Duplicate Each Element in a Matrix without using Repelem or Repmat

Is there a way I can make each element in a matrix duplicate into a 3 by 3? if
A=[1 0;0 1]
Is there a way i can turn it into
Aew=[1 1 1 0 0 0;1 1 1 0 0 0;1 1 1 0 0 0;0 0 0 1 1 1;0 0 0 1 1 1;0 0 0 1 1 1]
so basically
Anew=repelem(A,3,3)
but without repelem, repmat or any special functions?
Thanks

1 Comment

Yes, but the only reason I can see for wanting to do so is as a homework question. Like most things you can do it with a for loop.

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 Accepted Answer

a=eye(2);
k=3;
d=blkdiag(ones(k,1),ones(k,1));
b=d*a*d'
%%Or a more generic case
a=[1 2 3;4 5 6];
k=3;
[m,n]=size(a);
d1=ones(k,1)*ones(1,m);
d1=mat2cell(d1,k,ones(1,m));
d1=blkdiag(d1{:});
d2=ones(n,1)*ones(1,k);
d2=mat2cell(d2,ones(n,1),k);
d2=blkdiag(d2{:});
b=d1*a*d2

More Answers (1)

>> A = [1,0;0,1]
A =
1 0
0 1
>> V = ceil((1:6)/3);
>> B = A(V,V)
B =
1 1 1 0 0 0
1 1 1 0 0 0
1 1 1 0 0 0
0 0 0 1 1 1
0 0 0 1 1 1
0 0 0 1 1 1

5 Comments

This way only seems to produce a 2 by 2 everytime though no matter the size of the initial matrix
"This way only seems to produce a 2 by 2 everytime though no matter the size of the initial matrix"
Clearly it works, as the example in my answer shows.
Of course it can be trivially adjusted to replicate the elements of any sized array any number of times, to do so requires two very small changes to the V definition:
k = 3;
V = ceil((1:k*size(A,1))/k);
This is certainly a much simpler and more efficient solution than the one that you accepted.
Ahh ya I did realise your method is simpler but i was having big brainfart at the moment and the solution i accepted I seem to understand how it works better? But thanks for the help anyways :D
@Ayca Yigit: it generates a vector of indices V, where each index repeats k times. Consider what happens when you use those indices to select the elements of A(V,V).

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