# How is the number of bins chosen with the auto binning algorithm in histcounts?

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Magnus Brander on 22 Apr 2018
Commented: Magnus Brander on 13 Jul 2018
When I call the function histcounts as following:
[N,edges] = histcounts(data)
It returns a number of bins with the corresponding bin edges. How is the number of bins determined? At https://se.mathworks.com/help/matlab/ref/histcounts.html it is says "...uses an automatic binning algorithm...".
What is this for an algorithm and where can I find some documentation about it?

#### 1 Comment

Richard on 10 Jul 2018
I have the exact same question. This really is the sort of thing that should be in the documentation. At least a reference to the algorithm should be provided.

Anton Semechko on 10 Jul 2018
'histcounts' first estimates width of the histogram bins using 'scottsrule':
rawBinWidth = 3.5*std(data)/(numel(data)^(1/3));
It then passes this info along with the minimum and maximum values of input data (xmin and xmax, resp.) to the 'binpicker' function which first adjusts rawBinWidth depending on its order of magnitude:
powOfTen = 10.^floor(log10(rawBinWidth)); % next lower power of 10
relSize = rawBinWidth / powOfTen; % guaranteed in [1, 10)
if relSize < 1.5
binWidth = 1*powOfTen;
elseif relSize < 2.5
binWidth = 2*powOfTen;
elseif relSize < 4
binWidth = 3*powOfTen;
elseif relSize < 7.5
binWidth = 5*powOfTen;
else
binWidth = 10*powOfTen;
end
and then computes the total number of bins and positions of the left- and right-most histogram edges:
leftEdge = min(binWidth*floor(xmin ./ binWidth), xmin);
nbinsActual = max(1, ceil((xmax-leftEdge) ./ binWidth));
rightEdge = max(leftEdge + nbinsActual.*binWidth, xmax);
Remaining bin edges are distributed uniformly between leftEdge and rightEdge at binWidth intervals. You can inspect source code for the 'binpicker' function by typing:
edit histcounts
I imagine the reason Mathowrks does not describe how the bin edges are determined by 'hiscounts' is because the process involves several steps, and isn't based on a simple formula.

#### 1 Comment

Magnus Brander on 13 Jul 2018
Thanks a lot!