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Why NaN even though all elements considered

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Abhishek H P
Abhishek H P am 25 Mai 2016
Bearbeitet: Stephen23 am 25 Mai 2016
I am comparing every column with other columns but getting the answer in R(4,2) and R(3,1)
for i=1:nrow-1
a=data(1,i);
c=data(2,i);
e=data(3,i);
g=data(4,i);
for j=[1:i-1,i+1:nrow]
b=a-data(1,j);
d=c-data(2,j);
f=e-data(3,j);
h=g-data(4,j);
kU=b'*b;
kL=d'*d;
pU=f'*f;
pL=h'*h;
S=(kU'*kU+kL'*kL)/(kU+kL)+(pU'*pU+pL'*pL)/(pU+pL);
R(i,j) = S;
R(j,i) = S;
end
end
end
trial(A)
ans =
0 1.8299 NaN 1.8299
1.8299 0 1.8299 NaN
NaN 1.8299 0 1.8299
1.8299 NaN 1.8299 0
when the input
A = 2.0000 2.2000 2.0000 2.2000
3.0000 3.1000 3.0000 3.1000
2.0000 2.2000 2.0000 2.2000
3.0000 3.1000 3.0000 3.1000
  2 Kommentare
Stephen23
Stephen23 am 25 Mai 2016
Bearbeitet: Stephen23 am 25 Mai 2016
Your algorithm divides zero by zero. This happens for these pairs of (i,j): (1,3), (2,4), (3,1). What do you expect division of zero by zero to do?
Roger Stafford
Roger Stafford am 25 Mai 2016
Actually it is zero divided by zero that is producing the NaNs here. A non-zero divided by zero will produce inf or -inf,

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