Inefficient code - simple counter

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Joppy
Joppy on 15 Dec 2017
Commented: KL on 15 Dec 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 Comments
KL
KL on 15 Dec 2017
What are your conditions? Give an example.
Joppy
Joppy on 15 Dec 2017
Edited: Joppy on 15 Dec 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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Accepted Answer

KL
KL on 15 Dec 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 Comments
Joppy
Joppy on 15 Dec 2017
Oh. No the conditions aren't related to A, but I can adapt this approach anyway.. Silly me. Thanks!

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More Answers (1)

Birdman
Birdman on 15 Dec 2017
Edited: Birdman on 15 Dec 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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