Is there faster way to apply `det` function along the third dimension?
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wei zhang
el 17 de Jul. de 2020
Comentada: wei zhang
el 19 de Jul. de 2020
I am trying to calculate the det of many 4*4 matrix. I store the data in a matrix with shape 4*4*n. (n =4000000). I am using the for loop to get the result as below. Is there any way to accelerate the progress? Like bsxfun or arrayfun, or some simillar ideas?
v = zeros(length(m),1);
for i=1:size(m,3)
v(i)=det(m(:,:,i));
end
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Walter Roberson
el 18 de Jul. de 2020
Bruno's criticism of the precision problems and the high length of the formula for increasing n, are valid criticisms. Generally speaking, it is often the case that making code faster comes at the price of making it less accurate towards the margins. Theoretical definitions that suppose infinite precision get substituted for more nuanced checks that deal with floating-point realities, and time gets saved by not making the checks to figure out what compensation is needed for each case.
If you have some time you should look at how hypot() (finding the length of a hypotenuse) has to be implemented in practice in order to maintain accuracy. Consider sqrt(A^2 + B^2) under the circumstance that A or B is smaller than sqrt(realmin) and so squaring it might underflow to 0...
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Bruno Luong
el 17 de Jul. de 2020
Editada: Bruno Luong
el 17 de Jul. de 2020
I would note that Walter's solution that use recursive formal determinant formula might be fast but might be sensitive to umerical errors, that is how I was tough (for medium/large size matrix).
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