numpy.std和excel STDEV函数之间有什么区别吗?

问题描述

我有一个列表:

s = [0.995537725, 0.994532199, 0.996027983, 0.999891383, 1.004754272, 1.003870012, 0.999888944, 0.994438078, 0.992548715, 0.998344545, 1.004504764, 1.00883411]

在Excel中计算其标准偏差的地方，我得到了答案:0.005106477，我使用的函数是:=STDEV(C5:N5)

where I calculated its standard deviation in Excel, I got the answer: 0.005106477, the function I used was: =STDEV(C5:N5)

然后，我使用numpy.std进行以下计算:

Then I do the same calculation using numpy.std as:

import numpy as np

print np.std(s)

但是，我得到了答案:0.0048890791894

However, I got the answer: 0.0048890791894

我什至编写了自己的std函数:

I even wrote up my own std function:

def std(input_list):
        count = len(input_list)

        mean = float(sum(input_list)) / float(count)

        overall = 0.0
        for i in input_list:
            overall = overall + (i - mean) * (i - mean)

        return math.sqrt(overall / count)

和我自己的函数给出的结果与numpy相同.

and my own function gives the same result as numpy.

所以我想知道这样的区别吗?还是只是我犯了一些错误?

So I am wondering is there such a difference? Or it just I made some mistake?

推荐答案

有一个区别:Excel的STDEV计算样本标准偏差，而NumPy的std计算人口默认情况下为标准偏差(行为类似于Excel的STDEVP).

There's a difference: Excel's STDEV calculates the sample standard deviation, while NumPy's std calculates the population standard deviation by default (it is behaving like Excel's STDEVP).

要使NumPy的std函数的行为类似于Excel的STDEV，请传入值ddof=1:

To make NumPy's std function behave like Excel's STDEV, pass in the value ddof=1:

>>> np.std(s, ddof=1)
0.0051064766704396617

这将使用样本方差(即除以n-1而不是n)来计算s的标准偏差.

This calculates the standard deviation of s using the sample variance (i.e. dividing by n-1 rather than n.)

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