MATLAB if-else statement

Question

I'm trying to write an if-else statement that sets a value of an array to 0.1 at n=0 because the equation that computes that array has undetermined value at that input. The equation is:

So at 0 the value is infinity.

This is the code I wrote but it doesn't seem to work:

n = -(N-1)/2:(N-1)/2;
hlp = (1./n*pi).*(sin(0.1*pi.*n));

if n == 0
    hlp = 0.1;
else
    hlp = (1./n*pi).*(sin(0.1*pi.*n));
end

For refference, the issue you have in your code is that its not a loop, so n is never n==0, as n = -(N-1)/2:(N-1)/2;. Your code would work with a couple of changes: make a for loop for ii=n, and then access each variable per value ( n(ii) and hlp(ii)). That is what is flawed in your logic. The one liner solution is better, but sometimes you can't avoid loops. — Ander Biguri
– Ander Biguri, Commented Oct 31, 2022 at 14:34

Pursuit · Accepted Answer · 2022-10-31 15:39:44Z

3

I think that you are close. See if this works.

n = ( -(N-1)/2  ):(  (N-1)/2  );    %add some parentheses here
hlp = (1./n*pi).*(sin(0.1*pi.*n));  %this line is unchanged

When working on a subset of a vector, if/else isn't really the right construct. Try this instead.

%create a mask of the elements to replace
maskN0 = (n==0)
%then replace them
hlp(maskN0) = 0.1;

OR, after you've been doing this a while, it may be more natural to write (Thanks Ander Biguri)

hlp(n==0) = 0.1;
%When I read this, I hear: "hlp, where n=0, set to 0.1"

edited Oct 31, 2022 at 15:39

answered Oct 31, 2022 at 4:24

Pursuit

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3 Comments

Erik Ebrahimi Over a year ago

Are you sure help() will replace it? Because help in MATLAB is used to explain what the function does.

Pursuit Over a year ago

There was a critical.typo. help instead of hlp. Updated

Ander Biguri Over a year ago

You can even shorten this to hlp(n==0)=0.1; which makes it more readable, imho.

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