interp1 - [Matlab R2016a] - Speed Issues

조회 수: 18 (최근 30일)
Ilya Klyashtornyy
Ilya Klyashtornyy 2017년 6월 30일
답변: Nikolaus Koopmann 2020년 6월 3일
Heeeey guys,
as I already noticed in this community, I am not alone with that problem. I have tried several solutions proposed in here but it didnt help for reasons like: Old Matlab Versions (interp1 has changed significatly) or another kind of problem.
Let me first tell you what i am doing to make it easier for you to help me:
I a for-loop with approx. 30.000 iterations i use interp1 several time (5 times per iteration - see the profiler below).
Example of the call: [interp1( x, y, xq)]
Here, x is a vector (length 10.000) and y is also a vector (length 10.000) and xq is a vector (length 50.000 or more).
The problem is also that y changes in each iteration and therefore, I cannot precompute any operation (xq is also updated each iteration).
I would be very grateful if you had any suggestions to improve the computational time in this case.
Best regards, Ilya
  댓글 수: 1
Walter Roberson
Walter Roberson 2017년 6월 30일
Would x happen to be equally spaced? Is xq sorted?

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채택된 답변

Steven Lord
Steven Lord 2017년 6월 30일
Test if creating a griddedInterpolant once before the start of the for loop, replacing the Values property of the object each time you change y, is faster.
  댓글 수: 4
Ilya Klyashtornyy
Ilya Klyashtornyy 2017년 6월 30일
편집: Ilya Klyashtornyy 2017년 6월 30일
I replaced all
int=interp1(x,y,xq);
with
S=griddedInterpolant(x,y);
int=S(xq);
It gave me a speed up of almost 25%
Thank you! :)
Ilya Klyashtornyy
Ilya Klyashtornyy 2017년 7월 1일
편집: Ilya Klyashtornyy 2017년 7월 1일
Hey, thanks for this advice.
I found out that (mb only in my case) creating a new griddedInterpolant takes less time than updating the values.
S=griddedInterpolant(x,y_new): 1.03s
S.Values=y_new: 1.32s
Do you agree with me on that? Do you receive different results?

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추가 답변 (2개)

Walter Roberson
Walter Roberson 2017년 6월 30일
  댓글 수: 6
Walter Roberson
Walter Roberson 2017년 7월 1일
Under the assumption that xq might be "exactly" -pi but is less than +pi:
xmin = -pi; xmax = +pi; span = xmax - xmin;
idx = floor( (xq - xmin) * length(x) / span + 1 );
offset = xq - x(idx);
yq = y(idx) * (1-offset) + y(idx+1) * offset;
I would need to double-check that the rounding for the index calculation works out well enough.
The idea is that with x being equally spaced with known min and max, then algebraically you can calculate the index into the x vector through a simple scaling operation. Then you can use the index vector to look up the values to do linear interpolation.
Ilya Klyashtornyy
Ilya Klyashtornyy 2017년 7월 2일
Hey Walter,
thank you very much for your time!
I ve evaluated ther method and compared it to the griddedInterpolant:
%Time per iteration:
griddedInterpolant: average - 0.00328s
your suggestion: average - 0.00672s

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Nikolaus Koopmann
Nikolaus Koopmann 2020년 6월 3일
% Current date = June 03, 2020
% Matlab version = 9.6.0.1072779 (R2019a)
% User name = Nikolaus Koopmann
function [yq,p] = interp1_lin(x,y,xq)
validateattributes(x,{'double'},{'increasing','vector'}) % slow
%% lin. regr.
X = [ones(length(x),1) x(:)];
p = flipud(X\y(:)); % see https://www.mathworks.com/help/matlab/data_analysis/linear-regression.html
% flin = @(x_)p(1)*x_ + p(2); % slow
yq = p(1)*xq + p(2);
end

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