Precision, why do Matlab and Python numpy give so different outputs?

Question

I know about basic data types and that float types (float,double) can not hold some numbers exactly.

In porting some code from Matlab to Python (Numpy) I however found some significant differences in calculations, and I think it's going back to precision.

Take the following code, z-normalizing a 500 dimensional vector with only first two elements having a non-zero value.

Matlab:

Z = repmat(0,500,1); Z(1)=3;Z(2)=1;
Za = (Z-repmat(mean(Z),500,1)) ./ repmat(std(Z),500,1);
Za(1)
>>> 21.1694

Python:

from numpy import zeros,mean,std
Z = zeros((500,))
Z[0] = 3
Z[1] = 1
Za = (Z - mean(Z)) / std(Z)
print Za[0]
>>> 21.1905669677

Besides that the formatting shows a bit more digits in Python, there is a huge difference (imho), more than 0.02

Both Python and Matlab are using a 64 bit data type (afaik). Python uses 'numpy.float64' and Matlab 'double'.

Why is the difference so huge? Which one is more correct?

Should perhaps fit on Computational Science SE if asked nowadays — gerrit

rocksportrocker rocksportrocker · Accepted Answer · 2011-09-20T08:46:44

Maybe the difference comes from the mean and std calls. Compare those first.

There are several definitions for std, some use the sqaure root of

1 / n * sum((xi - mean(x)) ** 2)

others use

1 / (n - 1) * sum((xi - mean(x)) ** 2)

instead.

From a mathematical point: these formulas are estimators of the variance of a normal distributed random variable. The distribution has two parameters sigma and mu. If you know mu exactly the optimal estimator for sigma ** 2 is

1 / n * sum((xi - mu) ** 2)

If you have to estimate mu from the data using mu = mean(xi), the optimal estimator for sigma**2 is

1 / (n - 1) * sum((xi- mean(x))**2)

Precision, why do Matlab and Python numpy give so different outputs?

3 Answers