MATLAB Answers


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

Asked by Magnus Brander on 22 Apr 2018
Latest activity Commented on by 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 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

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.

Sign in to comment.

1 Answer

Answer by Anton Semechko on 10 Jul 2018
 Accepted Answer

'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;
binWidth = 10*powOfTen;
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
into your command prompt.
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.