Problem with logical indexing with arbitrary indices

I have the code:

 aarray(find(r(1,1)==1))   = bindices(end);

where aarray is a row vector, r is a matrix, and bindices is another matrix. If r(1,1)==1, I want

 aarray(1,end)=bindices(end)

Currently, however, if r(1,1)==1 the argument returns like:

 aarray(1,1)=bindices(end)

This makes sense, but I don't know how to modify the code to act on a different element without complications that are probably unnecessary.

Answer 1

[Iain]

 aarray(find(r(1,1)==1))   = bindices(end);

A. That's not logical indexing.

r(1,1) == 1 can only return a scalar logical 1 or 0. "find" then checks to see if there are any nonzero elements in the input (remember, this is a scalar) and returns the index of the nonzero elements (double format 1 or a double format empty, as the input is a scalar), so your code boils down to either:

 aarray(1) = bindices(end); % or
 aarray([]) = bindices(end);

The simple answer for what you want is:

 if r(1,1) == 1
  aarray(1,end) = bindices(end);
 end

[Christopher]

I know I can use a straightforward conditional, but I want to use vectorized operations. This is because the answer to my question will inform other operations on vectors and matrices, not just scalars. How can I put it all in one vectorized operation? or is that not possible?

[Iain]

Its possible. Say for example you have a curve:

 x = 0:0.01:2*pi;
 y = 5*sin(4*x);

And you want to know where y is negative:

 neg_y = find(y < 0);
 negs = y<0;

You can then use that to index into x & y to modify them:

 x(neg_y) = x(neg_y) *5;
 y(negs) = 0;

You can also combine the logic using matrices: (but it's harder to understand)

 identity = eye(3);
 y = [1 2 3 4 5 6 7 8 9];
 y(find(identity)) % gives the result of 1, 5, 9
 identity(mod(y,2) == 1) % gives the result of 1 0 1 0 1