vectorize a for loop

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Sam
Sam il 26 Ott 2012
I am finding a optimal way of vectorizing the following piece of code (of course the real data is much greater than this):
start_peak_hour = [3 10 20 50];
end_peak_hour = [5 14 27 70]; % always same size as start_peak_hour
n_peak = numel(start_peak_hour);
isPeak = zeros(1,80);
for j=1:n_peak,
isPeak(start_peak_hour(j):end_peak_hour(j)) = 1;
end
Your two cents are appreciated
-Sam
  1 Commento
José-Luis
José-Luis il 26 Ott 2012
Loops are not always evil, and in this case probably even faster than the alternatives.

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Risposta accettata

Sean de Wolski
Sean de Wolski il 26 Ott 2012
Modificato: Sean de Wolski il 26 Ott 2012
I highly doubt you'll find anything faster than that well-written for-loop. The JIT will pick that up and have wonders with it.
eval is evil and will be orders of magnitude slower and hideous/hard to follow.
start_peak_hour = [3 10 20 50];
end_peak_hour = [5 14 27 70]; % always same size as start_peak_hour
n_peak = numel(start_peak_hour);
t1 = 0;
t2 = 0;
for tt = 1:50 %50x
tic
isPeak = zeros(1,80);
for j=1:n_peak,
isPeak(start_peak_hour(j):end_peak_hour(j)) = 1;
end
t1 = t1+toc;
tic
eval(sprintf('isPeak(start_peak_hour(%d):end_peak_hour(%d)) = 1,',[1:n_peak;1:n_peak]))
t2=t2+toc;
end
[t1 t2] % 0.0010 0.0681
t1/t2
On my system the FOR-loop is 70x faster (for this small problem), it'll get better with bigger ones. And it doesn't use eval or any other impossible to understand syntax.
My $0.05 - a full nickel :) .
  6 Commenti
Sean de Wolski
Sean de Wolski il 26 Ott 2012
Modificato: Sean de Wolski il 26 Ott 2012
Well I guess it depends on how we define Sam's "optimal way"
Some possible definitions:
  • easy to read: goes to for-loop
  • fast: goes to for-loop
  • easy to debug: goes to for-loop
  • memory efficient: tie
  • Uses JIT features: goes to for-loop
So unless you want to confuse people and write cruddy code - avoid eval
Jan
Jan il 27 Ott 2012
@Azzi: I prefer to write fast code for all cases. A useful piece of code is used in larger programs also and then the total speed is limited by the most time consuming parts. A not useful piece of code is deleted soon.
A severe problem is the dependency to the problem size: while implementing a quicksort seems to be efficient, it is a bad idea for lists of less than 10 elements. Calculating a SUM using multiple threads and all cores is reasonable for large arrays, it is a waste of time for tiny arrays.
So I respond: We always need "speed", but the "speed" of an algorithm is not well defined.

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Più risposte (5)

Matt J
Matt J il 26 Ott 2012
Modificato: Matt J il 26 Ott 2012
Here's a way that avoid both EVAL and for-loops
isPeak = sparse(80,1,sum(end_peak_hour-start_peak_hour+1));
ispeak(start_peak_hour)=1;
ispeak(end_peak_hour+1)=-1;
ispeak=cumsum(ispeak);
  8 Commenti
Sean de Wolski
Sean de Wolski il 26 Ott 2012
@Matt, fair enough! That gets your method a 6x speedup over mine.
@Azzi, that isn't really a fair test since you're only calling it once so sinmple java things can throw the results way off and not get all of the JIT benefits:
N = 80;
start_peak_hour = [3 10 20 50];
end_peak_hour = [5 14 27 70]; % always same size as start_peak_hour
n_peak = numel(start_peak_hour);
% N=1e7;
% start_peak_hour = 1:20:N;
% end_peak_hour = start_peak_hour+5; % always same size as start_peak_hourn_peak = numel(start_peak_hour);
% n_peak = numel(start_peak_hour);
%
t1 = 0;
t2 = 0;
t3 = 0;
for tt = 1:10 %50x
tic
isPeak=zeros(1,N);
for j=1:n_peak,
isPeak(start_peak_hour(j):end_peak_hour(j)) = 1;
end
t1 = t1+toc;
tic
isPeak = zeros(1,80);
isPeak(cell2mat(arrayfun(@(x,y) x:y, start_peak_hour,end_peak_hour,'un',0)))=1;
t3=t3+toc;
end
[t1 t3]
I See a 27x speedup with the for-loop. Remember, converting to and from cells is slow and array/cellfun is typically slower than a for-loop too.
Azzi Abdelmalek
Azzi Abdelmalek il 26 Ott 2012
Yes, my tes is'nt adequate, like you said, I must make a test in a loop.

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Azzi Abdelmalek
Azzi Abdelmalek il 26 Ott 2012
eval(sprintf('isPeak(start_peak_hour(%d):end_peak_hour(%d)) = 1,',[1:n_peak;1:n_peak]))
  3 Commenti
Azzi Abdelmalek
Azzi Abdelmalek il 26 Ott 2012
If you want to avoid eval
isPeak(cell2mat(arrayfun(@(x,y) x:y, start_peak_hour,end_peak_hour,'un',0)))=1
Jan
Jan il 26 Ott 2012
And you should avoid EVAL...

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Jan
Jan il 26 Ott 2012
Modificato: Jan il 27 Ott 2012
[EDITED] See Matt J's answer, which I had overseen during typing.
start_peak_hour = [3 10 20 50];
end_peak_hour = [5 14 27 70]; % always same size as start_peak_hour
isPeak = zeros(1,80);
isPeak(start_peak_hour) = 1;
if end_peak_hour(end) == 80
end_peak_hour(end) = [];
end
isPeak(end_peak_hour + 1) = -1;
isPeak = cumsum(isPeak);
Please measure the timings with your original data set.
[EDITED] A new approach:
[EDITED 2]: Bugs fixed, now it compiles:
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[],
int nrhs, const mxArray *prhs[]) {
mwSize i, n, len;
double *ini, *fin, *r, *q, *qf;
ini = mxGetPr(prhs[0]);
fin = mxGetPr(prhs[1]);
n = mxGetNumberOfElements(prhs[0]);
len = (mwSize) mxGetScalar(prhs[2]);
plhs[0] = mxCreateDoubleMatrix(1, len, mxREAL);
r = mxGetPr(plhs[0]) - 1;
for (i = 0; i < n; i++) {
q = r + (mwSize) ini[i];
qf = r + (mwSize) fin[i];
while(q <= qf) {
*q++ = 1.0;
}
}
return;
}
In a real-world program checks of the inputs are OBLIGATORY.
Compile it, call it as:
myPeakFiller(start_peak_hour,end_peak_hour,80);
  3 Commenti
Jan
Jan il 26 Ott 2012
Modificato: Jan il 26 Ott 2012
Of course, sorry, Matt J. I've open the thread some hours ago, had something to work (problems with the Windows Update on the laptop...), and posted the answer without re-loading the thread. Therefore I haven't seen your answer before and because I obviously prefer this method, I'm going to vote your answer now.
At least I've considered the last element...
Matt J
Matt J il 26 Ott 2012
At least I've considered the last element...
True!

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Chris A
Chris A il 26 Ott 2012
start_peak_hour = [3 10 20 50]';
end_peak_hour = [5 14 27 70]'; % always same size as start_peak_hour
peak_size = end_peak_hour - start_peak_hour + 1;
ncols=max(peak_size);
nrows=numel(peak_size);
tmpmat=kron(start_peak_hour,ones(1,ncols))+(cumsum(ones(nrows, ncols), 2)-1),
lbmat=kron(start_peak_hour-1, ones(1,ncols));
ubmat=kron(end_peak_hour+1, ones(1,ncols));
u=tmpmat((tmpmat>lbmat)&(tmpmat<ubmat));
isPeak = zeros(1,80);
isPeak(u) = 1;
horzcat((1:80)',isPeak'),

Jan
Jan il 27 Ott 2012
Modificato: Jan il 27 Ott 2012
List of solutions:
function isPeak = Loop_(a, b, len)
isPeak = zeros(1, len);
for j = 1:length(a)
isPeak(start_peak_hour(j):end_peak_hour(j)) = 1;
end
%
function isPeak = CumSum_(a, b, len)
isPeak = zeros(1,80);
isPeak(a) = 1;
if b(end) == 80
b(end) = [];
end
isPeak(b + 1) = -1;
isPeak = cumsum(isPeak);
%
function isPeak = Array_(a, b, len)
isPeak(cell2mat(arrayfun(@(x,y) x:y, a, b, 'un', 0))) = 1;
%
function isPeak = Eval_(a, b, len)
eval(sprintf('isPeak(a(%d):b(%d)) = 1;' , [1:n_peak;1:n_peak]));
%
function isPeak = CumSum2_(a, b, len)
e = ones(1, n_peak);
nzmax = sum(b - a + 1);
isPeak = cumsum(sparse([a, b+1],1, [e, -e], len, 1, nzmax));
%
% And the C-Mex I've posted before
A larger test data set:
start_peak_hour = [3 10 20 50];
end_peak_hour = [5 14 27 70];
a = bsxfun(@plus, start_peak_hour', 0:100:100000);
a = a(:)';
b = bsxfun(@plus, end_peak_hour', 0:100:100000);
b = b(:)';
len = b(end) + 10;
Timings:
tic; for k = 1:1000, isPeak = FUNC(a, b, len); end; toc
Loop_: 4.75 sec
CumSum_: 1.19 sec
Array_: 37.08 sec
Eval_: 1224.83 sec
CumSum2_: 4.09 sec
Mex: 0.42 sec

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