Speeding up to find minimum value using min() function
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Hi,
I have to calculate the current distribution of two nonlinear resistors. To do that I’m searching for the closest index and value of an array.
For example.
[a,b]=min(abs(c-s))
c is an n x 1 array. c increases monotonically
c represents the characteristics of the resistors.
s is a 1 x m array and sinusoidal shaped
s represents the impressed current.
With index b I calculate the voltage drop and the current of each resistor.
Any suggestions to speed up the code?
5 Kommentare
Torsten
am 17 Mär. 2022
c-s will be an nxm matrix and b will be an 1xm vector.
Is it this what you want ?
Michael Walter
am 17 Mär. 2022
Bearbeitet: Michael Walter
am 17 Mär. 2022
Jan
am 17 Mär. 2022
What are typical sizes of the inputs? Do you have a C compiler installed?
Michael Walter
am 17 Mär. 2022
Bearbeitet: Michael Walter
am 17 Mär. 2022
Jan
am 17 Mär. 2022
The size matters, because the creation of the intermediate matrix c-s is time consuming. Then the RAM access is the bottleneck.
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Weitere Antworten (2)
Matt J
am 17 Mär. 2022
c=c(:)';
b=interp1(c,1:numel(c),s,'nearest');
a=abs(c(b)-s)
c = c(:); % Make it a column vector to be sure
nc = numel(c);
bin = [c(1); (c(1:nc-1) + c(2:nc)) * 0.5; c(nc)];
[~, idx] = histc(s, bin);
value = abs(c(idx) - s(:));
bin contains the start point, midpoints and final point of the intervals of c.
histc uses a binary search to find the index of the value in c for each element of s. This avoids to create the large matrix c - s. Unfortunately MathWorks decided to replace the working histc by histcounts, which has some performance problems.
See also: interp1('nearest')
c = linspace(0, 1, 1e4);
s = rand(1, 1e5);
tic;
[a, b] = min(abs(c(:) - s(:).'));
toc
tic;
c = c(:);
nc = numel(c);
bin = [c(1); (c(1:nc-1) + c(2:nc)) * 0.5; c(nc)];
[~, b2] = histc(s, bin);
% Alternative, seems to be faster:
% [~, ~, b2] = histcounts(s, bin);
a2 = abs(c(b2) - s(:));
toc
isequal(b, b2)
max(abs(a(:) - a2(:)))
4 Kommentare
Michael Walter
am 17 Mär. 2022
Bearbeitet: Michael Walter
am 17 Mär. 2022
Can you provide some input data, which reproduce the problem?
HISTC replies 0, if the data are not inside the bins. So you can expand the first an last bins to the maximal values of s:
% bin = [c(1); (c(1:nc-1) + c(2:nc)) * 0.5; c(nc)]; % Original
bin = [min(s); (c(1:nc-1) + c(2:nc)) * 0.5; max(s)];
If you know the values in advance, you can save the time for searching min and max.
Michael Walter
am 18 Mär. 2022
Bearbeitet: Michael Walter
am 18 Mär. 2022
Michael Walter
am 18 Mär. 2022
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