interp1 - problem with query points

Hi,

I am facing a really strange behavior of the function interp1.

I have the variables in the file .mat in attachment, where x is a vector , f is a matrix, and delta is a vector that is computed somewhere else in the code .

From the command windows, I can see that

>> delta =

    0.5000         0         0

I am confused because the first value of delta gives me NaN, but not if I simply interpolate directly for 0.5 (what I believe is the value of delta(1) !)

>> f_interp = interp1(x,f,delta(1))

f_interp =

   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN
   
>> f_interp = interp1(x,f,0.5)

f_interp =

  Columns 1 through 15

    0.9882    0.9882    0.9984    1.0221    1.0389    1.0482    1.0843    1.1814    1.2525    1.3288    1.3961    1.5075    1.6173    1.6866    1.7918

  Columns 16 through 21

    1.8649    1.8835    1.9232    1.9819    2.0135    2.0371  

I am assuming that this result has something to do with how the numbers are stored in delta. In fact, if I manually assign

    delta(1) = 0.5;

then it works. So this problem has to do with how the number is stored in the vector, and not with the vector itself.

How can I figure out what is wrong with the stored number? Can anyone help?

Answer 1

[Star Strider]

You have encountered ‘floating-point approximation error’. If you subtract ‘delta(1)’ from ‘x’:

Check = x-delta(1)

the result is:

Check =
    -5.000000000000001e-01
    -4.800000000000001e-01
    -4.500000000000001e-01
    -4.000000000000001e-01
    -3.000000000000001e-01
    -1.110223024625157e-16

so you are asking interp1 to extrapolate, without telling it how.

This worked when I tried it:

    f_interp = interp1(x,f,delta(1), 'linear','extrap');

and produced a numeric vector with no NaN values.

[Maria]

Thank you! I thought was something like that, so I tried to use the "long" format to see whether there was some last "1" that I could not see, but it did not help.

[Star Strider]

As always, my pleasure!

The long format is the correct appproach, however it’s best to see the difference, so using the long format with subtraction provides the necessary information.