How to fill 3D array using for loop with data z allocated to specific x and y values

Hi,

I want to create a logical 3D array with a 1 at each place where z (x,y) exists. My code is:

X=[0.5;0.5;0.6;0.6];                           
Y=[0.6;0.7;0.6;0.7];  
V=[1.0794;1.1794;1.1646;1.2646]; 
A=zeros(length(X),length(Y),length(V));
for i=1:length(X)
    for j=1:length(Y)
        for k=1:length(V)
            if  i==j==k & V(k)>0 
                A(i,j,k)=1;
            end
        end
    end

end

As you can see, the V values are already calculated and allocated to the varations of X and Y (first value of V belongs to 1st value of X and 1st value of Y, 2nd value of V belongs to 2nd values of X and Y, ...) So for i=1, j=1, k=1 there exists a value for V at the position k=i=j=1. At the position i=1, j=2, k=1 a wrong assignment arises, because in vector V at the position 1 there is of course a value. But this value does not belong to the combination of place 1 of vector X and place 2 of vector Y. My condition in the code does not work as I need it. I need as first condition that the correct position k of V is assigned to the corresponding positions i of X and j of Y. If the condition is met, the second condition checks if the value at location k in vector V is >0 and replaces the 0 at the location with a 1. Unfortunately there is no other way to display X and Y, because the data is calculated like this.

How can I write the first condition so that it works? At the moment I get the following array:

val(:,:,1) =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

val(:,:,2) =

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

val(:,:,3) =

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

val(:,:,4) =

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

I suppose the answer is simple and I just don't see it at the moment. Thank you for your help!

[Walter Roberson]

            if  i==j==k & V(k)>0 

That means the same as

            if  ((i==j)==k) & (V(k)>0)

which compares i to j producing a logical value (0 for false, 1 for true) and compares that logical value to k .

[Walter Roberson]

            if  i==j==k & V(k)>0 
                A(i,j,k)=1;
            end

If i must equal j and k then you can skip most of the loops:

for k=1:min([length(X), length(Y), length(V)])
    if V(k) > 0
      A(k,k,k) = 1;
    end
end

which could be further vectorized.

[Kati]

It works! Thanks!

Answer 1

[Matt J]

X=[0.5;0.5;0.6;0.6];                           
Y=[0.6;0.7;0.6;0.7];  
V=[1.0794;1.1794;1.1646;1.2646];

[I,J,K]=ndgrid(1:length(X), 1:length(Y), 1:length(V))

A =  (I==J) & (J==K) & V(K)>0;