How to colorize individual bar in bar3

Question

test 2011-4-13

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评论： Tyson Murray 2019-9-14

Hello.

I'm trying to plot 3D graph with bars, in which every bar is colored with color I choose. I found solution for 2D graphs:

Y=[ 1 2 3 ; 4 5 6 ; 3 4 5];
h = bar('v6',Y);
set(h(1),'facecolor','red') % use color name
set(h(2),'facecolor',[0 1 0]) % or use RGB triple
set(h(3),'facecolor','b') % or use a color defined in the help for PLOT

but i can't find out how to use this with bar3 function (use this on 3D). Has anyone any solution?

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bryce Howat 2017-8-1

I ran into this problem when I wanted to mark out "hits" on a 3d representation of a 16x24 plate during HTS. The solution I found works like this: first make a matrix of zeros the same size as the plate. Next use a for loop to change any parts you want marked to have the same value as the inital data but plus .0001 or some small number. Next graph both on the same axis using hold on, and for what was the row of zeros you can use set( name, 'facecolor', 'cyan').

The only downside is that if you have values below zero these columns will appear cyan from the top. This can be solved by also graphing a matrix of values near .0000001 on top of both other matrices.

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Answer 1

Matt Fig 2011-4-15

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在 MATLAB Online 中打开

Never say never!

Y=[ 1 2 3 ; 4 5 6 ; 3 4 5];
h = bar3(Y);
cm = get(gcf,'colormap');  % Use the current colormap.
cnt = 0;
for jj = 1:length(h)
    xd = get(h(jj),'xdata');
    yd = get(h(jj),'ydata');
    zd = get(h(jj),'zdata');
    delete(h(jj))    
    idx = [0;find(all(isnan(xd),2))];
    if jj == 1
        S = zeros(length(h)*(length(idx)-1),1);
        dv = floor(size(cm,1)/length(S));
    end
    for ii = 1:length(idx)-1
        cnt = cnt + 1;
        S(cnt) = surface(xd(idx(ii)+1:idx(ii+1)-1,:),...
                         yd(idx(ii)+1:idx(ii+1)-1,:),...
                         zd(idx(ii)+1:idx(ii+1)-1,:),...
                         'facecolor',cm((cnt-1)*dv+1,:));
    end
end
rotate3d

Now S has the handle to each surface so you can change the color of each (or set any other individual property) as you wish. I.e, set(S(1),'facecolor','red'). Also, if you knew ahead of time how many surfaces there would be, you could create a matrix of colors and index into that as S was created....

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Samar Singh 2019-4-25

Awesome !

Tyson Murray 2019-9-14

Thank you.

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Answer 2

Arnaud Miege 2011-4-13

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Does the following not do what you want or have I misunderstood your question?

h = bar3(Y);
set(h(1),'facecolor','red');
set(h(2),'facecolor','blue');
set(h(3),'facecolor','green');

Arnaud

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Arnaud Miege 2011-4-15

You can't, at least as far as I can tell. The handle is a 1x3 vector, so each group of bars is treated as a unit. You double-check this with plottools.

Ali Rezaei 2012-3-22

Well done! Great and easy to implement answer!

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Answer 3

Jan 2011-4-15

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编辑：Jan 2016-4-4

在 MATLAB Online 中打开

cm = get(gcf,'colormap');
cms = size(cm, 1);
Y = [ 1 2 3 ; 4 5 6 ; 3 4 5];
h = bar3(Y);
for i = 1:length(h)
   c = get(h(i), 'CData');
   set(h(i), 'CData', ceil(cms * rand(size(c))));
end

[EDITED - same color for all faces of a bar]

Y = [8 9 8; 4 5 6; 3 4 5; 1 2 3];  
h = bar3(Y);
[nBar, nGroup] = size(Y);
nColors  = size(get(gcf, 'colormap'), 1);
colorInd = randi(nColors, nBar, nGroup);
for i = 1:nGroup
   c     = get(h(i), 'CData');
   color = repelem(repmat(colorInd(:, i), 1, 4), 6, 1);
   set(h(i), 'CData', color);
end

This works at least in 2009a and 2015b.

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Ali Rezaei 2012-3-22

Well done! Great and easy to implement answer!

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Answer 4

Baha 2011-9-7

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Can you manipulate your code so that you can partition each column and colorize as you want? For example, Y = [ 1 2 3 ; 4 5 6 ; 3 4 5]; h = bar3(Y); and knowing that two variables are contributing this graph. Furthermore, each column can also be identified by another index. Such as, Y(2,2)=5, and there is an identifier Ei(i=1:3) that E1 contributes Y(2,2) as 1 point, E2 as 2.5 points and E3 1.5 points =total=5. Now, that is what I would like to show on bar graph. i.e. on each column, length(0-1)=red, length(1-3.5)= blue, length(3.5-5)= green. Any idea is appreciated.