vectorization of comparison against several intervals

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Cristian Berceanu on 13 Dec 2021
Commented: Jan on 13 Dec 2021
Hello,
I currently have the code below, which essentially checks if each element of the MainArray is within any of the intervals described in each row of CompArray; if yes, the corresponding position in ResultsArray is set to 0. In practice, I need to run this on mich bigger matrices and iteratively, so speed of execution becomes a problem. Is there a possibility to vectorize it? Note that CompArray can have a variable number of rows (comparison intervals), but always only two columns.
clc
clearvars
ResultsArray = ones(10,10);
MainArray = magic(10);
CompArray = [ 20 30;...
40 50;...
60 70];
for Row = 1:10
for Col = 1:10
if(any((MainArray(Row,Col) >= CompArray(:,1)) & (MainArray(Row,Col) <= CompArray(:,2))))
ResultsArray(Row,Col) = 0;
end
end
end
Best regards,
Cristian
Cristian Berceanu on 13 Dec 2021
Typical sizes for MainArray and ResultsArray would be 1000 x 1000 to 10.000 x 10.000.
CompArray could have anything between 20 and 200 rows.

Jan on 13 Dec 2021
Edited: Jan on 13 Dec 2021
Easier to run in the forum and maybe faster already:
N = 1000;
M = randi([1, 1000], N, N);
C = sort(randi([1, 1000], 100, 2), 2);
C1 = C(:, 1);
C2 = C(:, 2);
tic
R1 = ones(size(M));
for Row = 1:size(M, 1)
for Col = 1:size(M, 2)
if(any((M(Row,Col) >= C(:,1)) & (M(Row,Col) <= C(:,2))))
R1(Row,Col) = 0;
end
end
end
toc
tic
R2 = ones(size(M));
for k = 1:numel(M)
if any((M(k) >= C1) & (M(k) <= C2))
R2(k) = 0;
end
end
toc
% Alternatively:
tic
R3 = ones(size(M));
for k = 1:numel(M)
R3(k) = all((M(k) < C1) | (M(k) > C2));
end
toc
tic
R4 = true(size(M));
for k = 1:size(C1, 1)
R4(R4 & (M >= C1(k)) & (M <= C2(k))) = false;
end
toc
tic
R5 = true(size(M));
for k = 1:size(C1, 1)
R5 = R5 & (M < C1(k) | M > C2(k));
end
toc
isequal(R1, R2, R3, R4, R5)
% R2018b, Win10, i5 mobil:
% Elapsed time is 0.429244 seconds. original
% Elapsed time is 0.335571 seconds. simplilied
% Elapsed time is 0.351330 seconds. no IF
% Elapsed time is 0.247861 seconds. loop over intervals
% Elapsed time is 0.148806 seconds. loop over intervals 2
Jan on 13 Dec 2021
The vectorized version are much slower, therefore I did not copy them. Vectorisation is not efficient, if large intermediate arrays are required, which do not match into the CPU cache.

R2021b

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