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?
NumPy arrays are the equivalent to the basic array data structure in MATLAB. With NumPy arrays, you can do things like inner and outer products, transposition, and element-wise operations.
Question 3: Why is NumPy preferred to other programming tools such as IDL, Matlab, Octave, Or Yorick? NumPy is a high-performance library in the Python programming language that allows scientific calculations. It is preferred to Idl, Matlab, Octave, Or Yorick because it is open-source and free.
MATLAB® and NumPy have a lot in common, but NumPy was created to work with Python, not to be a MATLAB clone. This guide will help MATLAB users get started with NumPy.
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)
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