Efficient way to split a vector into Matrix

Hi,

I am looking for an efficient way to do the following:

Take an input vector, e.g B = [2;5;8;11;3;6;9;15]

and return the array

D = [2 5 8 11 3;5 8 11 3 6;8 11 3 6 9]

, ie each column is a rolling subvector of B

My attempt below works but it causes a bottleneck in my programme. I have been trying to vectorize it but am finding it difficult.

If somebody has thoughts I would appreciate a pointer.

Thanks!

function ret = MakeMatrix(inputVector, inputLookbackPeriod, numPeriodsToCalculate)

m_Array = zeros(inputLookbackPeriod, numPeriodsToCalculate);

for i=1:numPeriodsToCalculate
    m_Array(:,i) = inputVector(i:i+inputLookbackPeriod-1,1);
end
end

Answer 1

[the cyclist]

Here is an obscure way, using a Hilbert matrix. It is semi-coincidental that the indexing you need is exactly the (element-by-element inverse of the) Hilbert matrix.

For something less esoteric, you could probably use the filter command.

function m_Array = MakeMatrixFast(inputVector, inputLookbackPeriod, numPeriodsToCalculate)
    H = 1./hilb(max(inputLookbackPeriod,numPeriodsToCalculate));
    H = H(1:inputLookbackPeriod,1:numPeriodsToCalculate);
    m_Array = inputVector(H);
end

[RoJo]

Thanks. This does what I was asking. One point is that as I increase the argument to hilb, e.g 1./hilb(30) the resulting matrix contains 49.000.. which fails inputVector(H). So just needs a conversion before that line.

[the cyclist]

Yeah, I should have anticipated that. Can you just put in a rounding command?

Answer 2

[Andrei Bobrov]

variant

D0 = buffer(B,5,4);
D = D0(2:end-1,4:end)

[RoJo]

Thanks for the reply. I don't have the Signal Processing Toolbox to use the buffer function however.

[Teja Muppirala]

You learn something new every day...

Answer 3

[Matt Fig]

IDX = ones(3,5,'single');  
IDX(:,1) = 1:3;
D = B(cumsum(IDX,2))

Or, to match your functional form:

function m_Array = MakeMatrixFast2(inputVector, inputLookbackPeriod, numPeriodsToCalculate)
IDX = ones(inputLookbackPeriod,numPeriodsToCalculate,'single');  
IDX(:,1) = 1:inputLookbackPeriod;
m_Array = inputVector(cumsum(IDX,2));

[RoJo]

Sorry, I didn't refresh the page and see this answer. It works well and is efficient for large/small vectors and lookbacks. Thanks.

[Matt Fig]

That's o.k.! I tested all three algorithms and yours is actually faster than the cyclist's. Mine is somewhat faster than yours for certain inputs.