Scatter fails if all elements but one at index(1,1) are nan.

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I am trying to use a scatter plot to visualize a 2-D histogram. I set all points with 0 hits to nan to improve the visualization of the scattering patterns. I get the following errors if there is just 1 non-NaN value in the data set:
"The logical indices contain a true value outside of the array bounds."
"Error in matlab.graphics.chart.primitive.internal.AbstractScatter/doUpdate"
In an attempt to debug the issue, I simplified the code to the following 5 test cases where the first generates the error and the others do not.
x0 = 1:14;
y0 = 1:19;
hits = nan( 19, 14 );
[x,y] = meshgrid( x0, y0 );
test_case = 1;
if test_case == 1 % fails (single non-NaN point at 1,2)
hits(1,2) = 1;
scatter( y(:), x(:), 20, hits(:), 'filled', 's' )
elseif test_case == 2 % passes (single non-NaN point at 1,1)
hits(1,1) = 1;
scatter( y(:), x(:), 20, hits(:), 'filled', 's' )
elseif test_case == 3 % passes (two non-NaN points)
hits(1,1) = 1;
hits(1,2) = 1;
scatter( y(:), x(:), 20, hits(:), 'filled', 's' )
elseif test_case == 4 % passes (all points are NaN)
scatter( y(:), x(:), 20, hits(:), 'filled', 's' )
elseif test_case == 5 % passes (random values; identify those > 0)
hits = randi( 4, size( hits ) ) - 1;
hits(hits<1) = nan();
scatter( y(:), x(:), 20, hits(:), 'filled', 's' )
end
Any help would be much appreciated!
  2 Comments
David Goodmanson
David Goodmanson on 4 Nov 2024 at 5:50
Edited: David Goodmanson on 4 Nov 2024 at 5:59
Hi Joe, I tried this and all five cases worked including hits(1,1) = 1. I have Matlab 2023a Update 4
Joe Mueller
Joe Mueller on 4 Nov 2024 at 16:23
Thanks! Walter Roberson indicated that this was a bug and fixed in 2023b.

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Accepted Answer

Walter Roberson
Walter Roberson on 4 Nov 2024 at 6:04
Edited: Walter Roberson on 4 Nov 2024 at 6:06
This is fixed in R2023b (maybe slightly earlier)
There does not appear to be any public bug report about this.

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R2022a

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