Implicit expansion for griddedInterpolant

Hello,

I have to do a lot of interpolation on one data set and already found that griddedInterpolant is way faster than interp2;

It also allows element wise operation if two tensors of the same size are provided. As these are very big in my case, but repeat in some dimensions, I am wondering if something like implicit expansion (I hope I am using the correct terms here) can be used to speed up the code. Because for other functions using repmat is not advised.

Here I provide a minimum working example to show how I currently do it. Commented out are ways that I wish were possible to speed up the code but I can't get to run.

[X,Y] = ndgrid(0:10);

Z = rand([11,11]);

J = griddedInterpolant(X,Y,Z);

xq = sort(rand([4 1 3])*10);
yq = sort(rand([1 3 3])*10);

zq = J(repmat(xq,1,length(yq),1),repmat(yq,length(xq),1,1));

% zq = J(xq,yq); %implicit expansion?
% zq = bsxfun(@J,xq,yq);

% zq=J({xq,yq}) %Matt J's suggestion

If you have any input ( that can be generalized to different sizes of grids and lookups etc) I would be really thankful!

edit: changed xq, yq in the eaxmple to higher tensor to represent my problem more accurately

Answer 1

[Matt J]

Use the grid vector syntax of griddedInterpolant:

zq=J({xq,yq})

[Manuel Deuerling]

This would be perfect! Sorry, I overlooked this in the documentation-

However xq and yq might be tensors ( eg. size 4x1x3, and 1x3x3). Does this still work?

( I will be updating my example)

[Bruno Luong]

Just loop on top of it

[X,Y] = ndgrid(0:10);
Z = rand([11,11]);
J = griddedInterpolant(X,Y,Z);
xq = rand([4 1 3])*10;
yq = rand([1 3 3])*10;

zq = zeros(max(size(xq),size(yq)));
for k=1:size(zq,3)
    zq(:,:,k) = J({squeeze(xq(:,:,k)),squeeze(yq(:,:,k))});
end
zq

zq = 
zq(:,:,1) =

    0.2847    0.4967    0.5708
    0.4310    0.3534    0.5933
    0.2953    0.5418    0.6691
    0.2063    0.5544    0.3659


zq(:,:,2) =

    0.6066    0.4763    0.8624
    0.4667    0.6433    0.4922
    0.4370    0.5760    0.4852
    0.5364    0.2695    0.4939


zq(:,:,3) =

    0.3236    0.6992    0.5391
    0.4699    0.4195    0.4336
    0.6179    0.2786    0.3980
    0.2377    0.4675    0.3985

[Matt J]

@Manuel Deuerling Whether implicit expansion would help you depends on the dimensions of the tensors. For example, if you had size 2x1x3000, and 1x2x3000, there just wouldn't be any point pursuing a solution other than repmat. You basically have scattered query locations at that point.

[Manuel Deuerling]

@Matt J Yeah thats true. To be even more specific in my current problem it is like 100x1x100x100 and 1x100x1x100 so I would expect it to be useful. But this might change. So I guess there just isnt a generic solution I wished for. Thanks for the input I will work with that!

@Bruno Luong This actually seems to be faster. Not quite what i hoped for, as programming does not seem to be so clean (especially with multiple dimensions etc). But thanks too!

[Bruno Luong]

Not too much messy with 4+D

zq = zeros(max(size(xq),size(yq)));
for k=1:size(zq(:,:,:),3)
    zq(:,:,k) = J({squeeze(xq(:,:,k)),squeeze(yq(:,:,k))});
end

Answer 2

[Bruno Luong]

You can do your own extension as showed here

https://www.mathworks.com/matlabcentral/answers/2043017-bilinear-interpolation-faster-than-interp2?s_tid=srchtitle

[Manuel Deuerling]

Thanks for the suggestion & link.

If there is no other solution I will try this. I dont expect that an implementation by me could beat an efficient MathWorks algorithm in terms of memory/speed usage but I should probably just listen to you and try it.

[Bruno Luong]

It's probably depends on the size of your data. The strip down version like the one in this thread might beat TMW generic implementation.