Help With Conditonally variant Anonymous function.

Okay, this question is more abstract than specific, ergo, I'll be using more pseudocode and less specifics.

Say I have a very important variable:

B2 = +/- val;

And three arrays:

z1=[1:1:500];
z2=[501:1:1500];
z3=[1501:1:2000]; 

Which in turn exist in larger array z such that:

z=[z1 z2 z3];

The value of B2 is given by the general relationships:

B2(z1) = -val;
B2(z2) = +val;
B2(z3) = -val;

Now, im well aware ill need an if/else loop, but I must program this in an anonlymous function:

B2=@(X)(...)

I think the following solution is something similar to what I'd like:

B_2 = @(X)(if X<=z(500)&&X>=z(1500) B_2 = +val, else  B_2 = -val)

Before some genius suggest an anonymous function is unecaserry, stupid, superflous etc etc etc, theres no ather way i can implement it. It HAS to be a anonymous function, I'm sorry.

If anyone could help I'd be really apprecative. I've been doing coursework all day, my brains quite fried, i think im like 90% of the way to a solution, im just being buggered by the execution. Hopfully i wake up with some assistance :)

Ty guys.

[Walter Roberson]

If the same value appears in both z1 and z2, then what is the value of B2 to be?

[Mark Dawson]

Shouldnt happen. Consider z (the overall array) to be:

z=[1:1:2000]

[Mark Dawson]

I eddted the original post. Hopefully should be clearer.

Answer 1

[Mark Dawson]

Thanks once again for the help guy. As always, I solved it on my own. Would've been neater with an anonymous function through. Ah well. If anyone from the future cares:

function beta_2 = b_2(Z)

z = [0:(40000)/1999:40000];

if Z < z(501) && Z > z(1501)

beta_2 =6.8e-03;

else

beta_2 = -6.8e-03;

end

end

Answer 2

[Walter Roberson]

B_2 = @(z) ((z>=z(501)&z<=z(1500)) * 2 - 1) .* val

The logical test that is satisfied by the range will return true (1) in the middle range, and false (0) outside the range. 1 * 2 - 1 is 1, so a match (true) will become 1. 0*2 -1 is -1 so a non-match (false) will become -1 . That all is multiplied by val.

Answer 3

[Steven Lord]

Let's say val is 2. So if Z is between 1 and 500 B2 should be -2, if it is between 501 and 1500 B2 should be 2, and if it is between 1501 and 2000 B2 should be -2. You're trying to discretize your data.

edges = [1, 500, 1500, 2000]

edges = 1×4

1 500 1500 2000

●

values = [-2, 2, -2]
randomZ = randi([1 2000], 10, 1);

B2 = @(x) discretize(x, edges, values);
B2vals = B2(randomZ);
results = table(randomZ, B2vals)

results = 10x2 table
    randomZ    B2vals
    _______    ______

      771         2  
     1590        -2  
      731         2  
      901         2  
      384        -2  
      222        -2  
      345        -2  
     1754        -2  
     1471         2  
     1153         2