Regsrding for loop with array

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Hi all, A=[1;3;5; 6]; for 1 to 100 iterations, if any element in array matches with the iteration,e.g .for 1st iteration A's 1st data is matching. else it should come out from the loop. Can u suggest me any idea?
Thanks
  1 Commento
Muruganandham Subramanian
Muruganandham Subramanian il 26 Mag 2012
for lm=1:100
if (~isempty(find(A == lm)))
%calcultion
end
end

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Risposta accettata

Wayne King
Wayne King il 26 Mag 2012
for nn = 1:8
if(any(A==nn)),
disp('hi');
else
disp('bye');
end
end
Obviously, replace disp('hi') with your calculation and I've just shown this up to an index of 8.

Più risposte (1)

Muruganandham Subramanian
Muruganandham Subramanian il 26 Mag 2012
Hi wayne, This also works!!! for lm=1:100 if (~isempty(find(A == lm))) %calcultion end end
  1 Commento
Wayne King
Wayne King il 26 Mag 2012
Hi, sorry! I didn't see your comment above :)

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