Why do I get undefined variable in an if statement?

Hi everyone, I get an error for undefined function or variable in the if statement, when I have already assigned the equalities.

l_min = nan(372,1);
A = randn(372,2);
B= randn(372,3);
for t=1:372
    min_ct = min( A(t,:));

      if min_ct == A(t,1);
        l = B(t,1);
        if min_ct == A(t,2);
            l = B(t,2);
        elseif min_ct == A(t,3);
            l = B(t,3);
        end
    end
    l_min(t) = l;
end

Could anyone help with this one?

[Adam]

What is the actual error message and which line does it point to? Your indenting of the code is a bit confusing, but 'l' is not defined on all paths so far as I can see - i.e. if min_ct ~= A(t,1) then it is undefined because the other if statements sit inside the first one.

[gsourop]

The error points the line before the last 'end'

   Undefined function or variable 'l'.

    Error in GenerateVariables (line 394)
      l_min(t,1) = l;

What I would like to do is, compare the minimum value of a vector and make the following, if this minimum value is the first element then go to another vector and pick the first element, if it not then do the same process with the second element and so on.

Answer 1

[KL]

You create l inside if statements, so if none of the condition is fulfilled, then l goes undefined and when you try to access it outside that block, it throws an error.

But anyway, I noticed some other things, A only has 2 columns but you say,

 min_ct == A(t,3); % I suppose A also has 3 columns

and to find minimum along columns, you could simply use (no need for loop),

 [min_ct, ind_min_ct] = min(A,[],2);

now ind_min_ct has all the indices you can simply say use it to find l and l_min

[gsourop]

My apologies, A is an 372 by 3 matrix. Finding the minimum of A is not the point of this exercise. It is to retrieve the respective value of B. For example if the minimum of A (1,:) is A(1,2) then get the value of B(1,2).

[Robert]

I think KL got it right. ind_min_ct is now the column index of the minimum value in each row. It is simple to use this to extract the corresponding values of B.

For your data

A = randn(372, 3);
B = randn(372, 3);

You could use

[~, idx] = min(A, [], 2);
l_min = nan(372, 1);
for t = 1:372
  l_min(t) = B(t, idx(t));
end

or better still

[~, idx] = min(A, [], 2);
l_min = B(sub2ind(size(B), (1:372)', idx));

That last snippet runs around 10x faster than your original code (which I interpreted as the following)

l_min = nan(372, 1);
for t = 1:372
  min_ct = min( A(t, :));
  if min_ct == A(t, 1)
      l = B(t, 1);
  elseif min_ct == A(t, 2)
      l = B(t,2);
  elseif min_ct == A(t, 3)
      l = B(t, 3);
  end
  l_min(t) = l;
end

[Stephen23]

@gsourop: your code is an extremely inefficient way to do that. Just use the second output from min (which is an index), and use that index to get the value from B.