# efficient way of transforming vector

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Rien am 16 Jul. 2013
Hi,
I have a rather simple question on vector manipulation. I want to do some matrix manipulation as efficiently as possible and I think I found a solution, but therefore I have to do some vector manipulation. For example, I want to transform vector A into vector B:
A =
2
1
0
3
B =
1
1
2
4
4
4
So each element of A shows how many times the number 1,2,3,4... should be repeated. Is there a way to do this transformation in an efficient way without setting up a loop myself?
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Jos (10584) am 16 Jul. 2013
Just for your information, this is an example of run-length decoding. There are very efficient contributions available on the Matlab File Exchange for encoding and decoding run lengths (see, e.g., RUDE).
the cyclist am 16 Jul. 2013
@Jos, good comment. My first thought on seeing this question is that I was pretty sure that it was covered in the classic "tips and tricks" document of Peter Acklam (still available on the web here: http://home.online.no/~pjacklam/matlab/doc/mtt/doc/mtt.pdf).
I was correct in my memory, but it turned out that the three algorithms he offers do not correctly handle cases where there is a zero run-length.

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### Akzeptierte Antwort

Andrei Bobrov am 16 Jul. 2013
A =[2
1
0
3];
n = A ~= 0;
k = cumsum(A);
k1 = k(n);
s = (1:numel(A))';
s1 = s(n);
k2 = k1 - A(n) + 1;
ii = zeros(k(end),1);
ii(k2) = 1;
idx = cumsum(ii);
out = s1(idx)
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### Weitere Antworten (2)

Azzi Abdelmalek am 16 Jul. 2013
Bearbeitet: Azzi Abdelmalek am 16 Jul. 2013
B=cell2mat(arrayfun(@(x) x*ones(A(x),1),1:numel(A),'un',0)')
%or a for loop which is much faster
B=[]
for k=1:numel(A)
B(end+1:end+A(k),1)=k*ones(A(k),1)
end
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Lokesh Ravindranathan am 16 Jul. 2013
b is the resultant vector
b = [];
j = 1;
for i = 1:numel(a)
b(j: j + a(i) - 1) = i;
j = j+ a(i);
end
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