Vectorize the following loop

Hi all,
I'm trying to vectorize the following loop to speed it up
c = cumsum(weights);
A = ones(1,n);
x = rand(1,n);
for i = 1:n
j = find(c > x(i) ,1,'first');
A(i) = j;
end
where weights is an array of doubles which sum to 1 and n <= size(weights).
Any help would be grand!
B

 Akzeptierte Antwort

Oleg Komarov
Oleg Komarov am 13 Apr. 2011

3 Stimmen

n = 3e4;
weights = rand(n,1);
c = cumsum(weights/sum(weights));
A = zeros(1,n);
x = rand(1,n);
tic
for i = 1:n
j = find(c > x(i) ,1,'first');
A(i) = j;
end
toc
% WARNING: only if weights are monotonically increasing
tic
[B1,B1] = histc(x,c);
toc
tic
B2 = n-sum(bsxfun(@gt,c,x))+1; % Goes fast in out of memory
toc
isequal(A,B1+1,B2) % Don't forget to add 1 to histc result
LOOP : Elapsed time is 2.247998 seconds.
HISTC : Elapsed time is 0.005381 seconds.
BSXFUN: Elapsed time is 1.770875 seconds.

4 Kommentare

Sean de Wolski
Sean de Wolski am 13 Apr. 2011
Nice!
Andrew Newell
Andrew Newell am 13 Apr. 2011
In recent versions of MATLAB you can use
[~,B1] = histc(x,c)
(but it doesn't make it any faster).
Matt Fig
Matt Fig am 13 Apr. 2011
Well-played, Oleg!
Jan
Jan am 28 Jun. 2012
accepted by JSimon

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Weitere Antworten (1)

Sean de Wolski
Sean de Wolski am 13 Apr. 2011

0 Stimmen

Note sure if it'll be faster but:
[row col] = find(bsxfun(@gt,c(:)',x(:)));
A2 = accumarray(row,col,[],@min)';

2 Kommentare

Matt Fig
Matt Fig am 13 Apr. 2011
Your intuition is correct. On my machine this is 50 times slower than the simple loop, using:
A = magic(1000);
weights = A(1,:)/sum(A(1,:));
n = 1000;
Sean de Wolski
Sean de Wolski am 13 Apr. 2011
It would only get slower as the matrices get bigger. I think Ben's elementary, but properly constructed, FOR-loop is probably optimal.
I just realized and am kind of surprised FIND doesn't have a dimensional argument.

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