A problem with function handle

5 Ansichten (letzte 30 Tage)
Mohammad Shojaei Arani
Mohammad Shojaei Arani am 6 Nov. 2022
Hello,
I have a simple problem but I do not understand why function handle behaves like this!!! I explain by a simple example in bellow:
>> x1=linspace(-pi,pi,20);y1=sin(x1);f=@(x)interp1(x1,y1,x);f(5)
ans =
NaN
which makes sense. But, when I type
>> x1=linspace(-pi,pi,20);y1=sin(x1);f=@(x)interp1(x1,y1,x).*(x>=-pi & x<=pi)+0.*(x<-pi | x>pi);f(5)
ans =
NaN
and this does not make sense to me.
Any idea?
Thanks in advance,
Babak

Akzeptierte Antwort

John D'Errico
John D'Errico am 6 Nov. 2022
Bearbeitet: John D'Errico am 6 Nov. 2022
NaNs are like wire coathangars, they multiply. Or perhaps evil zombies is a better description. They slime everything they touch, turning everything into new NaNs, propagating NaNs almost everywhere. Evil indeed. ;-)
For example, NaN plus ANYTHING = NaN.
NaN + 2
ans = NaN
NaN times ANYTHING = NaN. Even zero.
NaN*0
ans = NaN
So what did you do? You were hoping the trick of putting a test inline, that a multiply by zero would kill off the NaNs? Yeah, right. The zombies still survived.
Instead, you do have an option. Read the help for interp1!
x1=linspace(-pi,pi,20);
y1=sin(x1);
extrapval = 42;
f=@(x)interp1(x1,y1,x,[],extrapval);
f(5)
ans = 42
I cannot understand why interp1 uses NaN as the default for extrapolation, as 42 has been proven to be the answer to all questions. But if you really want zero to be returned, you can tell interp1 to do that too.
  2 Kommentare
Mohammad Shojaei Arani
Mohammad Shojaei Arani am 6 Nov. 2022
Oh! I got the problem. Am trying to dix it. Thanks John. Indeed, they are like zombies. It is a headache
Mohammad Shojaei Arani
Mohammad Shojaei Arani am 6 Nov. 2022
Thanks John!
I came up by an idea to solve it. But, your way of solving the problem is much more elegant than mine!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by