Myujth

I have a binary vector like this x = [0 0 1 1 1 1 1 1 1]. I want to find the index of lets say the 7th 1, which is 9.

I know I can do this:

y = find(x);
index = y(7);

But what if the vector is huge and I want to conserve memory usage? Wouldn't y = find(x) use alot of memory? If so, is there any way around this?

I am using this as an alternate way of storing indexes for nonbasis and basis elements in a linear programming problem. So I would like to avoid storing the indices as numerical values.

Would the following be a good solution?

basis = [0 0 1 1 1 1 1 1 1];
basisIndex = 7;
correctIndex = getIndex(basisIndex, basis); % should be 9 when basisIndex = 7

function ret = getIndex(basisIndex, basis) 
 counter = 1;
 for value = find(basis) % iterate through [3, 4, 5, 6, 7, 8, 9]
 if counter == basisIndex
 ret = value;
 break;
 end
 counter = counter + 1;
 end
end

Just iterate through x. First, it will not create a new vector y=find(x) (save memory). Second, if basisIndex is small, it will be more efficient.

Suppose x is a 1e8 by 1 vector. Let's compare find with just iteration.

basis = randi(2,1e8,1) - 1;
basisIndex = 7;

tic % your first method
y = find(basis);
index = y(basisIndex);
toc

tic % iterate through base
index = 1;
match = 0;
while true
 if basis(index)
 match = match + 1;
 if match == basisIndex
 break
 end
 end
 index = index + 1; 
end 
toc

Output

Elapsed time is 1.214597 seconds.
Elapsed time is 0.000061 seconds.

Even if the basisIndex is large

basisIndex = 5e7;

The result from iteration is still more efficient

Elapsed time is 1.250430 seconds. % use find
Elapsed time is 0.757767 seconds. % use iteration