Effacer les filtres
Effacer les filtres

Inefficient code - simple counter

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Joppy
Joppy le 15 Déc 2017
Commenté : KL le 15 Déc 2017
I have the following inefficient bit of code.
for i = 1:n %arb n
if something
A(i,1) = A(i,1) + 1;
elseif somethingelse
A(i,2) = A(i,2) + 1;
end
end
There are actually 5 if/else clauses inside a function which is called inside the loop, but I've reduced it to the above for simplicity. For large n, this is very slow. Is there a faster way to do this? It's supposed to be a simple counter that retains information about the counts.
Thanks
edit: I have filled A with zeros before the loop also.
  2 commentaires
KL
KL le 15 Déc 2017
What are your conditions? Give an example.
Joppy
Joppy le 15 Déc 2017
Modifié(e) : Joppy le 15 Déc 2017
I'm checking if one number is smaller than another. While they do contribute to the overall time, it isn't nearly as problematic as the counter (I've timed with and without).

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KL
KL le 15 Déc 2017
I suppose your condition is something like checking the range of the specific element. Depending on what range they are in you want to perform something on that specific element.
A = rand(50,1);
ind1 = A<0.25; %first if
ind2 = A>=0.25&A<0.5; %elseif
ind3 = A>=0.5&A<0.75;
ind4 = ~ind1&~ind2&~ind3; %else
B = (A+1).*ind1+(A+2).*ind2+(A+3).*ind3+(A+4).*ind4;
  2 commentaires
Joppy
Joppy le 15 Déc 2017
Oh. No the conditions aren't related to A, but I can adapt this approach anyway.. Silly me. Thanks!
KL
KL le 15 Déc 2017

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Plus de réponses (1)

Birdman
Birdman le 15 Déc 2017
Modifié(e) : Birdman le 15 Déc 2017
One approach:
Consider a B vector with dummy data.
B=randi([1 10],20,1);A=zeros(numel(B),2);
ind1=find(B<5);%condition1
ind2=find(B>=5);%condition2
A(ind1,1)=1;A(ind2,2)=1;

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