How can I avoid loop for code optimization?

3 vues (au cours des 30 derniers jours)
Bhromzara
Bhromzara le 28 Avr 2021
Commenté : Walter Roberson le 29 Avr 2021
y = linspace(1,10,1000);
for m = 1:length(Y)
Z = [1 sin(Y(m)); 2 cos(Y(m))];
X(m) = det(Z);
end
  4 commentaires
Bruno Luong
Bruno Luong le 29 Avr 2021
X = -Y; % det of Z
Bhromzara
Bhromzara le 29 Avr 2021
@Bruno Luong assume : Z = [1 sin(Y(m)); 2 cos(Y(m))];

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Réponses (2)

Bruno Luong
Bruno Luong le 29 Avr 2021
Modifié(e) : Bruno Luong le 29 Avr 2021
Use MultipleQR FEX (C compiler for MEX build is required, unless for Windows platform)
y = linspace(1,10,1000);
Y = reshape(y,1,1,[]);
z = zeros(size(Y));
Z = [1+z sin(Y);
2+z cos(Y)];
n = size(Z,1);
[~,R]=MultipleQR(Z);
R = reshape(R,n*n,[]);
X = prod(R(1:n+1:end,:),1)*(-1)^n;
plot(y,X)

Jan
Jan le 29 Avr 2021
Modifié(e) : Jan le 29 Avr 2021
Start with pre-allocating the output. Letting an array grow iteratively consumes a lot of resources, because Matlab has to create a new vector in each iteration.
Y = linspace(1,10,1000);
X = zeros(1, length(Y)); % Pre-allocation!!!
for m = 1:length(Y)
Z = [1, sin(Y(m)); 2, cos(Y(m))];
X(m) = det(Z);
end
The code can be simplified and no loop is required:
Y = linspace(1, 10, 1000);
X2 = cos(Y) - 2 * sin(Y);
isequal(X, X2) % TRUE
It is not useful to optimize a code, which is a massive simplification of the real code. Calling DET() for a 2x2 matrix is a waste of time, because DET([a,b; c,d]) is simply: q*c - b*d. So either your problem can be solved in a cheap way without a loop, or you have hidden the important parts by posting an example which has been simplified too much.
  5 commentaires
Bruno Luong
Bruno Luong le 29 Avr 2021
Modifié(e) : Bruno Luong le 29 Avr 2021
No it does not exeeding 2^53 terms
>> log2(factorial(16))
ans =
44.2501
And using Laplace expansion is very bad idea, the cancellation between terms make it very bad numerical mehod. It can only used few very small dimension, let say <= 5.
Walter Roberson
Walter Roberson le 29 Avr 2021
Hmmm, I must have been misremembering.
I see it is 18! that is the last one before 2^53.
Let me see... If you do a number of up-front assignments along the lines of
AA = in(:,1); AB = in(:,2);
for each of the 256 different elements of a 16 x 16 matrix, then you can use two characters per entry (with 256 elements of 16 x 16 matrix, that is too many for one character variable names); and each term would be 2 characters occuring 16 times, plus 15 .* operators = 47 characters. There would be 16! = 20922789888000 terms, and one '+' character between each term so we get 16! * 48 - 1 = 1004293914623999 characters, plus the overhead of breaking out the inputs into variables. It works out to just under one petabyte.

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