# How to tell an multi-output anonymous function to vectorize its output components which are constant

1 view (last 30 days)
Mohammad Shojaei Arani on 16 Oct 2021
Hello friends,
I have an apparently simple but very annoying problem. Consider the function handle
f=@(x,y) [x+y;2;x*y;x^2-1;3];
When I want to calculate the expression f([1 2],3) then matlab gives an error message ('mismatch between dimensions'). Of course, I can easily fix this problem by re-defining the above function handle as
f=@(x,y) [x+y;2+x-x;x*y;x^2-1;3+x-x];
However, the problem is that my anonymous function is defined whithin a for-loop where I cannot mannually add x-x to the constants plus that
I do not know beforehand which outputs are constant and which are not. The natural way to fix this problem seems to be to update f like
f=@(x,y) f(x,y)+[0;x-x;0;0;x-x];
But, this trick does not work, unfortunately.
It is greatly appreciated if anybody has an idea how to fix this really annoying problem.
Babak Shojaei Arani

Matt J on 16 Oct 2021
Edited: Matt J on 16 Oct 2021
One way,
f=@(x,y) completeRows({x+y;2;x*y;x.^2-1;3});
f([1 2],3)
ans = 5×2
4 5 2 2 3 6 0 3 3 3
function out=completeRows(C)
n=max(cellfun('length',C));
for i=1:numel(C)
if isscalar(C{i}), C{i}=repelem(C{i},1,n); end
end
out=cell2mat(C);
end
##### 1 CommentShowHide None
Mohammad Shojaei Arani on 13 Nov 2021
Thanks a lot!