How do I compare values within a for loop using if/else?

I am trying to compare the values of Vx and Vy as they are generated and display a statement based on whether Vx is greather than, less than, or equal to Vy. The first value returns correctly as equal, but the rest of the values return as X > Y, even though many should return as X < Y.

clc
clear all
close all

x=0:5:20;
y=0:10:40;
z=0:20:80;
[X,Y,Z]=meshgrid(x,y,z);

Vx=zeros(5,5,5);
Vy=zeros(5,5,5);
Vz=zeros(5,5,5);

for i=1:length(x)
    for j=1:length(y)
        for k=1:length(z)
            Vx(i,j,k)=(((i-1)*5).^2)+(((j-1)*10).^2)+(((k-1)*20).^2);
            Vy(i,j,k)=2*((i-1)*5)*((j-1)*10)*((k-1)*20);
            Vz(i,j,k)=((i-1)*5)+((j-1)*10)+(2*((k-1)*20));
            if Vx==Vy
                disp('X = Y')
            elseif Vx<Vy
                disp('X < Y')
            else
                disp('X > Y')
            end
        end
    end
end

figure;
quiver3(X,Y,Z,Vx,Vy,Vz)
xlabel('X Axis')
ylabel('Y Axis')
zlabel('Z Axis')
set(gca,'FontSize',12)
set(gca,'ZDir','reverse')

Answer 1

[Voss]

Seems like you want to compare one element of Vx and Vy at a time:

if Vx(i,j,k) == Vy(i,j,k)
    disp('X = Y')
elseif Vx(i,j,k) < Vy(i,j,k)
    disp('X < Y')
else
    disp('X > Y')
end

When you use a non-scalar array in a comparison, such as Vx == Vy, the result is considered true only if it's true for all elements.

[Timothy Nix]

That works! Thank you very much.

Answer 2

[Walter Roberson]

            Vx(i,j,k)=(((i-1)*5).^2)+(((j-1)*10).^2)+(((k-1)*20).^2);
            Vy(i,j,k)=2*((i-1)*5)*((j-1)*10)*((k-1)*20);
            Vz(i,j,k)=((i-1)*5)+((j-1)*10)+(2*((k-1)*20));

Those statements are building arrays.

            if Vx==Vy

That statement is comparing all of Vx to all of Vy. In MATLAB, an "if" is considered true if all of the values computed are non-zero, same as if you hard written

            if all(Vx(:)==Vy(:))

If even a single element of Vx is not equal to its corresponding Vy then the test would be considered false.

Perhaps you should only be comparing the array elements you just computed.

[Walter Roberson]

Also, remember that the == operator between the same data type is bit-for-bit exact comparison (with the exception that negative 0 is considered to compare equal to 0, and nan is not equal to nan). See for example

format long g
A = 0.3 - 0.2

A = 
                       0.1

B = 0.1

B = 
                       0.1

A == B

ans = logical
   0

A - B

ans = 
     -2.77555756156289e-17

The same algebraic value calculated two different ways can end up rounded differently. Using == might not be appropriate. Consider comparing with tolerance, such as using ismembertol

[Timothy Nix]

Ah I can see how that would cause an issue now. I will keep that in mind in the future, thank you!