I want to copy some data

I want to copy some data

if i have a data(matrix) like this ,,

2 3 4
6 2 3
9 8 3

I would like to copy this data like this ,,

2 3 4
2 3 4
2 3 4
2 3 4
6 2 3
6 2 3
6 2 3
6 2 3
9 8 3
9 8 3 
9 8 3
9 8 3

This is easy to solve if I use the 'for loop' But I will work with very large data ,so speed is very important for me... Is there any way to solve this problem without 'for loop'??

[Adam]

Where speed is critical I would always recommend putting together a script/function to test the various different alternatives on different data sizes because results can often be unexpected with regard to how fast certain implementations are.

Creating a quick function with subfunctions that implement each of the alternatives to use with

doc timeit

is very simple. All you need then is the ideas for alternative implementations, of which you now have 4 (always include the for loop option for timing too - it isn't guaranteed not to be fastest)

Answer 1

[Adam]

a = [2 3 4; 6 2 3; 9 8 3];
b = repelem( a, 4, 1 );

If you are using R2015a or later. I haven't tested it for speed against a for loop though. As a built-in Matlab function it should be faster, but no guarantees.

[Jan]

+1. This is the job of repelem. I cannot compare the speed currently with the less nice:

Result = M(repmat(1:size(M,1), N, 1), :);

Answer 2

[José-Luis]

a = [2 3 4; 6 2 3; 9 8 3];
idx = ndgrid(1:3,1:4)';
result = a(idx(:),:)

Edit: Felt the nagging need for a one-liner, not repeating existing answers:

 result = a(idivide(int32(0:11),4)+1,:)

Answer 3

[Jan]

N = 4;
M = [2,3,4;6,2,3;9,8,3];

index  = repmat(1:size(M,1), N, 1);
Result = M(index, :);

Or:

kron(M, ones(M, 1))

The indexing should be efficient, while I expect kron to be remarkably slower. repelem is the command explicitly defined for this job, so I prefer Adam's solution if you have a modern Matlab version. If this is the bottleneck of the code, it is worth to check, which solution is the fastest.