MATLAB: Understanding repmat(6,,(1:n*N),1)

repmatvector

If i compute

repmat((1:n*N),6,1)

with n=3 and N=3 the complier gives me a 6X9 matrix.

Why if a create a

repmat (6,(1:n*N),1)

it gives me a 9-D double?

9-D is nine dimensions?
What the compiler is doing?

Best Answer

Have you even considered looking at the documentation? It really helps answering your question.

Looking at the documentation, I could find how supplying a vector as the second argument would work, but I get a warning. "repmat(A,M,N) where M or N is a row vector will return an error in a future release. Use repmat(A,[M,N]) instead." This hints at what is happening: it concatenates all inputs into 1 vector.

The first argument is the scalar/vector/matrix that you want to replicate. Then you have 3 options:

using 1 value, which will result in that many replications in the first two dimensions (rows and cols)
using 1 vector, which specifies the number of replications in each dimension (e.g. [1 1 3] will replicate a value/vector/matrix into the 3rd dimension, useful for going from greyscale to RGB images)
splitting the vector over separate arguments

Knowing this it is obvious why your results are what they are. The first replicates a 1x9 vector (1:9) 6 times in the first dimensions (the rows), so you get a 6x9. The second replicates a scalar (so 1x1), using [1:9,1] as the repetition scheme, so the result is a 1x2x3x4x5x6x7x8x9x1 matrix, which is an 9-dimensional matrix (as any trailing x1 can be ignored).

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Are you sure it's repmat that you want? The calling syntax looks like you were trying to use reshape() instead. If you're sure repmat was intended, then this might be what you were trying to do,

   imseg(idx, :) = repmat(color(i, :), [length(idx),1]);

MATLAB: How does Squeeze work

In addition to what James said, just look at an example or two. Look what happens to the dimensions of A with squeeze, particularly what happens to the dimension with the 1 (so-called singleton dimensions).

A = rand(6,7,1,2);  % size(A) = [6 7 1 2]
size(squeeze(A))
A = rand(1,6,7,2);
size(squeeze(A))
A = rand(6,7,1,2);
size(squeeze(A))

Now try it with two (or more) 1 (singleton) dimensions.

A = rand(6,7,1,2,1,7);  % size(A) = [6 7 1 2 1 7]
size(squeeze(A))
A = rand(1,6,1,7,2,1,8); % size(A) = [1 6 1 7 2 1 8]
size(squeeze(A))
A = rand(6,1,7,1,2);
size(squeeze(A))

I think you see the pattern. Now that you see what is going on with the dimensions, look at the data (using simpler examples):

A = reshape(randperm(6),2,1,3)  % Look at this data
squeeze(A) % Now look at it.
A = reshape(randperm(6),1,2,3)
squeeze(A)

So wee see that SQUEEZE reads off the columns of A while creating an array with no singleton dimensions.

Best Answer

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