How to assign values to an array with broadcasting

Is there any way to do broadcasting while assigning values to an array?
I know how to broadcast a computation:
G = zeros(3)
G + [1 2 3]
ans =
1 2 3
1 2 3
1 2 3
But I want to assign the above values to G directly:
G(:,:) = [1 2 3]
Unable to perform assignment because the size of the left side is 3-by-3 and the size of the right side is 1-by-3.
I know how to broadcast a simple scalar value:
G(:) = 1
G =
1 1 1
1 1 1
1 1 1
And I realise I could do this instead:
G(:,:) = repmat([1 2 3], 3, 1)
G =
1 2 3
1 2 3
1 2 3
But I wondered if there is a simpler way to assign a vector of values to all the rows using automatic broadcasting.
Judging by the answers to this question, I am guess not, but I wanted to check.

 Accepted Answer

madhan ravi
madhan ravi on 28 Jul 2020
Edited: madhan ravi on 28 Jul 2020
Your way of using repmat() is the easiest way.

9 Comments

Why not simply
G = zeros(3) + (1:3) % ?
Or
kron(repelem(1,3,1),1:3)
The reason I want to avoid G = zeros(3) + (1:3) is for performance reasons—to avoid the possibility that it destroys the existing memory for G and reallocates it. I'm used to Python where you always do: G[:] = [1,2,3] (assign new values) if G is very big rather than: G = np.repeat([[1, 2, 3]],3,axis=0) (create a new array in memory).
Seems like G = repmat([1 2 3], 3, 1) is the best solution. As far as I can see, none of these methods assign new values directly to the original array without first creating a whole new array of the same size for the right-hand-side (so there is no point in doing G[:,:] = ...). I ran some speed tests with a large G array and the repmat method is the fastest, followed by kron(repelem(n,10,1),1:10). Thanks!
Ah perhaps the implicit expansion:
zeros(3, 1) + (1:3)
Interesting result I just discovered:
G = zeros(500000,10)
is about twice as fast as
G(:) = 0
(for the same size G). So I guess, assignment of new values is slower in MATLAB than building a new array anyway!
It's surprising* and it means that my idea of avoiding allocating new memory by somehow assigning to the original array probably wouldn't have speeded up the code even if it were possible (* Example: allows a 1.6x speedup in Python).
For any dimension matrix,you can do
X=repmat(fading,20,1)
where fading is a matrix of any size. For me it was fading coefficient matrix
repmat was discussed above. Bill was hoping for something more efficient.

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