How to compare array to a number for if statement?

Question

H0 is an array ([1:10]), and H is a single number (5).

How to compare every element in H0 with the s

Answer 1

Anyway take a look at this: using ismemeber() function. Frankly not sure how do you expect to compare. Either greater than, smaller , equal or within as a member. So my answer might not be yet satisfying to you. But just giving you an idea anyway.

H0 = [0 2 4 6 8 10 12 14 16 18 20];
H  = [10];
ismember(H,H0)
IF (ans = 1) then
// true
else
//false
end 

Update Answer

This is super bruteforce method - just use it explain. You are better off with any other answers given here than what I present. Ideally what you need is to rip off greater/lower values into two different vectors with ^3 processing - I assume... :)

H0 = [0 2 4 6 8 10 12 14 16 18 20];
H  = [10];

H0(:)
ans = 
     0
     2
     4
     6
     8
     10
     12
     14
     16
     18
     20

Function find returns indices of all values in H0 greater than 10 values in a linear index.

X = find(H0>H)
X = 
     7
     8
     9
     10
     11

Function find returns indices of all values in H0 lower than 10 in a linear index.

Y = find(H0

If you want you can access each element of H0 to check greater/lower values or you can use the above matrices with indices to rip the values off H0 into two different arrays with the arithmetic operations.

G = zeros(size(X)); // matrix with the size = number of values greater than H
J = zeros(size(Y)); // matrix with the size = number of values lower than H

for i = 1:numel(X)
     G(i) = H0(X(i)).^3
end

G(:)
ans =

      1728
      2744
      4096
      5832
      8000

for i = 1:numel(Y)
     J(i) = H0(Y(i)).^2
end

J(:)
ans =

      0
      4
     16
     36
     64
    100