Apply pattern to Matrix Indexes in For Loop

3 vues (au cours des 30 derniers jours)
Olivia Licata
Olivia Licata le 20 Déc 2017
Commenté : Star Strider le 23 Déc 2017
Hello, I have a list of values in groups of three. Imagine a column [x1 x2 x3... x112146] I would like to plot x versus y of
(x1,x2) (x2, x1), (x1, x3), (x3,x1), (x2, x3), (x3, x2)
The next group of three would be
(x4, x5), (x5,x4), (x6, x4), (x4, x6), (x5, x6), (x6,x5)
The order within groupings doesn't really matter. I was able to create a pattern for my previous plot with groupings of two by establishing indices and then sorting by even or odd. (code below)
xy = zeros(length(za_twos),2);
xy(:,1)=za_twos(:,1);
for i = 2:(length(za_twos)+1)
if mod(i,2)==0
xy(i-1,2)=za_twos(i,1);
else
xy(i-1,2)=za_twos(i-2,1);
end
end
This returned
(x1, x2) (x2, x1) (x3, x4) (x4, x3)
(in two columns representing x and y)

Réponse acceptée

Star Strider
Star Strider le 20 Déc 2017
See if this does what you want:
V = (1:18)'; % Original Vector
Vr = reshape(V, 3, []); % Reshape To (3xN) Matrix
idx = nchoosek(1:3,2); % Create Index Permutations - Half
idx = [idx; fliplr(idx)]; % Create Index Permutations - All
hax = axes('NextPlot', 'add');
for k1 = 1:size(Vr,2)
plot(hax, Vr(idx(:,1), k1), Vr(idx(:,2), k1), 'p')
end
I used a linear vector to test the code, so after you run this to be certain if it produces the result you want, substitute your own vector for ‘V’. (The length of your vector is evenly divisible by 3, so my code here should work with your column vector.) I plotted them all on the same axes.
  2 commentaires
Olivia Licata
Olivia Licata le 23 Déc 2017
Thank you! This did work, it took a little long since the plotting went stepwise through the loop, but the index permutations were very helpful and I was able to adjust this for groupings 4-10 as well. I changed the plot to a scatter plot and adjusted all the points to one color for personal preferences. Thank you so much!
Star Strider
Star Strider le 23 Déc 2017
As always, my pleasure!

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