how can i get a matrix from an array like this:
2 vues (au cours des 30 derniers jours)
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Starting with
a = [12 21 32];
I would like to get this:
M = [12 21 32; 21 32 12; 32 12 21];
with size and length fun. only
Is it some kind of loop???
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Réponse acceptée
Azzi Abdelmalek
le 2 Jan 2013
Modifié(e) : Azzi Abdelmalek
le 2 Jan 2013
a = [12 21 32 42];
n=length(a)
M=a
for k=1:n-1
c=a(1)
a(1:end-1)=a(2:end)
a(end)=c
M=[M;a]
end
2 commentaires
Walter Roberson
le 2 Jan 2013
This does not satisfy your stated requirements, as assignment and subscripting and [] are all functions in MATLAB: in particular, those operations call upon subsasgn(), subsref(), and vertcat (and horzcat).
Plus de réponses (5)
Roger Stafford
le 2 Jan 2013
M = hankel(a,fliplr(a));
or if you can't use the 'hankel' function do this
n = length(a);
M = reshape(a(mod(floor(((n+1)*(0:n^2-1))/n),n)+1),n,n);
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Azzi Abdelmalek
le 2 Jan 2013
Modifié(e) : Azzi Abdelmalek
le 2 Jan 2013
M=[a ;circshift(a,[0 -1]); circshift(a,[0 -2])]
%or more general
a=[12 21 32 42 52]
M=cell2mat(arrayfun(@(x) circshift(a,[0 -x]),(0:numel(a)-1)','un',0))
Image Analyst
le 2 Jan 2013
I don't know what "with size and length fun. only" means, but the simplest solution is
M = [a(1,1), a(1,2), a(1,3); a(1,2), a(1,3), a(1,1); a(1,3), a(1,1), a(1,2)];
Since you haven't given any other requirements, like it needs to work for other sizes of matrices, this will work exactly as you have said. No functions like toeplitz() or circshift() needed. If you needed other requirements and flexibility, I assume you would have said so (ok, that's probably a bad assumption given the types of posts people make).
4 commentaires
Image Analyst
le 2 Jan 2013
Modifié(e) : Image Analyst
le 2 Jan 2013
OK - I didn't know the acronym HW. Others probably didn't recognize that as homework either, because it looks like three people have already done your homework for you and given you an answer that should work. I'll add the "homework" tag for you. Next time you can add it yourself so you won't be caught copying.
Sean de Wolski
le 2 Jan 2013
M = a(sortrows(bsxfun(@(x,y)mod(x+y,numel(a))+1,1:numel(a),(1:numel(a))')))
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