Why does numpy.histogram (Python) leave off one element as compared to hist in Matlab?

Question

I am trying to convert some Matlab code to Python, and the Matlab code looks like:

[N,X] = hist(Isb*1e6, -3:0.01:0)

where Isb is a 2048000

Answer 1

As said above, in matlab -3:0.01:0 specifies the bin-centers, namely 301. What your doing in numpy specifies bin-edges, so this will be one bin less than in matlab.

Hence, you could either move from hist to histc within matlab, or make sure you apply the same bin-edges in numpy. In this special case (equidistant bins) you could also use numpy like this:

N,X = np.histogram(x, bins = n_bins, range = (xmin, xmax))

In this case: n_bins of 301 and (xmin, xmax) being (-3.005,0.005) should be equivalent to matlab's hist.

See also:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.histogram.html

Answer 2

Note that in matlab's hist(x, vec), vec difines the bin-centers, while in matlab histc(x, vec) vec defines the bin-edges of the histogram. Numpy's histogram seems to work with bin-edges. Is this difference important to you? It should be easy to convert from one to the other, and you might have to add an extra Inf to the end of the bin-edges to get it to return the extra bin you want. More or less like this (untested):

import numpy as np

def my_hist(x, bin_centers):
    bin_edges = np.r_[-np.Inf, 0.5 * (bin_centers[:-1] + bin_centers[1:]), 
        np.Inf]
    counts, edges =  np.histogram(x, bin_edges)
    return counts

For sure it does not cover all the edge-cases that matlab's hist provides, but you get the idea.