Alexander
请注意,'df.groupby('Category')。apply(np.std)'将std返回为'0.0',正如您所期望的那样。 – dhrumeel