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Use interp2 for many simultanious 1D operations?

Asked by Gustaf Lindberg on 21 Feb 2013

I have a loop where I do many extrapolations using interp1 with the flags 'linear' and 'extrap' so I guess it's technically an extrapolation. Now I want to use the same code but do many pairs of extrapolations at the same time. I have two matrices, X and Y where each row is a matching set to be extrapolated for a value xi = 0 but I can't wrap my head around how to make it work. I've been told that it is possible.

Basically, what I want to do is a faster version of this:

for i = 1:n
yi = interp1(X(i,:),Y(i,:),0,'linear','extrap')
end

But without the loop using interp2. each row n in the matrices X and Y are a matching set to be evaluated independently.

4 Comments

Gustaf Lindberg on 21 Feb 2013

I know nothing in advance. There are many, many iterations done with this check and the values can vary widely and there are no real sorting. Since this is for a CFD solver, stability is more important than speed. I've been told that interp2 could do what I want but I can't figure out how. All examples just uses it as an look up table for a 2D surface but I want multiple, independent 1D vectors.

José-Luis on 21 Feb 2013

I don't think interp2() is suited for what you want. If stability is more important than speed then I would stick to interp1() and a loop. If I really want to speed up things and I know nothing about the data, then I would write my own routine, preferably as a mex file.

You could also use parfor() when evaluating in a loop to speed things along.

Gustaf Lindberg on 21 Feb 2013

I was more or less told to use interp2 by my supervisor so I don't have that much of a choice right now but your suggestions are appreciated.

Gustaf Lindberg

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2 Answers

Answer by Matt J on 21 Feb 2013
Edited by Matt J on 21 Feb 2013
    [M,N]=size(X);
    yi=zeros(M,1);
    map=X>=0;
    [maxval,idx]=max(map,[],2);
    a =  idx==1 & maxval==1;  %All X(i,:) are non-negative, extrapolate left
    b =  idx==N | maxval==0;  %All X(i,:) are nonpositive,  extrapolate right
    c=   ~(a|b);              %Interpolation needed
    %Extrapolate left
    e1=X(a,1);
    c1=Y(a,1);
    e2=X(a,2);
    c2=Y(a,2);
    slopes=(c2-c1)./(e2-e1);
    yi(a) = c1-slopes.*e1;
    %Extrapolate right
    e1=X(b,end-1);
    c1=Y(b,end-1);
    e2=X(b,end);
    c2=Y(b,end);
    slopes=(c2-c1)./(e2-e1);
    yi(b) = c1-slopes.*e1;
    %Interpolate (linearly)
    j = sub2ind([M,N],find(c),idx(c));
    e1=X(j-M);
    c1=Y(j-M);
    e2=X(j);
    c2=Y(j);
    slopes=(c2-c1)./(e2-e1);
    yi(c) = c1+slopes.*abs(e1);  

2 Comments

Matt J on 21 Feb 2013

Had to make a few fixes, but it should work now. Here's some sample input/output data from my tests

    X =
        1.0000    2.0000    3.0000    4.0000    5.0000
      -10.0000   -9.0000   -8.0000   -7.0000   -6.0000
       -2.2500   -1.2500   -0.2500    0.7500    1.7500
    Y =
         3     4     5     6     7
        -5    -4    -3    -2    -1
        -2    -1     0     1     2
    yi =
        2.0000
        5.0000
        0.2500
Matt J on 21 Feb 2013

Some more tests, this time a speed comparison with for-looped interp1. My version shows a factor of 300 speed-up, but of course I don't know what data sizes you're actually working with.

%Fake data
  n=3e4;
  Y=rand(n,300);
  X=bsxfun(@plus, sort(Y,2), 2*rand(n,1)+1 );
    tic;
      % My code
    toc

Elapsed time is 0.017832 seconds.

    yi2=yi;
    tic;
      for i = 1:n
       yi2(i) = interp1(X(i,:),Y(i,:),0,'linear','extrap');
     end
    toc;

Elapsed time is 5.419498 seconds.

Matt J
Answer by Teja Muppirala on 21 Feb 2013

As others have also pointed out, this is REALLY not what INTERP2 is meant to do. In fact if anything, since there are 3 variables: X, Y, and i, you'd have to use INTERP3. But I think that would be rather inefficient.

Why do you think the FOR loop is a problem? How do you think doing something else will make it any more "stable"?

I can think of ways that might make it faster, but not really any more stable:

[n,m] = size(X)
[XS,Xi] = sort(X,2);
YS = Y(bsxfun(@plus,(Xi(:,1:2)-1)*n,(1:n)'));
y = YS(:,1) + diff(YS,1,2)./diff(XS(:,1:2),1,2).*(-XS(:,1));

2 Comments

Gustaf Lindberg on 21 Feb 2013

Removing the FOR loop is not meant to make it stable, it's meant to keep the same level of stability but increase speed. I've understood that this is not the proper way to use interp2 but I was told by my supervisor/examiner that it's what I should do.

José-Luis on 21 Feb 2013

With interp2() you are interpolating to a plane, not a line.

I guess you could assign increasing y values for each row, call them y =[ 1:number of rows] and then interpolate for x and y, where x = zeros(nRows,1), using interp2(). That could work.

Having said that, it is a very roundabout way of going about it and probably very inefficient. It takes more calculations to interpolate to a plane than to a line.

Teja Muppirala

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