How can i change color of a specific bin using histogram2 or anything similar

suppose i have some data which i want to cluster into 3D histogram

x = 100*randn(10000,1);
y = 100*randn(10000,1);

h2 = histogram2(x,y,-100:20:100,-100:20:100,'FaceColor','flat')

h2 = 
  Histogram2 with properties:

             Data: [10000×2 double]
           Values: [10×10 double]
          NumBins: [10 10]
        XBinEdges: [-100 -80 -60 -40 -20 0 20 40 60 80 100]
        YBinEdges: [-100 -80 -60 -40 -20 0 20 40 60 80 100]
         BinWidth: [20 20]
    Normalization: 'count'
        FaceColor: 'flat'
        EdgeColor: [0.1500 0.1500 0.1500]

  Show all properties

i want to change the color of a specific bin lets say bin which belongs to row and column 5

i would want something like this:

h2.Cdata(5,5,:) = [1 0 0] % set bin color to red

Unrecognized method, property, or field 'Cdata' for class 'matlab.graphics.chart.primitive.Histogram2'.

but theres no Cdata property

Answer 1

[Adam Danz]

One way to do this is to plot a histogram for each group.

rng('default') % For reproducibility of this example
x = 100*randn(10000,1);
y = 100*randn(10000,1);

[N, xEdges, yEdges] = histcounts2(x,y,-100:20:100,-100:20:100);

% Plot group 1
N1 = N;
N1(5,5) = 0; 
h0 = histogram2('XBinEdges',xEdges,'YBinEdges',yEdges,'BinCounts',N1,'FaceColor','Flat');

% Plot group 2 
N2 = N(5,5);
hold on
h1 = histogram2('XBinEdges',xEdges(4:5),'YBinEdges',yEdges(4:5),'BinCounts',N2,'FaceColor','r');

Another solution may be to use bar3 with a width of 1 but that has the following issues

• bar3 does not have an x input. Solution is to adjust the x values after plotting.
• the y input to bar3 defines the center of each bin, not the edge. Solution is to compute the centers.
• defining colors to individual bars might be tricky since a handle is returned for each row along the x axis

[daniel mitchell]

thank you this workes perfectly.

Probably it will be a good suggestion for MATLAB to add a property for histogram2 class which will enable modifying colors directly.

[Adam Danz]

Thanks for the suggestion, Daniel.