pandas vs numpy中的不同std [英] Different std in pandas vs numpy

问题描述

熊猫和numpy之间的标准差有所不同.为什么以及哪一个是正确的? (相对差异为3.5％，不应四舍五入，在我看来，这是很高的水平.)

The standard deviation differs between pandas and numpy. Why and which one is the correct one? (the relative difference is 3.5% which should not come from rounding, this is high in my opinion).

示例

import numpy as np
import pandas as pd
from StringIO import StringIO

a='''0.057411
0.024367
 0.021247
-0.001809
-0.010874
-0.035845
0.001663
0.043282
0.004433
-0.007242
0.029294
0.023699
0.049654
0.034422
-0.005380'''


df = pd.read_csv(StringIO(a.strip()), delim_whitespace=True, header=None)

df.std()==np.std(df) # False
df.std() # 0.025801
np.std(df) # 0.024926

(0.024926 - 0.025801) / 0.024926 # 3.5% relative difference

我使用以下版本:

熊猫: '0.14.0' numpy的: '1.8.1'

pandas: '0.14.0' numpy: '1.8.1'

推荐答案

简而言之，都不是不正确的".熊猫使用无偏估计量(分母中的N-1)，而Numpy默认情况下不使用.

In a nutshell, neither is "incorrect". Pandas uses the unbiased estimator (N-1 in the denominator), whereas Numpy by default does not.

要使它们的行为相同，请将ddof=1传递给 numpy.std() .

To make them behave the same, pass ddof=1 to numpy.std().

有关进一步的讨论，请参见

For further discussion, see

• 有人可以解释偏差/无偏差的总体/样本标准偏差吗?
• 人口差异和样本差异.
• 为什么要除以n-1?

• Can someone explain biased/unbiased population/sample standard deviation?
• Population variance and sample variance.
• Why divide by n-1?

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