Is there a way to use hist3 without the height of each bar indicating the number of elements in a bin?

I'm trying to create a bivariate histogram plot, but I have a third z variable other than x and y. Since each x and y pair corresponds with a specific z value, I am trying to create a bivariate histogram plot whose height will not be the number of elements in that bin, but instead the cumulative z values in that bin. Is there any way to do this with hist3? If not, what else can I use?

Answer 1

[Walter Roberson]

discretize on the X and Y recording the indices in each case. Then

totals = accumarray([Yindices(:), Xindices(:)], Zvalues());

The result will be max(Yindices) by max(Xindices) of total Z, and you can then bar3() .

(In practice it would be a good idea to pass in the expected output size of the array as the third parameter.)

[Eric]

Will this still work if x is longitude and y is latitude? I forgot to mention this detail in my post.

[Eric]

Also, I'm using hist3 to make a graph like this:

Is there anything other than bar3 that I can use? Or can bar3 replicate the same kind of image?

[Walter Roberson]

rng(655321)
data = sort(randi(9, 10, 10));
imagesc(0.5, 0.5, data);
grid on
set(gca, 'GridAlpha', 1.0)

You might also want to increase GridLineWidth

This example online seems to imply that vertical grid lines are not drawn, but that is an artifact: they are drawn.

[Eric]

Sorry to bother but would it be possible for you to provide an example of using discretize in this context? I am trying to bin the data into longitude and latitude values so the values need to be put into a 2d array.

[Walter Roberson]

Example with bins that are 15 degrees by 15 degrees.

rng(655321)
N = 100;
lats = rand(1, N) * 180 - 90;
lons = rand(1,N) * 360 - 180;
values = rand(1,N) * 50;
geoscatter(lats, lons)

latedges = -90:15:90;
lonedges = -180:15:180;
nlatbins = numel(latedges)-1;
nlonbins = numel(lonedges)-1;

latbin = discretize(lats, latedges);
lonbin = discretize(lons, lonedges);

bincounts = accumarray([latbin(:), lonbin(:)], 1, [nlatbins, nlonbins]);

imagesc(latedges, lonedges, bincounts); colorbar(); title('counts per bin'); set(gca, 'YDir', 'normal')

bintotals = accumarray([latbin(:), lonbin(:)], values(:), [nlatbins, nlonbins]);
imagesc(latedges, lonedges, bintotals); colorbar(); title('totals per bin'); set(gca, 'YDir', 'normal')

lats(1:5)

ans = 1×5

-33.4773 -20.4551 -56.6026 67.4535 51.1198

latbin(1:5)

ans = 1×5

4 5 3 11 10

latedges(1:6)

ans = 1×6

-90 -75 -60 -45 -30 -15

-33.4773 degrees is between -45 degrees and -30 degrees, so it falls into the 4th bin-- the bin that covers [-45, 30). Likewise, -20.4551 is in [-30, -15) which is the 5th bin so it gets index 5.

In the case of edges that are equidistant, you can calculate the indices in more efficient ways -- but not necessarily clearer ways than using discretize()

[Eric]

Sorry for the late comment, but thank you so much! You are literally a lifesaver.