YTickLabel in colormap will be shown with some specified value (not show all)

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I have code and the result as below:
%%How to plot each matrix in a cell in 3d plot(1 matrix with 1 color) ?
% Generate Sample data cell A(1x10 cell array)
clear; clc;
A = cell(1,10); % cell A(1x10 cell array)
for kk = 1:numel(A)
z = 10*rand()+(0:pi/50:10*rand()*pi)';
x = 10*rand()*sin(z);
y = 10*rand()*cos(z);
A{kk} = [x,y,z];
end
A_6 = A(1:6); % generate a cell with the first 6 matrices in "A" cell array
% The numer "6" can be changed to be any number which you want to plot them by colormap
newA = vertcat(A_6{:}); %Concatenating all matrices inside A vertically
numcolors = numel(A_6); %Number of matrices equals number of colors
colourRGB = hsv(numcolors); %Generating colours to be used using hsv colormap
colourtimes = cellfun(@(x) size(x,1),A_6);%Determining num of times each colour will be used
colourind = zeros(size(newA,1),1); %Zero matrix with length equals num of points
colourind([1 cumsum(colourtimes(1:end-1))+1]) = 1;
colourind = cumsum(colourind); %Linear indices of colours for newA
scatter3(newA(:,1), newA(:,2), newA(:,3), [], colourRGB(colourind,:),'filled');
%if you want to specify the size of the circles, use the following line instead:
% scatter3(newA(:,1), newA(:,2), newA(:,3), colourind , colourRGB(colourind,:),'filled');
grid on;
view(3); %view in 3d plane
colormap(colourRGB); %using the custom colormap of the colors we used
%Adjusting the position of the colorbar ticks
caxis([1 numcolors]);
colorbar('YTick',[1+0.5*(numcolors-1)/numcolors:(numcolors-1)/numcolors:numcolors],'YTickLabel', num2str([1:numcolors]'), 'YLim', [1 numcolors]);
I have the image like this:
How can I shown "YTickLabel" with some specified value (not show all) as below figure?
  5 comentarios
ha ha
ha ha el 13 de Abr. de 2018
what do you mean? "save handle for colorbar, cb=colorbar(...)" I add the additional code as follow:
cb=colorbar;
tix=cb.Ticks;
cb.Ticks=tix([1 3 5 6])
cb.TickLabels={'1' '3' '5' '6'};
But, the result is not same as my expect?

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Respuesta aceptada

ha ha
ha ha el 13 de Abr. de 2018
Additional code:
cb = colorbar('YTick',[1:numcolors],'YTickLabel', num2str([1:numcolors]'), 'YLim', [1 numcolors]);
tix=cb.Ticks;
cb.Ticks=tix([1 3 5 6]);
cb.TickLabels={'1' '3' '5' '6'};

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