repeating elements of a vector

I have two vectors, say lambda=[1; 2] and ng=[3;4]. I want to have a long vector where the elements of lambda are repeated the corresponding times of the values of ng, i.e., I want to get
lambda_long=[1;1;1;2;2;2;2]
I can do it with some for loop, but is there any other efficient way of doing this?

 Respuesta aceptada

Oleg Komarov
Oleg Komarov el 16 de Mayo de 2012
Fully vectorized run-length decoding:
% Example inputs (as vector columns)
lambda = [2; 4;9];
ng = [3;4;7];
% Preallocate output
out = zeros(sum(ng),1);
% Distribute starting point of sequences
csng = cumsum(ng);
out([1; csng(1:end-1)+1]) = [lambda(1); diff(lambda)];
% Propagate
out = cumsum(out)

Más respuestas (3)

Andrei Bobrov
Andrei Bobrov el 16 de Mayo de 2012

3 votos

x = arrayfun(@(a,b)a(ones(b,1)),lambda,ng,'un',0)
lambda_long = cat(1,x{:})
add
eg:
lambda = [14 5 89 47]'
ng=[3 4 2 4]'
% solution
id = zeros(sum(ng),1);
id(cumsum(ng)-ng+1)=1;
lambda_long = lambda(cumsum(id));
Jan
Jan el 16 de Mayo de 2012
Similar to Oleg's encoding, but avoid DIFF/CUMSUM of data to prevent rounding errors:
lambda = [2; 4; 9];
ng = [3; 4; 7];
index(cumsum(ng)+1) = 1;
index(1) = 1;
index(end) = [];
out = lambda(cumsum(index));
Untested!
Wayne King
Wayne King el 16 de Mayo de 2012

0 votos

One way
lambda=[1; 2];
ng=[3;4];
A = repmat(lambda(1),ng(1),1);
B = repmat(lambda(2),ng(2),1);
C = [A;B];
Of course, you could put this on one line.
C = [repmat(lambda(1),ng(1),1); repmat(lambda(2),ng(2),1)];

1 comentario

Rabeya
Rabeya el 16 de Mayo de 2012
this is fine for such low length lambda, but not when you have 50 values for lambda!

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el 16 de Mayo de 2012

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