can you add probability to a for loop?

I have a very long and complex fitness function that I want to add even more complexity to, and I'm wondering if I can shortcut it.

The basic idea is that an individual has to choose between two patches to forage in and the patches can now have a varying probability, either high or low, of finding food.

I want the combination of probabilities to be different from each other, that is

p1 = prob of finding food in patch 1 = hi or lo
p2 = prob of finding food in patch 2 = hi or lo
(p1,p2) combinations
30% chance of (hi,lo)
25% chance of (hi,hi)
25% chance of (lo,lo)
20% chance of (lo,hi)

The most straight forward way would just be adding it up

total fitness Fd = 0.3(F1) + 0.25(F2) + 0.25(F3) + 0.2(F4)

My code for fitness is already very long with only one combination of finding food probabilities. Is there a way I could do a for loop with these combinations and add in that probability or do I have to go the long way around?

[Matt J]

What is "the long way around"?

[Kitt]

Fdd(i,j,k,tt)= (1-u1)*( ...
            b(tt)*(... 
                m1(tt)*(...
                   p(z1(j),k)*(n1(tt)*state1 + (1-n1(tt))*state2)+...
                   (1-p(z1(j),k))*(n2(tt)*state3 + (1-n2(tt))*state4) ...
                )+...
                (1-m1(tt))*(... 
                    p2(z2(j),k)*(n1(tt)*state5 + (1-n1(tt))*state6)+...
                    (1-p2(z2(j),k))*(n2(tt)*state7 + (1-n2(tt))*state8) ...
                ) ...
             )+...
            (1-b(tt))*( ...
                m2(tt)*(...
                   p3(j,y1(k))*(n1(tt)*state9 + (1-n1(tt))*state10)+ ...
                   (1-p3(j,y1(k)))*(n2(tt)*state11 + (1-n2(tt))*state12) ...
                )+ ...
                (1-m2(tt))* ...
                   (p4(j,y2(k))*(n1(tt)*state13 + (1-n1(tt))*state14)+...
                   (1-p4(j,y2(k)))*(n2(tt)*state15 + (1-n2(tt))*state16) ...
                ) ...
            ) ...
            );

This is the fitness with p1 (represented here as n1) and p2 (represented here as n2) being static. I would do this 4 times with a combination of n1=0.9 or 0.3 and n2=0.9 or 0.3

I know that's a totally valid way of doing it, but I want to try and learn how to do this in a more efficient way if possible.

Answer 1

[dpb]

Are F1, F2, ..computationally different other than the two probabilities I gather from the above?

If not, encapsulate the calculation in a function and just call the function with the combinations and return in an array

HL=[0.9 0.3];
P=unique(combnk([HL HL],2),'rows');
W=[0.25; 0.20; 0.30; 0.25];
N=size(W,1);
F=zeros(N,1);
for i=1:N
    F(i)=functionF(P(i,:));
end
Fd=dot(W,F);

[Kitt]

That's something I wanted to do, to see if there's a way to turn my long equation into a function. It's difficult because there's a lot of little different things. I already have a function for calculating the "state" the individual is in (of which there are 20), which is making it hard for me to conceptualize how to do this in a function?

%a snippet of the equation: 
p(z1(j),k)*(n1*state1 + (1-n1)*state2)...
     
 %state1 and state2 are completely different
 %in one fitness function there are 16 states all at the same time, not
 %just running through 1 at a time
 %my equation is basically the law of total expectation embedded like 4
 %times

I'm still new to learning matlab, but this gives me a starting point, I think, in how I can add this in!

[dpb]

%a snippet of the equation: 
p(z1(j),k)*(n1*state1 + (1-n1)*state2)...
...

Variables like state1 and state2 ... stateN are likely indicators that state should be an array and then things could be written with vector/matrix algebraic expressions -- or at least turned into loops.

[Kitt]

Okay, I'll ask my PI if this is something that I could do. Thanks, I never would've thought that was possible!

[dpb]

Isn't guaranteed that it can be done; I've not read the code thoroughly enough to fully follow the logic, but there's at least a chance.

I'd be pretty sure it could be turned into the looping construct with the function first shown, however, which would go a long way towards getting you to the point you need...