How to avoid counting elements in each row of matrix multiple times?

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I have written this function to calculate the location distribution in a form of a grid pattern(2D map of distribution).
x&y which are coordinates of each points are as follows:
X: is a matrix that each row contins a x position of points which creats an individual trajectory.(number of rows)
Y: is a matrix that each row contins a y position of points corresponding to x positions.
Note: each rows of X or Y contains a trajectory with a specified length and the rest of row is filled with NaNs:
X=[0.3 0.18 1.2 2.3 nan nan nan ,....,nan ;
0.1 0.12 0.5 2.6 nan nan nan ,....,nan ;
.......]
the same thing for Y.
=====================================
Question: the N matrix calculated in "for loop" below counts points multiple times if they are belong to one trajectory(one row of X) but pass through one grid multiple times. how can I modify the code to avoid counting points multiple times if they belong to the same trajectory but pass through the grid multiple times?
=====================================
function [binx,biny,dx,dy,N] = myfunc(X,Y,Nbin,[Xmin Xmax Ymin Ymax])
%Nbin=number of bins to create 2D matrix of N
lim(1)=Xmin;
lim(2)=Xmax;
lim(3)=Ymin;
lim(4)=Ymax;
% change to 1D arrays
X = X(:);
Y = Y(:);
%exclude nans
X = X(find(~isnan(X)));
Y = Y(find(~isnan(Y)));
%variable range
if isempty(lim)
in1 = min(X(:));
en1 = max(X(:));
in2 = min(Y(:));
en2 = max(Y(:));
else
in1 = lim(1);
en1 = lim(2);
in2 = lim(3);
en2 = lim(4);
end
%initialise bin sizes
dx = abs(en1-in1)/Nbin;
dy = abs(en2-in2)/Nbin;
binx = [in1+dx/2:dx:en1-dx/2];
biny = [in2+dy/2:dy:en2-dy/2];
%compute N matrix of 2D-map
N = zeros(Nbin,Nbin);
for ii = 1:length(X)
%check if variables fall within the domain
if X(ii) >= binx(1)-dx/2 && X(ii) <= binx(end)+dx/2 && Y(ii) >= biny(1)-dy/2 && Y(ii) <= biny(end)+dy/2
%find bin
indx = find(binx+dx/2 >= X(ii));
indx = indx(1);
indy = find(biny+dy/2 >= Y(ii));
indy = indy(1);
N(indy,indx) = N(indy,indx) + 1;
end
end
  12 Comments
Ham Man
Ham Man on 14 Mar 2024
Edited: Ham Man on 14 Mar 2024
Thank you Matt, yes that should be counted once as well.

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Answers (1)

Matt J
Matt J on 14 Mar 2024
Edited: Matt J on 14 Mar 2024
Xedges=linspace(Xmin,Xmax, Nbin+1);
Yedges=linspace(Ymin,Ymax, Nbin+1);
N=0;
for i=1:height(X)
deltaN = histcounts2(X(i,:),Y(i,:),Xedges,Yedges);
N=N+logical(deltaN);
end

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