How to merge function handles in matlab?

Hi!

I have a function handle of this form:

fk = @(a, b, k) a+b+k;

I would like to obtain another function handle that is actually a vector where every element is the first function handle evaluated in k, for k between 1 and a given n.

f = @(a, b) [fk(a, b, 1) fk(a, b, 2) ... fk(a, b, n)]

Example: for n =3, I would like to obtain

f = @(a,b) [fk(a, b, 1) fk(a, b, 2) fk(a, b, 3)]

How can I do this?

P.S.: I would like something that looks like a for loop, that can work for every given n.

Answer 1

[Stephen23]

You could use arrayfun for this:

fun = @(a,b,n) arrayfun(@(nn)fk(a,b,nn), 1:n);

[Dina Irofti]

Actually this is what I was looking for

fun = @(a, b) arrayfun(@(k)fk(a,b,k),1:n);

Thank you for your answer. It was very useful!!

[Guillaume]

Stephen's answer is just a more generic version of yours, and more importantly safer, if you're not aware of the closure behaviour of anonymous functions. Consider the result of

fk = @(a, b, k) a+b+k;
n = 2;
fun = @(a, b) arrayfun(@(k)fk(a,b,k),1:n); %your fun
fun(5, 6)  
n = 5;
fun(5, 6) %for fun n is still 2

versus

fk = @(a, b, k) a+b+k;
fun = @(a, b, n) arrayfun(@(k)fk(a,b,k),1:n); %Stephen's fun
n = 2;
fun(5, 6, n)  
n = 5;
fun(5, 6, n) %fun uses the new n

Answer 2

[Sean de Wolski]

You can store those function handles in a cell using a for-loop or just run the for-loop over the k values.

for ii = 1:10
  Cfun{ii} = @(a,b)f(a,b,ii);
end

[Dina Irofti]

Try this out

fk = @(a, b, k) a+b+k;
for ii = 1:10
  Cfun{ii} = @(a,b)f(a,b,ii);
end
disp(Cfun)

The output is

Cfun = 

Columns 1 through 7

    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)

Columns 8 through 10

    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)    @(a,b)f(a,b,ii)

What I would like to obtain is

Cfun = 

Columns 1 through 7

    @(a,b)f(a,b,1)    @(a,b)f(a,b,2)    @(a,b)f(a,b,3)    @(a,b)f(a,b,4)    @(a,b)f(a,b,5)    @(a,b)f(a,b,6)    @(a,b)f(a,b,7)

Columns 8 through 10

    @(a,b)f(a,b,8)    @(a,b)f(a,b,9)    @(a,b)f(a,b,10)

[Sean de Wolski]

They're actually equivalent, ii is a static copy of the integer that it is equal to at anonymous function creation, you can see its values by looking at the workspace values in functions:

q = 7
f = @(x)x*q
fv = functions(f)
fv.workspace{1}