Feeding a vector to a cost function

Hello

I am wondering how I can feed a random (10,2) matrix to a cost function that get 2 inputs

My cost function is

   function [f] = CostFunction(x,y)
  f=4*(1-x).^2.*exp(-(x.^2)-(y+1).^2) -15*(x/5 - x.^3 - y.^5).*exp(-x.^2-y.^2) -(1/3)*exp(-(x+1).^2 - y.^2)-1*(2*(x-3).^7 -0.3*(y-4).^5+(y-3).^9).*exp(-(x-3).^2-(y-3).^2);
  end

and I have created a random matrix like

vector=rand(10,2)

Now I want to be able to say something like this

CostFunction(Vector)

So I can get the value of Cost function for all those inputs

thank you

Answer 1

[Andrei Bobrov]

function [f] = CostFunction(v)
x = v(:,1);
y = v(:,2);
f=4*(1-x).^2.*exp(-(x.^2)-(y+1).^2) -15*(x/5 - x.^3 - y.^5).*exp(-x.^2-y.^2) -(1/3)*exp(-(x+1).^2 - y.^2)-1*(2*(x-3).^7 -0.3*(y-4).^5+(y-3).^9).*exp(-(x-3).^2-(y-3).^2);

solution

vectors = rand(10,2);

out = CostFunction(vectors);

[Pourya saber]

Thank you andrei, works well.

Answer 2

[David Young]

How about

CostFunction(vector(:,1), vector(:,2))

though whether this is useful depends on what you want to achieve.

[Pourya saber]

Yes this one do work as well, thank you

Answer 3

[bym]

function f = CostFunction(vector) % brackets not necessary
x = vector(:,1);
y = vector(:,2);
f=4*(1-x).^2.*exp(-(x.^2)-(y+1).^2) -15*(x/5 - x.^3 - y.^5).*exp(-x.^2-y.^2) -(1/3)*exp(-(x+1).^2 - y.^2)-1*(2*(x-3).^7 -0.3*(y-4).^5+(y-3).^9).*exp(-(x-3).^2-(y-3).^2);
end

[Pourya saber]

thanks