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an Alternative funtion which is faster than "ismember"

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Hello everybody,
I was using ismembertol with XY(Nx2) and xy(Mx2). However code never ends due to the enormous amount of data(N=400million M=80mil.).
Is there any way that I can speed this function. (The matrices are not unique)
[LIA,~]= ismembertol(XY,xy,0.00001,'ByRows',true,'OutputAllIndices',true);
Thank you for your support
Ahmet Hakan UYANIK
Ahmet Hakan UYANIK on 24 Nov 2022
I deeply appreciate the time and effort you took Bruno. It works perfectly without a loop which is amazing. I also would like to accept this answer but since it is in the comments, the botton does not apper.
Bruno Luong
Bruno Luong on 24 Nov 2022
I post a short code as answer so you can accept it. Thanks

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Accepted Answer

Bruno Luong
Bruno Luong on 24 Nov 2022
If XY is gridded coordinates, then you can use discretize or simple division if they are uniform to determine which grid the river point belong to.
% Generate some toy fake data
xgrid = cumsum(randi(5,1,10))
x = min(xgrid)+rand(1,10)*(max(xgrid)-min(xgrid))
midpoints = (xgrid(1:end-1)+xgrid(2:end))/2;
x_edges = [-Inf midpoints Inf];
iclosest_x = discretize(x, x_edges)
xgridclosest = xgrid(iclosest_x);
d = abs(xgridclosest-x)
Do the same for y, then
LIA = false(length(ygrid),length(xgrid));
LIA(sub2ind(size(LIA), iclosest_y, iclosest_x)) = true;

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