Azzera filtri
Azzera filtri

Another way instead of for loop

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muhammad ismat
muhammad ismat il 17 Giu 2015
Modificato: Walter Roberson il 17 Giu 2015
I have this code
for i = 1:34
for j = 1:i % <-- Note the 1:i instead of 1:n
s(i,j) = abs(z(i,ind(j))-z(j,ind(i)))/(z(i,ind(j))+z(j,ind(i)));
s(j,i) = s(i,j)
end
end
but it is very slow, so i want another way instead of for loop.
  2 Commenti
Christiaan
Christiaan il 17 Giu 2015
Dear Muhammad, What would you like to acchieve? Kind regards
muhammad ismat
muhammad ismat il 17 Giu 2015
i have a matrix 10 x 10
then i make a cluster to previous matrix then i want calculate eq
(z(i,ind(j))-z(j,ind(i)))/(z(i,ind(j))+z(j,ind(i)))
to each point in the matrix
where ind cluster no that a point belong to, z is the distance from point to a cluster

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