Python numpy.var returning wrong values

Question

I'm trying to do a simple variance calculation on a set of 3 numbers:

numpy.var([0.82159889, 0.26007962, 0.09818412])

which returns

0.09609366366174843

However, when you calculate the variance it should actually be

0.1441405

Seems like such a simple thing, but I haven't been able to find an answer yet.

Answer 1

As the documentation explains:

ddof : int, optional
    "Delta Degrees of Freedom": the divisor used in the calculation is
    ``N - ddof``, where ``N`` represents the number of elements. By
    default `ddof` is zero.

And so you have:

>>> numpy.var([0.82159889, 0.26007962, 0.09818412], ddof=0)
0.09609366366174843
>>> numpy.var([0.82159889, 0.26007962, 0.09818412], ddof=1)
0.14414049549262264

Both conventions are common enough that you always need to check which one is being used by whatever package you're using, in any language.

Answer 2

np.var by default calculates the population variance.

The Sum of Squared Errors can be calculated as follows:

>>> vals = [0.82159889, 0.26007962, 0.09818412]
>>> mean = sum(vals)/3.0
>>> mean
0.3932875433333333
>>> sum((mean-val)**2 for val in vals)
0.2882809909852453
>>> sse = sum((mean-val)**2 for val in vals)

This is the population variance:

>>> sse/3 
0.09609366366174843
>>> np.var(vals)
0.09609366366174843

This is the sample variance:

>>> sse/(3-1)
0.14414049549262264
>>> np.var(vals, ddof=1)
0.14414049549262264

You can read more about the difference here.