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Filling an array with color points

2 vues (au cours des 30 derniers jours)
Ursula Trigos-Raczkowski
Ursula Trigos-Raczkowski le 1 Avr 2022
Commenté : Les Beckham le 1 Avr 2022
I am trying to create an array (matrix?)
I have J curves I am plotting and I want to create a vector of color values.
The matrix will have J rows and 3 entries per row. the First entry will be 0 the third entry will be 1. The second middle entry will be (i-1)*1./J. where i is my counter in the for loop.
```
j = 34; g = 3;
A = zeros(j,g);
for i= 1:j
A(:,i) = 0 (i-1)*1./j 1;
end
disp(A)
```
So I want something like, if j=34 , A=[0 0 1; 0 1/34 1; 0 2/34 1; ... ; 0 33/34 1;]
Thank you for your time and help.

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Réponse acceptée

Les Beckham
Les Beckham le 1 Avr 2022
Modifié(e) : Les Beckham le 1 Avr 2022
Ran in:
j = 34; g = 3;
A = zeros(j,g);
A(:,3) = 1; % put ones in the third column
A(:,2) = ([0:j-1]/j).'; % fill in the second column (' makes it a column)
format rat
disp(A)
0 0 1 0 1/34 1 0 1/17 1 0 3/34 1 0 2/17 1 0 5/34 1 0 3/17 1 0 7/34 1 0 4/17 1 0 9/34 1 0 5/17 1 0 11/34 1 0 6/17 1 0 13/34 1 0 7/17 1 0 15/34 1 0 8/17 1 0 1/2 1 0 9/17 1 0 19/34 1 0 10/17 1 0 21/34 1 0 11/17 1 0 23/34 1 0 12/17 1 0 25/34 1 0 13/17 1 0 27/34 1 0 14/17 1 0 29/34 1 0 15/17 1 0 31/34 1 0 16/17 1 0 33/34 1
  2 commentaires
Ursula Trigos-Raczkowski
Ursula Trigos-Raczkowski le 1 Avr 2022
Thank you!
Les Beckham
Les Beckham le 1 Avr 2022
You are quite welcome.

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Plus de réponses (1)

Voss
Voss le 1 Avr 2022
Ran in:
I think this is what you were going for:
j = 34; g = 3;
A = zeros(j,g);
for i= 1:j
A(i,:) = [0 (i-1)/j 1];
end
disp(A)
0 0 1.0000 0 0.0294 1.0000 0 0.0588 1.0000 0 0.0882 1.0000 0 0.1176 1.0000 0 0.1471 1.0000 0 0.1765 1.0000 0 0.2059 1.0000 0 0.2353 1.0000 0 0.2647 1.0000 0 0.2941 1.0000 0 0.3235 1.0000 0 0.3529 1.0000 0 0.3824 1.0000 0 0.4118 1.0000 0 0.4412 1.0000 0 0.4706 1.0000 0 0.5000 1.0000 0 0.5294 1.0000 0 0.5588 1.0000 0 0.5882 1.0000 0 0.6176 1.0000 0 0.6471 1.0000 0 0.6765 1.0000 0 0.7059 1.0000 0 0.7353 1.0000 0 0.7647 1.0000 0 0.7941 1.0000 0 0.8235 1.0000 0 0.8529 1.0000 0 0.8824 1.0000 0 0.9118 1.0000 0 0.9412 1.0000 0 0.9706 1.0000
And you can do it without the loop like this:
A = [zeros(j,1) (0:j-1).'/j ones(j,1)];
disp(A)
0 0 1.0000 0 0.0294 1.0000 0 0.0588 1.0000 0 0.0882 1.0000 0 0.1176 1.0000 0 0.1471 1.0000 0 0.1765 1.0000 0 0.2059 1.0000 0 0.2353 1.0000 0 0.2647 1.0000 0 0.2941 1.0000 0 0.3235 1.0000 0 0.3529 1.0000 0 0.3824 1.0000 0 0.4118 1.0000 0 0.4412 1.0000 0 0.4706 1.0000 0 0.5000 1.0000 0 0.5294 1.0000 0 0.5588 1.0000 0 0.5882 1.0000 0 0.6176 1.0000 0 0.6471 1.0000 0 0.6765 1.0000 0 0.7059 1.0000 0 0.7353 1.0000 0 0.7647 1.0000 0 0.7941 1.0000 0 0.8235 1.0000 0 0.8529 1.0000 0 0.8824 1.0000 0 0.9118 1.0000 0 0.9412 1.0000 0 0.9706 1.0000
  2 commentaires
Ursula Trigos-Raczkowski
Ursula Trigos-Raczkowski le 1 Avr 2022
Thank you!
Voss
Voss le 1 Avr 2022
You're welcome!

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