Problem with logical indexing with arbitrary indices

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Christopher
Christopher 2014-10-6
编辑： Iain 2014-10-6
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.

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回答（1 个）

Iain
Iain 2014-10-6
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
  2 个评论
Christopher
Christopher 2014-10-6
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
Iain 2014-10-6
编辑：Iain 2014-10-6
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

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