Filter löschen
Filter löschen

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

1 Ansicht (letzte 30 Tage)
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?

Akzeptierte Antwort

Andrei Bobrov
Andrei Bobrov am 1 Nov. 2012
Bearbeitet: Andrei Bobrov am 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 Kommentare
Sean de Wolski
Sean de Wolski am 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 am 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 )

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by