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Inefficient code - simple counter

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Joppy
Joppy 2017-12-15
评论： KL 2017-12-15
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 个评论
KL
KL 2017-12-15
What are your conditions? Give an example.
Joppy
Joppy 2017-12-15
编辑：Joppy 2017-12-15
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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采纳的回答

KL
KL 2017-12-15
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 个评论
Joppy
Joppy 2017-12-15
Oh. No the conditions aren't related to A, but I can adapt this approach anyway.. Silly me. Thanks!

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更多回答（1 个）

Birdman
Birdman 2017-12-15
编辑：Birdman 2017-12-15
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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