How to change plot marks with each step

Hello everyone,

So I have the following code. What I want is to plot the same colors at marks under the same "i" value. So for example if i =200 then the marks are blue, if i= 300 all marks are red etc. Does anybody know how to do that? I tried several things but didn't work.

R = 83.1446; 
b = 58; 
for i = 200:100:1000 
  T = i + 273.15; 
  d = (9380 - (8.53.*T)) 
  c = (28.31 + (0.10721.*T)) 
  e = (-368654 + (715.9.*T)) 
  for  v = 150:1:250;
     y = b./(4*v);
     a = c + d./v + e./(v.^2);
     P = (R*T.*(1 + y + y.^2 - y.^3))./(v.*(1-y).^3)- a./(sqrt(T)*v.*(v+b));    
     Z = (1+y+y.^2-y.^3)./((1-y).^3) - a./(R*T^1.5.*(v+b));
     lnfi =(8*y-9*y.^2+3*y.^3)
     fi=exp(lnfi);
     f=abs(fi.*P); 
     plot(P,f,'.')
    drawnow
    hold on
  end
end
hold off

[dpb]

In plot(P,f,'.') there's no reference to color and so you'll just get default. Make up a color list for each i and use it:

If you use plot, each plot call is a new line handle; and a line can only have one color. You could simplify that process by rearranging to save the values for each T and plot all at one time with the chosen color. That could also eliminate one loop by using the vectorized form of the equations you've written as

  v=150:250;
  y = b./(4*v);
  a = c + d./v + e./(v.^2);
  P = (R*T.*(1 + y + y.^2 - y.^3))./(v.*(1-y).^3)- a./(sqrt(T)*v.*(v+b));    
  Z = (1+y+y.^2-y.^3)./((1-y).^3) - a./(R*T^1.5.*(v+b));
  lnfi =(8*y-9*y.^2+3*y.^3)
  fi=exp(lnfi);
  f=abs(fi.*P); 
  hL(i)=plot(P,f,'.',clr(i,:));

Where clr(i,:) is the defined color triplet for ith T you'll have defined before beginning the loop.

[Dimitris Moutzouris]

I get an error which says "Index exceeds matrix dimensions.". Do you maybe know why is that?

[dpb]

Because you buried data in the for...end loop on i and both Ameer and I fell into the trap...

Instead of

for i = 200:100:1000 
  T = i + 273.15; 
  ...

use something like

T0=[200:100:1000];   % put the temperature data out of code so easily changed
for i = 1:numel(T0)
  T = T0(i) + 273.15;       % do the units conversion
  ...

Would be even better to vectorize, but above is the least editing of existing code to make data independent.

[Dimitris Moutzouris]

Thank you so much dpb !! I was planning to make that change as well but I was really frustrated that i couldn't solve the color problem haha!

I still have many things to learn.

[Ameer Hamza]

dbp, that correct. I didn't notice that at first, too. :D

Answer 1

[Ameer Hamza]

See the twi lines i changed in your code. You need to input colors as RGB values.

R = 83.1446; 
b = 58;
colors = rand(9,3); % <---- RGB value of colors. Each row contain one color.
count = 1;
for i = 200:100:1000 
  T = i + 273.15; 
  d = (9380 - (8.53.*T)) 
  c = (28.31 + (0.10721.*T)) 
  e = (-368654 + (715.9.*T)) 
  for  v = 150:1:250;
     y = b./(4*v);
     a = c + d./v + e./(v.^2);
     P = (R*T.*(1 + y + y.^2 - y.^3))./(v.*(1-y).^3)- a./(sqrt(T)*v.*(v+b));    
     Z = (1+y+y.^2-y.^3)./((1-y).^3) - a./(R*T^1.5.*(v+b));
     lnfi =(8*y-9*y.^2+3*y.^3)
     fi=exp(lnfi);
     f=abs(fi.*P); 
     plot(P,f,'.', 'Color', colors(count,:)); % <---- change here
    drawnow
    hold on
  end
  count = count + 1;
end
hold off

[Dimitris Moutzouris]

I get an error,like the code in the comment above which says that the index exceeds matrix dimensions

[Ameer Hamza]

Can you paste your code here? Note that number of rows in the colors matrix should be equal to the iterations of for loop.

[Dimitris Moutzouris]

Well, I used the code that you gave me above in the answer. The i iterations are 9 in total And you have inserted 9 rows in the colors matrix. So I don't get why it doesn't work :( .

[Ameer Hamza]

Ok. I found a mistake in my code. Please check the updated code.

[Dimitris Moutzouris]

Worked perfectly! So what you did is that you entered one color in each row and then you moved to the next one with the "count +1" .

[Ameer Hamza]

Yes, thats correct.