Effacer les filtres
Effacer les filtres

creating a vector from a(:) : b(:) , but one for each index without using loops.

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kjetil87
kjetil87 le 1 Nov 2012
I have two vectors a=[1,2,3] and b=[6,4,9]. My goal is to generate this(see below) without the use of foor loops (a single command line);
result=[a(1):b(1),a(2):b(2),a(3):b(3)];
The problem is that a and b is varying in size (but size(a)=size(b)=[1,N]).
I`ve tried
result=[a(:):b(:)];
and other variations, but it doesnt do the job. Any idea of how i can achieve this?

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 1 Nov 2012
Modifié(e) : Andrei Bobrov le 1 Nov 2012
a=[1,2,3];
b=[6,4,9];
result = cell2mat(arrayfun(@(x,y)x:y,a,b,'un',0));
ADD
a = randi(150,32,4,4,4); % your 4-D array
idx = {1:32,3,2,1}; % your array with indexs
[i1,i2,i3,i4] = ndgrid(idx{:});
i1 = cell(1,4);
[i1{:}] = ndgrid(idx{:});
result = a(sub2ind(size(a),i1{:}));
  8 commentaires
Sean de Wolski
Sean de Wolski le 1 Nov 2012
@Kjetil87, probably not, ind2sub is slow and and gets much slower for big data.
Use the for-loop just make pull out as many computations as you can which it looks like you've done already.
kjetil87
kjetil87 le 1 Nov 2012
for those interested i found a way that works.
siz=size(psi_table);
Xn(:,3)=(Xn(:,3)-1)*prod(siz(1:3)); % (l-1)(d3)(d2)(d1)
Xn(:,2)=(Xn(:,2)-1)*prod(siz(1:2)); % (k-1)(d2)(d1)
Xn(:,1)=(Xn(:,1)-1)*siz(1); % (j-1)(d1)
idx=sum(Xn,2)+1; % +1 to start at 4D_matrix(1,j,k,l)
idx_V=repmat(0:31,length(idx),1)+repmat(idx,1,32);
idx_V=idx_V';
idx_V=idx_V(:)';
psi=psi_table(idx_V);
% all operations but the repmat line is basicly 0 cost. And repmat is considerably faster then the for loop indexing ( total of aprox 0.6secs vs 6secs for 2263 calls with a big matrix )

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