If you use N{3} instead of N3 an additional loop would be easy. At least I assume that it would. Unfortunately I cannot understand the first sentence already:
N3 uses N2 and one "k" variable(k is 3 for N3).
What is "one k variable" and where does this appear in the code? What is rr?
You could avoid the inner loop:
N = zeros(rr-3, rr-3);
for j = 1:rr-3
A1 = X(2:rr-1) - X(1:rr-3);
A2 = X(4:rr) - X(2:rr-2);
% A1==0 && A2==0 : N(j,i)=0; Is the default already
index = find(A1==0 & A2~=0);
N(j, index) = ((X(index + 3) - t(j)) ./ A2(index)) * N2(j,index + 1);
index = find(~A1~=0 & A2==0);
N(j, index) = ((t(j) - X(index)) ./ A1(index)) * N2(j,index);
index = find(A1~=0 & ~A2~=0);
N(j, index) = ((t(j) - X(index)) ./ A1(index)) * N2(j,index) + ...
((X(index + 3) - t(j)) / A2(index)) * N2(j, index+1);
end
N{3} = N;
Another approach would be to calculate both matrices and replace NaNs by zeros:
% !For demonstration only! Indices are most likely wrong!
A1 = X(2:rr-1) - X(1:rr-3);
A2 = X(4:rr) - X(2:rr-2);
c1 = bsxfun(@minus, (t ./ A1)' - X ./ A1) .* N2;
c2 = bsxfun(@minus, (X ./ A2)' - t ./ A2) .* N2;
c1(insan(c1)) = 0;
c2(insan(c2)) = 0;
N{3} = c1 + c2;
I did not care about the pile of indices here and I cannot run the versions due to the absence of test data. So treat this as an idea only and implement it by your own.
