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pythonexcelnumpystandard-deviation
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Reputation: 7408

Is there any difference between numpy.std and excel STDEV function?

I have a list:

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]

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

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

import numpy as np

print np.std(s)

However, I got the answer: 0.0048890791894

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)

and my own function gives the same result as numpy.

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

Upvotes: 10

Views: 11488

Answers (1)

Alex Riley
Alex Riley

Reputation: 176810

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).

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

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

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

Upvotes: 23

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