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

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Eric 2023-8-21

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评论： Eric 2023-9-6

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?

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Walter Roberson 2023-8-21

在 MATLAB Online 中打开

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.)

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Walter Roberson 2023-8-28

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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 2023-9-6