Python: Numpy standard deviation error

Question

This is a simple test

import numpy as np data = np.array([-1,0,1]) print data.std()  >> 0.816496580928 

I don't understand how this result been generated? Obviously:

( (1^0.5 + 1^0.5 + 0^0.5)/(3-1) )^0.5 = 1 

and in matlab it gives me std([-1,0,1]) = 1. Could you help me get understand how numpy.std() works?

Answer 1

The crux of this problem is that you need to divide by N (3), not N-1 (2). As Iarsmans pointed out, numpy will use the population variance, not the sample variance.

So the real answer is sqrt(2/3) which is exactly that: 0.8164965...

If you happen to be trying to deliberately use a different value (than the default of 0) for the degrees of freedom, use the keyword argument ddofwith a positive value other than 0:

np.std(data, ddof=1) 

... but doing so here would reintroduce your original problem as numpy will divide by N - ddof.

Answer 2

It is worth reading the help page for the function/method before suggesting it is incorrect. The method does exactly what the doc-string says it should be doing, divides by 3, because By default ddofis zero.:

In [3]: numpy.std?  String form: <function std at 0x104222398> File:        /System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/numpy/core/fromnumeric.py Definition:  numpy.std(a, axis=None, dtype=None, out=None, ddof=0, keepdims=False) Docstring: Compute the standard deviation along the specified axis.  ...  ddof : int, optional     Means Delta Degrees of Freedom.  The divisor used in calculations     is ``N - ddof``, where ``N`` represents the number of elements.     By default `ddof` is zero.