MATLAB Answers

Generic Anonymous Function Setup

5 views (last 30 days)
dsmalenb on 26 Jul 2018
Commented: dsmalenb on 26 Jul 2018
I was wondering if the following is possible using anonymous functions. I am fairly new to them and this is not for a class - just FYI.
Let N >= 2 specify the number of mixture components I would like to build into my anonymous function. Let's say I want each component to be a Normal Distribution. I have three vectors for their weights (w), means (m), and standard deviations (s).
If N = 2 then the anonymous function would look like:
fun = @(x) w(1)/(s(1) *sqrt(2*pi))*exp(-(x-m(1))^2/(2*s(1)^2)) + w(2)/(s(2) *sqrt(2*pi))*exp(-(x-m(2))^2/(2*s(2)^2));
This becomes exceedingly arduous as N grows. It would be preferable to be able to define this function generically so it does not have to get written out.
Is this possible? From research the literature I am only finding examples where the author "hard coded" the function.
However, I could be searching for it incorrectly or missing a specific term.
Any insights would be greatly appreciated.


Sign in to comment.

Accepted Answer

Stephen Cobeldick
Stephen Cobeldick on 26 Jul 2018
Edited: Stephen Cobeldick on 26 Jul 2018
Learn how to write vectorized code, then your task is easy!
Your function:
>> w = 1:2;
>> m = 2:3;
>> s = 3:4;
>> fun = @(x) w(1)/(s(1)*sqrt(2*pi))*exp(-(x-m(1))^2/(2*s(1)^2)) + w(2)/(s(2)*sqrt(2*pi))*exp(-(x-m(2))^2/(2*s(2)^2));
>> fun(1)
ans = 0.30183
Vectorized function (simpler):
>> foo = @(x)sum(w./(s.*sqrt(2*pi)).*exp(-(x-m).^2./(2*s.^2)));
>> foo(1)
ans = 0.30183
Vectorized function, with N==4:
>> w = 0:3;
>> m = 3:6;
>> s = 6:9;
>> foo = @(x)sum(w./(s.*sqrt(2*pi)).*exp(-(x-m).^2./(2*s.^2)));
>> foo(1)
ans = 0.25397
When you write vectorized code you can give it arrays of any size and it will work. Any time you find yourself copy-and-pasting code then you are doing something wrong.

  1 Comment

dsmalenb on 26 Jul 2018
Not all heroes wear capes I see. Thank you.
Been programming since '90. Old habits take some time to unlearn!

Sign in to comment.

More Answers (0)

Sign in to answer this question.