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if NaN then...

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Clarisha Nijman
Clarisha Nijman 2019 年 1 月 2 日
コメント済み: Clarisha Nijman 2019 年 1 月 4 日
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.

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採用された回答

Steven Lord
Steven Lord 2019 年 1 月 2 日
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

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Clarisha Nijman
Clarisha Nijman 2019 年 1 月 3 日
Thanks a lot!!!!,
This works for me the best! I agree really straight forward!
Steven Lord
Steven Lord 2019 年 1 月 3 日
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
Clarisha Nijman 2019 年 1 月 4 日
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.

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その他の回答 (1 件)

GT
GT 2019 年 1 月 2 日
編集済み: per isakson 2019 年 1 月 3 日

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Clarisha Nijman
Clarisha Nijman 2019 年 1 月 3 日
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?

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