Discrepancy in sparse matrix math, when NaN's present

I expect result1 and result2 below to be identical, but they aren't. The discrepancy must be a bug, right? I'm working in R2024b, but the Run output below shows the issue exists as well in whatever the Matlab online engine is now running.

n = 5; m = 3;

S = sparse(n, m);
v=nan(n, 1); 
D=sparse(diag(v));

result1=full(v.*S)  %correct

result1 = 5×3

NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN

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result2=full(D*S) %incorrect

result2 = 5×3

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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[Matt J]

A possibly related observation. Below, I get different results depending only on whether S is in full or sparse form:

n = 5;
m = 3;
S = sparse(n, m);

v=nan(n, 1); 
D=sparse(diag(v));

result3 = full(D*full(S))

result3 = 5×3

NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN

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result4 = full(D*S)

result4 = 5×3

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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[Walter Roberson]

Are you sure that you want I and J to be empty, making S all-zero ?

randperrm(something, something_bigger) gives the empty array.

[Matt J]

Yes, I am sure I want S to be all zero. I have now simplified the code to reflect that.

[Walter Roberson]

It looks like the key is that S is sparse in D*S

n = 2; m = 1;

S = sparse(n, m);
v = zeros(n,1);
v(1) = nan;
D=sparse(diag(v))

D = 2×2 sparse double matrix (1 nonzero)
   (1,1)      NaN

result1=full(v.*S)  %correct

result1 = 2×1

NaN 0

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result2a=full(D)*(S) %incorrect

result2a = 2×1

0 0

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result2 = full(result2a)

result2 = 2×1

0 0

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[Matt J]

Yes, it seems that when S is sparse, all NaNs in the left operand D of D*S are treated as finite, nonzeros.

n = 5;
S = sparse(n, n);

v=nan(n, 1); 
D=sparse(diag(v));
 
testOps(D,S)

ans = "D=full  S=full"

DmtimesS = 5×5

NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN

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ans = "D=full  S=sparse"

DmtimesS = 5×5

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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ans = "D=sparse  S=full"

DmtimesS = 5×5

NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN

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ans = "D=sparse  S=sparse"

DmtimesS = 5×5

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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function testOps(D,S)

  ops={@full,@sparse};

  for i=1:2

      L=ops{i};

      for j=1:2

          R=ops{j};
          
          "D="+func2str(L)+"  S="+func2str(R)

          DmtimesS=full( L(D)*R(S) ),

      end
  end


end

[dpb]

As an observation point, the same behavior exists at least as far back as R2017b; one might guess it would go back much further.

[Matt J]

@dpb Yes, I think there's little doubt that it has always been there. It's clear, from a performance optimization point of view, why you mgiht want sparse mtimes() to ignore NaNs. The whole idea of sparse type is to exploit the fact that 0*somethingFinite=0.If you have to worry about "somethingFinite" being non-finite, I wonder how much of the intended performance can be salvaged.

[dpb]

There was a similar thread on sparse performance here just a week or so ago where it ended up being full for similar reasons. I looked but didn't find it in a cursory search, iirc, it was @Steven Lord who pointed out the issue about timing.