Using NaN in if and interp1 commands

Hi everbody, I have 6 values and some values are NaN. If interp1 command finds the final value as NaN for giving ' c' value, MATLAB must give me error message.

a=[1 2 3 4 5 6];
b=[10 15 20 NaN NaN NaN];
c=[1.5 4 3.5 4.5 5.1 5.9];
for g=1:1:numel(a)
if interp1(a,b,c(g))==NaN;
        disp('There is a mistake here')
   end
end

But MATLAB does't run this command with NaN values.

 *In interp1 at 178
Warning: NaN found in Y, interpolation at undefined values
   will result in undefined values.*

How can i achieve this problem? How can i use NaN values in interp1 and if command? I get the problems when i use isnan.

Thanks...

[per isakson]

c(g) is a scalar. Why not replace

if interp1(a,b,c(g))==NaN;

by

if isnan( interp1(a,b,c(g)) )

?

[Volkan Yangin]

Thank you per isakson. I am using the command you sent which includes isnan.

Answer 1

[Jan]

It is tzhe definition of NaN, that any comparison with it is false, even NaN == NaN. Therefore

if interp1(a,b,c(g))==NaN;

will most likely not do, what you expect. I assume you want:

for g = 1:numel(c)   % Not numel(a) !??
  cg = interp1(a, b, c(g));
  if isnan(cg)
    disp('There is a mistake here')
  end
end

Or without a loop:

cg = interp1(a, b, c);
if any(isnan(cg))
  disp('There is a mistake here')
end

I do not get a warning in R2016b.

[Volkan Yangin]

Yes, it works on my MATLAB, too, but

Warning: NaN found in Y, interpolation at undefined values will result in undefined values. > In interp1 at 178

warning message is still on my command window.