Efficient way to split a vector into Matrix
显示 更早的评论
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
采纳的回答
the cyclist
2011-4-14
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
2 个评论
RoJo
2011-4-14
the cyclist
2011-4-14
Yeah, I should have anticipated that. Can you just put in a rounding command?
更多回答（2 个）
Andrei Bobrov
2011-4-14
variant
D0 = buffer(B,5,4);
D = D0(2:end-1,4:end)
Matt Fig
2011-4-14
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));
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