Cannot plot a color gradient (Data must be a single matrix Y or a list of pairs X,Y.)

Hi, i want to plot on the same graph 30 curves with colour gradient decreasing from the first to the last curve (from yellow to red). I have tried this code :

c = colormap(jet(1000))

figure

plot((1:num_images),augm(1:30,1:num_images),'Color',c(600:10:880,:)','-')

• My variable augm is a matrix of num_images*30.
• When I run the program without the color part ('Color',c(600:10:880,:)'), it runs perfectly fine but i do not have my gradient of colours...

I think I did not understand completely the 'Color' function.

Could you help me with this ?

Thank you !

Answer 1

[Walter Roberson]

When you use the 'color' name/value pair, it applies to all lines that you drew in the single plot() call. It is not possible to use 'color' to give different colors for different lines.

If you draw multiple lines with one plot() call, then to give them different RGB colors, you need to loop through the graphics handles setting the Color property.

num_plot = 30;
h = plot(augm(1:num_plot,1:num_images) .', '-');    %make the different plots into columns
color_idx = 600 + 10 * (0:num_plot-1);
for K = 1 : length(h)
  set(h(K), 'Color', c(color_idx(K),:) );
end

[Stephen23]

"then to give them different RGB colors, you need to loop through the graphics handles setting the Color property."

Not true. See my answer.

Answer 2

[Stephen23]

You can easily set the line ColorOrder property of the axes, then you just need one plot call:

>> map = [0,0,1;1,0.4,0.6]; % e.g. cool(2)
>> axes('ColorOrder',map,'NextPlot','replacechildren')
>> plot([0,1],[1,2],'-',[0,1],[2,1],'*-')

[Walter Roberson]

Good point.

Some more information on ColorOrder is at https://www.mathworks.com/help/matlab/graphics_transition/why-are-plot-lines-different-colors.html

Answer 3

[fish minox]

Hi Stephen, hi Walter,

Thank you both for your answers, i did not know that the color feature was applied to all curves at the same time. thank you again for the great and fast answers !