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Re: Alternative solution for NAN
Posted:
Feb 27, 2013 6:08 AM


"Torsten" wrote in message <kgko7b$e4d$1@newscl01ah.mathworks.com>... > "Carl S." wrote in message <kgkmon$aeo$1@newscl01ah.mathworks.com>... > > "Torsten" wrote in message <kgklbh$6ph$1@newscl01ah.mathworks.com>... > > > "Carl S." wrote in message <kgki2g$rie$1@newscl01ah.mathworks.com>... > > > > The following code gives NAN (Not a Number) values > > > > [U,D]=eig(N); > > > > > > > > To solve this problem, I wrote that > > > > while(det(N) == 0) > > > > N=(1e10.*randi(1,size(N)))*eye(size(N)); > > > > end > > > > > > > > But, the loop does not stop > > > > > > Your matrix N within the loop always has determinant (1e10)^(size(N)) > > > which may become very small if N is large. > > > > > > :( Are there any alternative solution instead of this loop to solve the NAN problem ? > > > > > > Depends on the original matrix N. > > > > > > Best wishes > > > Torsten. > > > > Dear Torsten, > > The matrix N has standard deviation values of grayscale images. So, it changes for each image. How to solve the NAN problem in this case ? > > Did you check whether the matrix N already contains NaN values ? > > Best wishes > Torsten.
Yes, I have checked with this codes
if isnan(N)==1 h=1; else h=0; end
This code return that h=0



