if NaN then...

Dear all,

I have a function that returns me a PMatrix. But sometimes that matrix is full with nan's. The cause is known, but the best solution I found is to allow the function to compute such kind of "nan" matrices and to replace that "NaN matrix in the end by the identity matrix.

This is my actual code that works fine, uptill now.

 Som=sum(sum(PMatrix));
    Som(isnan(Som))=1;    
    if Som==1
        PMatrix=eye(k,k);
    end

I find this code clumsy and would prefer a more straight forward code saying:

if sum(sum(PMatrix))=NaN
    PMatrix=eye(k,k)
end

But I can't figure out how to compose such a straight forward code. Does anyone has a suggestion for me?

Thank you in advance.

Answer 1

[Steven Lord]

Which release are you using? For release R2018b or later:

if any(isnan(PMatrix), 'all')
    
end

For release R2018a or earlier:

if any(isnan(Pmatrix(:)))
    
end

[per isakson]

Don't fight Matlab. Trust Steven Lord. His code looks "straight forward" enough to me.

The code D==nan just doesn't work the way you want.

>> D=nan;
>> D==nan
ans =
  logical
   0
>> 

[Clarisha Nijman]

Thanks a lot!!!!,

This works for me the best! I agree really straight forward!

[Steven Lord]

As per isakson said, NaN is not equal to any value, not even another NaN. That's by design.

NaN == NaN % returns false

[Clarisha Nijman]

Oh, now I see, it can not be seen as a number nor a string. By the way, your second suggestion worked for me! I accidently omiited the colon (:). It should be right in the place where you put it.

Answer 2

[Gareth]

Have you tried using the function isnan?

https://nl.mathworks.com/help/matlab/ref/isnan.html?searchHighlight=isnan&s_tid=doc_srchtitle

[Clarisha Nijman]

Yes, but I do not want to transfer the entries in ones, It would imply a matrix full of ones. I just need the identity matrix. I figured already out that sum(sum(P))==nan. So an if loop saying:

D+sum(sum(P))==nan
if D==nan
P=eye(k)
end

but it does not work. The if statement is the problem. How should it be corrected?