dlgradient of a subset of variables

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Matt J
Matt J on 9 Feb 2024
Edited: Matt J on 14 Feb 2024
I would like to apply dlgradient to a parametrized function, which would allow me in some cases to get the complete gradient of the function, but in others would take only the derivative with respect to the i-th variable. My attempt below fails. Is there a way that both succeeds and which would be efficient in the more general case when numel(x0) is large?
x0 = dlarray( [1; 2] );
i=[];
y_grad=dlfeval(@(x)oneDeriv(x,i), x0) %works
y_grad =
2×1 dlarray 1 2
i=2;
dy_dxi=dlfeval(@(x)oneDeriv(x,i), x0) %fails
dy_dxi =
1×1 dlarray 0
function out = oneDeriv(x,i)
y=sum(x.^2)/2;
if isempty(i) %take complete gradient
out=dlgradient(y,x);
else %take gradient only w.r.t x(i)
out=dlgradient(y,x(i));
end
end

Accepted Answer

Matt J
Matt J on 13 Feb 2024
Edited: Matt J on 14 Feb 2024
This seems to work:
X0 = dlarray( [1; 2; 3; 4]*10 );
subset=[2,3];
gradTotal = getGradient(X0) %total gradient
gradTotal =
4×1 dlarray 10 20 30 40
gradSubset = getGradient(X0,2:3) %gradient on subset of x
gradSubset =
2×1 dlarray 20 30
function grad = getGradient(Xall,subset)
if nargin<2, subset=':'; end
grad = dlfeval( @(xsub)theFunction(X,xsub,subset), Xall(subset));
end
function grad = theFunction(Xall,xsub,subset)
Xall(subset)=xsub;
y=sum(Xall.^2)/2;
grad=dlgradient(y,xsub);
end

More Answers (1)

Ben
Ben on 13 Feb 2024
This is a subtle part of the dlarray autodiff system, the line dlgradient(y,x(i)) returns 0 because it sees the operation x -> x(i), and only knows that y depends on x, not x(i).
You can work around this by computing all the derivatives and do indexing afterwards:
function out = oneDeriv(x,i)
y = sum(x.^2)/2;
out = dlgradient(y,x);
if ~isempty(i)
out = out(i);
end
end
  1 Comment
Matt J
Matt J on 13 Feb 2024
Edited: Matt J on 13 Feb 2024
Thanks, Ben. So the answer is no?

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