Octave barh with colors based on third value - bar-chart

Question

A is a 16x124 matrix with the values that plot in the bar chart. I guess the default is to have the colors associated with the index (1 to 124). The trouble I'm having is that I want each section of the bar chart to be color coded based on the corresponding value within matrix B. If B(1,1) = 3, then the box for A(1,1) is green, and so forth.
A = {3,5,6,8,1,3,6,3;
6,3,8,2,3,7,8,4;
2,7,8,3,8,2,1,3;
4,9,6,4,3,6,1,2
};
B = {3,2,3,2,1,0,1,3;
3,1,2,1,2,0,3,1;
1,3,2,3,2,3,1,3;
3,1,2,3,2,3,2,3
};
P=[2016,2017,2018,2019];
For i = 1:4
For j = 1:8
If B(i,j) = 3
C(i,j) = 'g'
Elseif B(i,j) = 2
C(i,j) = 'y'
Elseif B(i,j) = 1
C(i,j) = 'r'
Elseif B(i,j) = 0
C(i,j) = 'k'
End
End
End
H = barh(Positions,A,'stacked');
H.facecolor = C
EDIT:
Tasos, thank you! I got it all worked out. Took me a bit to understand the proper syntax for changing this line:
facevertexcdata(m, :) = Colormap(B(m,n),:);
I also ended up using Prism so that the green to red is intuitive with the intent of the figure.
Final Version of figure

Answer 1

Here's an example of what I think you're trying to do
M = 4;
N = 8;
A = randi( 100, M, N );
P = [ 10, 20, 35, 60 ];
H = barh( P, A, 0.9, 'stacked' );
% define colours
B = { [0, 0, 0], [0, 0, 0.1], [0, 0, 0.2], [0, 0, 0.3], [0, 0, 0.4], [0, 0, 0.5], [0, 0, 0.6], [0, 0, 0.7], [0, 0, 0.8] ;
[0, 0, 0], [0.1, 0, 0], [0.2, 0, 0], [0.3, 0, 0], [0.4, 0, 0], [0.5, 0, 0], [0.6, 0, 0], [0.7, 0, 0], [0.8, 0, 0] ;
[0, 0, 0], [0, 0.1, 0], [0, 0.2, 0], [0, 0.3, 0], [0, 0.4, 0], [0, 0.5, 0], [0, 0.6, 0], [0, 0.7, 0], [0, 0.8, 0] ;
[0, 0, 0], [0.1,0,0.1], [0.2,0,0.2], [0.3,0,0.3], [0.4,0,0.4], [0.5,0,0.5], [0.6,0,0.6], [0.7,0,0.7], [0.8,0,0.8]
};
for n = 1 : N
h = get( H, 'children' ){n};
facevertexcdata = zeros(M,3);
for m = 1 : M
facevertexcdata(m, :) = B{m,n};
end
set( h, 'facevertexcdata', facevertexcdata );
end
UPDATE: If you effectively have a matrix B of linear indices (note, no zeros), you can use that to index an array or cell array of predefined colours accordingly (including one generated by a colormap).
E.g. assume
Colormap = { [1,0,0], % red
[0,1,0], % green
[0,0,1], % blue
[1,0,1] } % violet
B = [ 1,2,3,4;
2,3,4,1;
3,4,1,2; ] % ... etc
Then you can use this to generate a cell array that contains the right color triplet in the appropriate position, as:
C(B)
Alternatively, since you only try to access one element at a time, you can define a normal colormap, and use elements of B to obtain the right color from the colormap at the point in the code where you need it (i.e. for the example above, it would be inside the for loop). E.g. if
> Colormap = jet(4)
Colormap =
0 0 1
0 1 1
1 1 0
1 0 0
% B as above
Then, say, for B(2,2) (= 3), you can obtain the third color in the generated colormap, i.e.
> Colormap( B(2,2), : )
ans =
1 1 0