Repeat element of a vector n times without loop.

Say I have a column vector x=[a;b;c]. I want to repeat each element n times to make a long length(x)*n vector. For example, for n=3, the answer would be:

ans=
a
a
a
b
b
b
c
c
c

Can anyone think of an elegant way to do this without looping?

Thanks,

Justin

[John]

U can use repmat it not exactly elegant but it will do the job

x=[a;b;c]; n=3;

newx = [repmat(x(1),n,1);repmat(x(2),n,1);repmat(x(3),n,1)]

Answer 1

[Azzi Abdelmalek]

 n=3 ; x=(1:3)' % example
 r=repmat(x,1,n)';
 r=r(:)'

[Jan]

This is more efficient than KRON. Small improvement: r = repmat(x, n, 1); without tranposing.

[Azzi Abdelmalek]

%you mean

r = repmat(x', n, 1)

[Jan]

I guess, you are right. repmat(1:3, 1, 2) = [1,2,3,1,2,3] but the OP wants [1,1,2,2,3,3]. Then r = repmat(1:3, 2, 1); r = r(:) avoid the expensive transposition of the matrix. Well, I admit that even reading this message will waste more time then millions of matrix transpositions will cost...

Answer 2

[jack]

I would use

repelem(X,3,1)

[Yuzhen Lu]

Very neat answer!

[Arif Billah]

This should be chosen as the best 'correct' answer, thanks!

[DGM]

This is probably the more accepted answer today (hence the upvotes), but repelem() was not available until after the question was originally answered (R2015a).

Answer 3

[Walter Roberson]

kron(x, ones(n,1))

[Vinay Chakravarthi]

Thx man.......

[Phat Nguyen]

Very nice man

[Abdelrahman Abdeltawab]

Dear Walter Roberson,

why you did not use outer product and you chosen kronecker ( just curious ) because the guy's question was having vectors ?

[Walter Roberson]

The * matrix multiplication operator cannot by itself repeat elements. You would need something like

(x.' * repmat(eye(length(x)), 1, n)).'

if you wanted to use the * operator to duplicate elements -- forcing you to call upon repmat() to duplicate elements.

Using the kronecker is a known idiom for duplicating data. It can be used for non-vectors too.

>> kron([1 2;3 4], ones(3,1))
ans =
     1     2
     1     2
     1     2
     3     4
     3     4
     3     4

Answer 4

[Kevin Moerman]

There is several others ways of doing it which in some cases are more efficient. Have a look at what the size of your vector is and compare the methods. Below I compare speeds and it appears that on my computer the third and fourth methods are mostly faster for large arrays.

n=100000; x=1:3; 
a=zeros(n,numel(x)); b=a; c=a; d=a; %memory allocation
tic; a=repmat(x, n, 1); t1=toc; %Repmat method
tic; b=kron(x, ones(n,1)); t2=toc; %kron method
tic; c=x(ones(1,n),:); t3=toc; %indexing method
tic; d=ones(n,1)*x; t4=toc; %multiplication method

Kevin

[Vinay Chakravarthi]

Thx Man..

[Walter Roberson]

format long g
n=100000; x=1:3; 
a=zeros(n,numel(x)); b=a; c=a; d=a; %memory allocation
tic; a=repmat(x, n, 1); t1=toc %Repmat method

t1 = 
                  0.000543

tic; b=kron(x, ones(n,1)); t2=toc %kron method

t2 = 
                  0.006106

tic; c=x(ones(1,n),:); t3=toc %indexing method

t3 = 
                  0.002276

tic; d=ones(n,1)*x; t4=toc %multiplication method

t4 = 
                  0.001798

Answer 5

[Justin Solomon]

Thanks guys, these all work perfectly!

Answer 6

[Jianshe Feng]

y = repmat(x,1,3); y = transpose(y); y = y(:);

Answer 7

[Jianshe Feng]

ind = [1;1;1;2;2;2;3;3;3]; x(ind)

[Walter Roberson]

Ah, but how do you construct the ind vector for general length n repetitions ?