Change color of diagonal bars (1,1), (2,2), (3,3)... in bar3

I am plotting a 2D histogram where there are 7 bars in each axis. An example of the data is attached as N.

On this plot, I would like to be able to color the diagonal bars [e.g., (1,1), (2,2) ... (7,7)] green, while leaving the rest of the bars the the base color.

I've built the chart using:

[N, Xedges, Yedges] = histcounts2(Xvals, yVals, [7, 7]);
h = bar3(N);  

And I'm trying to color the bars using something like this, but I can't seem to figure out how to select the only the kth bar in each row:

for k = 1 : length(h) % operate on each row of bars
    % Set FaceColor to 'flat' to use CData for coloring
    h(k).FaceColor = 'flat';

    set(h(k),'facecolor',[0 1 0]) % green        
end

Ideally, I'd like to color both the top and sides of the selected bars green.

Answer 1

[dpb]

load N
%[N, Xedges, Yedges] = histcounts2(Xvals, yVals, [7, 7]);
h = bar3(N)

h = 
  1×7 Surface array:

    Surface    Surface    Surface    Surface    Surface    Surface    Surface

h(1)

ans = 
  Surface (bar3) with properties:

       EdgeColor: [0.1294 0.1294 0.1294]
       LineStyle: '-'
       FaceColor: 'flat'
    FaceLighting: 'flat'
       FaceAlpha: 1
           XData: [42×4 double]
           YData: [42×4 double]
           ZData: [42×4 double]
           CData: [42×4 double]

  Use GET to show all properties

There is only one surface per row so there is no way to address the individual surfaces by the top-level handles; you would have to delve into the underlying CData arrays and set the correct set.

The colors are color order, not rgb triplets by default as well so

[h(1).CData(1,:);h(2).CData(1,:);h(6).CData(1,:);h(7).CData(1,:)]

ans = 4×4

1 1 1 1 2 2 2 2 6 6 6 6 7 7 7 7

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You can poke at the CData array; there are six faces per bar so

figure
subplot(1,2,1)
h = bar3(N);
subplot(1,2,2)
h = bar3(N);
for i=1:7
  offset=6*(i-1);
  h(i).CData([1:6]+offset,:)=5;
end

changes all the faces of the diagonal bar to the color of the 5th bar initially which happens to be green. One would have to delve into the depths of the X|YData to figure out just which faces are which in the sections; I'm not sure of the storage order.

I'd have to do a lot of digging into the colormap and all to figure out how, precisely, one could manage to get to an rgb color or specific color in the mapping -- that's been an area I've never really figured out the inards of.

ADDENDUM

I've complained "since forever" that the bar and bar3 functions are weak sisters -- not being able to address each bar individually at the user level as desired here is a real weakness as is the fact you can't even set the X dimension vector.

[Jim McIntyre]

Thank you! That is helpful.

I agree with you that the coloring here is hard to understand weak and fraught with peril.

Below is code that basically does what I want using the techniques you showed me. A couple of notes are in order:

• If you change the FaceColor setting to explicit values for any column in the array, for example:

set(h, 'FaceColor', [0 0 1]);

or

set(h(3), 'FaceColor', [0 0 1])

it overrides the CData settings for those columns.

• The use of Colormap is a little bit buggy here, too. I defined three values for the colormap. Unless the color limits are explicitly defined, Matlab will display the highest value in CData using the upper value in the colormap.

load('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1839710/N.mat');

% Create a 3D bar plot
h = bar3(N);

CharMap = [1,0,0; 0,1,0; 0,0,1]; % red, green, blue
colormap(CharMap);
clim([1 3]);

% Do not set FaceColor to a value other than flat. 
% If FaceColor has a numeric value then CData values are ignored

for col = 1 : size(N, 2) 
  h(col).CData = ones(6 * size(N, 2), 4);  % Set base color of all bars      
  offset = 6 * (col - 1);
  h(col).CData([1 : 6 ] + offset, :) = 2;  % Set the color of the diagonal bars
end
h(3).CData([1 : 6 ] + 30, :) = 3;          % Set the color of some other bar

●

[dpb]

Another approach that is perhaps somewhat simpler to color the diagonal to any specific color at the expense of creating a "veritable plethora" of new objects uses the time-honored "trick" of setting unwanted to be displayed values to NaN --

load N
subplot(1,2,1)
bar3(N)
hold on
for i=1:height(N)
  N1=nan(size(N));
  N1(i,i)=N(i,i);
  bar3(N1,'g')
end
subplot(1,2,2)
bar3(N)
hold on
for i=1:height(N)
  N1=-ones(size(N));
  N1(i,i)=N(i,i);
  bar3(N1,'g')
  zlm=zlim; zlm(1)=0; zlim(zlm)
end

The two are slightly different in that the one using -1 has one more locations displayed in the original color than the NaN version -- why the green shows for bars with height <0 with lower zlim set to 0 is a mystery to me.

[dpb]

Glad to have pointed out at least some way forward; I agree again that bar graphs are very "user belligerent" in MATLAB although there have been some improvements in bar(), bar3() is still horrible user interface and extremely limited in functionality.

Perhaps @Adam Danz will chime in with some clever ideas, Adam?

[dpb]

"I defined three values for the colormap. Unless the color limits are explicitly defined, Matlab will display the highest value in CData using the upper value in the colormap. "

That is expected behavior. From the doc for clim

"The Axes object interpolates data values between cmin and cmax across the colormap. Values outside this range use either the first or last color, whichever is closest."

The scaling is on the data values, so in your sample data, there is a large disparity between the data values

load N
N

N = 7×7

261278 0 727 490 2178 354 101 3 265 0 0 0 0 0 3851 5 80 31234 527 379 65 1093 2 1 11351 170 42 19 18857 77 17376 5418 213807 2484 4510 0 0 40568 3092 8630 237426 2325 22 0 6177 6 28496 1155 74303

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so if you group by only three groups,

histcounts(N,3)

ans = 1×3

46 0 3

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the middle group is empty.

histcounts(log10(N),3)

ans = 1×3

10 19 11

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while on the other hand, the data are almost exactly log-distributed.

However, I don't know precisely how bar3 does the scaling when clim is set since it doesn't color the individual bars by data height, but only by row.

However, using the linear or the log10 scaling seems to change its internal decision...

CharMap = [1,0,0; 0,1,0; 0,0,1]; % red, green, blue
colormap(CharMap);
subplot(1,2,1)
clim(([min(N,[],'all') max(N,[],'all')]))
bar3(N)
subplot(1,2,2)
bar3(log10(N))
clim(log10([min(N,[],'all') max(N,[],'all')]))

●

The linear case is 2:2;3 while the log case is 2:1:4.

What's such a bugger is the encapsulation of the handles for each row into the one blob--granted, it is undoubtedly more efficient internal coding to do that, but it makes for very difficult customization if want anything but the out-of-the-box results.

[Jim McIntyre]

Agreed on all counts.

In this case, I am comparing how two algorithms assign classifications to data points in a diferent array. There are seven classifications and this is the array of classifications. So the data are the number of points that one algorithm classifies each way versus the how the second algorithm classifies the same data.

The most important to me are the first row and the first column. Classification 1 could be thought of as "good." The other classifications could be thought of as "not good because..."

So in this case, the absolute number of points is less important than the percentage of points in each case. Ideally, all points would lie on the diagonal since the diagonal is the set of points that both algorithms classify the same. Off-axis points are differences, some of which are more concerning than others. To present these data, I will probably color the diagonal points green, the first row and column (excluding 1,1) red, and the other points blue.

[Jim McIntyre]

dpb, you say: why the green shows for bars with height <0 with lower zlim set to 0 is a mystery to me.

Regarding that, see a related question that I asked about a month ago:

Plotting a 2D array of values as discrete squares without interpolation - MATLAB Answers - MATLAB Central

In these bar3 plots, Matlab appears to be creating a "floor" value at 0 and I haven't been able to find a way to recolor that or turn it off. Even filling cell values with NaN still shows this "floor."

[dpb]

I don't have time to play further at the moment, but I suspect if there is a way it will be to dig into the X|YData to figure out which CData vector is the bottom face of each bar in the six-vector that describes each bar.

However, the floor is something else I expect because it isn't part of the actual X|Y|Z data and whatever handles there are to it are hidden.

If there's any access at all, it'll probably have to be found by "handle-diving" into the hidden properties.

[dpb]

What about an alternative approach, maybe????

load N
heatmap(N,'CellLabelFormat','%d')
colormap jet

figure
edges=[0.5:1:7.5];
subplot(1,2,1)
histogram2('XBinEdges',edges,'YBinEdges',edges,'BinCounts',N)
subplot(1,2,2)
histogram2('XBinEdges',edges,'YBinEdges',edges,'BinCounts',N,'DisplayStyle','tile')

[Jim McIntyre]

Thank you for your continued interest. Here's a little more background.

I understand that this is effectively a heatmap. However, some points in the map are far more important than others, so to explain this to management I'm using colors to indicate the importance and then just looking at the heights of the bars. In this case, a question I'm trying to answer is why did method 1 classify some points as 4, when method 2 called the same points 3.

In essence, I'm reverse-engineering something that my company did about 30 years ago.

These algorithms each include about a dozen separate identification routines and algorithm 1 only exists in a compiled state within a larger piece of code. So decompiling really isn't an option. I also don't have the ability to look at the results from each individual routine of algorithm 1. Having that is actually one of the advantages of redoing this in Matlab.

I have the "source code," but I'm not confident that that code is exactly what was finally compiled, and based on the results I'm getting I believe that there were probably some modifications made to it that I don't have.

So, thank you for your interest and help. Although this solution is clunky, it's gotten me where I need to be for now.

[dpb]

OK, glad to have been able to help...as we have noted, unfortunately there's only so much one can do with bar3 itself; getting more would require writing one's own replacement.