N-dimensional histogram

Hi, Bruno:

Your coding is really elegant and excellent. ;-)
I had one question on line 66 " sz(d) = length(ed)-1; ". So here, why sz(d) is not instead length(ed)?

Due to this, when I provide, .i.e, 3 edges of size mx1, for a 3d matrix, I would get *count* as a (m-1)^3 rather than m^3 matrix. Am I misunderstanding sth here, or there is some special reason for this?

Thanks,
Bin

Works great! I'm using it to make a 2-dimensional histogram with an additional third dimension of "weights."

Will, you better post the plotting problem on the newsgroup. I suggest to take a look at BAR3 function for the moment.

Jaclyn, to put y-axis increase from bottom to top call

>> set(gca, 'Ydir', 'normal')

This is question is generic and better posted in the newsgroup. Thanks for the comment.

Great. 100,000 XY float points in a 512x512 bin matrix takes ~40 ms, and I still get the list of bin assignment!

About plotting the results:

Jaclyn, if you just want bars plotted for the pairwise densities of two variables (i.e., a plot with x and y axes corresponding to values of your variables, and vertical bars on the x,y grid, corresponding to the counts for each 2D bin), you can get that with hist3().

Will, if you want to plot weights for 3 variables as varying sizes of bubbles on an x,y,z grid, you can do something like this:

% get the histogram
[count edges mid loc] = histcn(vals);

% make a grid for plotting
[X Y Z]=ndgrid(edges{1}, edges{2}, edges{3});
X=X(:); Y=Y(:); Z=Z(:);

% calculate sizes so the most dense cell gets a value of 100
% also convert from volume to "area" (as if drawing a sphere with
% the right volume and cross-sectional area s)
s_scale = 100/(max(count(:))^(2/3));
s = count(:).^(2/3) * s_scale;
% convert any zeros to small numbers for scatter3
s(s==0)=realmin;

% plot the densities
fh=figure();
set(fh, 'Renderer', 'OpenGL'); % faster drawing
scatter3(X, Y, Z, s, 'filled');