problem with vector logic

Why doesn't the final logical operation work?

Oneigh=zeros(10,10);
r  = round(0.5+rand(10,10)*8);
Oneigh(r==1)=r;

For the elements where r==1, Oneigh should adopt the corresponding value in r.

By the way for the operation:

Oneigh(r==1)=r;

should actually be

Oneigh(r==1)=O1;

where O1 is another matrix of the same size, but the above is simpler and not working.

Answer 1

[Geoff Hayes]

Christopher - if you run the following code, the line

 Oneigh(r==1)=r;

is probably generating the

 In an assignment  A(I) = B, the number of elements in B and I must be the same.

error message. Look closely at what is happening

 >> r==1

 ans =

     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     1     0     0     0     0     0     1     0     0
     1     0     0     0     0     0     0     0     0     0
     1     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     1     1     0     0     0     1     0     1     0
     0     0     0     0     0     0     0     0     0     0

is a 10x10 matrix. Now look at

 >> Oneigh(r==1)

 ans =
     0
     0
     0
     0
     0
     0
     0
     0
     0

is a 9x1 vector corresponding to the elements of Oneigh indexed on the logical matrix generated by r==1. So

 Oneigh(r==1)=r;

tries to assign a 10x10 matrix (r) to the 9x1 column vector, and the error makes sense.

As you say, for the elements where r==1, Oneigh should adopt the corresponding value in r, then this could be simplified to

 Oneigh(r==1)=1;

since the corresponding value in r is 1.

Else, if you know the indices in r that are identical to 1, then we can use that information to update Oneigh given O1 as

 idcs = find(r==1);
 Oneigh(idcs) = O1(idcs);

Answer 2

[Star Strider]

If you want ‘Oneigh’ to be a matrix with only any particular value of ‘r’ existing and with zeros elsewhere, It’s a bit more involved but still relatively straightforward:

Oneigh=zeros(10,10);
r  = round(0.5+rand(10,10)*8);
iv = 1;                         % Integer To Match
q = r==iv;                      % Logical Array
Oneigh = r;                     % Assign ‘Oneigh’ As ‘r’
Oneigh(~q) = 0;                 % ‘Oneigh’ Now Has ‘iv’ Matching ‘r’, Zero Elsewhere

Note that in this method there is actually no need to preallocate ‘Oneigh’. I retained it for consistency. You can also use the randi function to generate ‘r’.