How to get rid of a loop that depends on its previous iterations

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JPL
JPL 2021 年 10 月 19 日
コメント済み: JPL 2021 年 10 月 25 日
I have an old c++ program that I want to translate to Matlab. It has a loop in which every iteration depends on a decision in the previous iteration. I have a lot of data, so it takes a long time. Is there a way to do this with matrix operations instead of a loop? Here's the code:
data = rand(1,105);
for k = 100 : -1 : 6
lowerAverage = mean(data(k-5:k-1));
upperAverage = mean(data(k+1:k+5));
surroundingAverage = mean([lowerAverage,upperAverage]);
if data(k) > surroundingAverage
data(k) = upperAverage
end
end
Thanks in advance.

採用された回答

Jan
Jan 2021 年 10 月 19 日
編集済み: Jan 2021 年 10 月 20 日
Start with calculating the average manually instead of the slower mean() function:
% Timings: R2018b, Win10, i7
data0 = rand(1, 100005);
data = data0;
tic;
for k = 100000 : -1 : 6
lowerAverage = mean(data(k-5:k-1));
upperAverage = mean(data(k+1:k+5));
surroundingAverage = mean([lowerAverage,upperAverage]);
if data(k) > surroundingAverage
data(k) = upperAverage;
end
end
toc % Elapsed time is 1.050304 seconds.
data_orig = data;
data = data0;
tic;
for k = 100000 : -1 : 6
lowerAverage = sum(data(k-5:k-1)) / 5;
upperAverage = sum(data(k+1:k+5)) / 5;
surroundingAverage = (lowerAverage + upperAverage) / 2;
if data(k) > surroundingAverage
data(k) = upperAverage;
end
end
toc % Elapsed time is 0.050326 seconds.
isequal(data, data_orig) % true
20 times faster already. Mayby this is some percent faster:
data = data0;
tic;
for k = 100000 : -1 : 6 % 1 multiplication, 1 division:
low = sum(data(k-5:k-1));
up = sum(data(k+1:k+5));
if 10 * data(k) > (low + up)
data(k) = up / 5;
end
end
toc % Elapsed time is 0.049661 seconds.
The lower average is not influenced by overwriting the data. So this can be vectorized:
data = data0;
tic;
lowv = conv(data, ones(1, 5), 'same');
for k = 100000 : -1 : 6
low = lowv(k - 3);
up = sum(data(k+1:k+5));
if 10 * data(k) > (low + up)
data(k) = up / 5;
end
end
toc % Elapsed time is 0.025510 seconds.
The last method takes significantly more time when I run it in the forums MATLAB online 2021b. So try this locally on your computer.
[EDITED] If replacing mean() increases the speed, try this with sum() also:
data = data0;
tic;
avg = conv(data, ones(1, 5), 'same');
for k = 100000 : -1 : 6
low = avg(k - 3);
up = data(k+1) + data(k+2) + data(k+3) + data(k+4) + data(k+5);
if 10 * data(k) > (low + up)
data(k) = up / 5;
end
end
toc % Elapsed time is 0.007511 seconds.
A speedup of factor 140 on my R2018b and the random test data.
  5 件のコメント
Jan
Jan 2021 年 10 月 22 日
A mex version needs about half of the processing time compared to the last M-version of my asnwer:
#include "mex.h"
mvoid mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
double *X, low, up;
mwSize i, n;
n = mxGetNumberOfElements(prhs[0]);
plhs[0] = mxDuplicateArray(prhs[0]);
X = mxGetDoubles(plhs[0]);
for (i = n - 6; i > 4; i--) {
low = X[i-5] + X[i-4] + X[i-3] + X[i-2] + X[i-1];
up = X[i+1] + X[i+2] + X[i+3] + X[i+4] + X[i+5];
if (10 * X[i] > low + up) {
X[i] = up / 5;
}
}
return;
}
JPL
JPL 2021 年 10 月 25 日
Thanks,
I think this will work out well.
John

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