Regsrding for loop with array

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Muruganandham Subramanian
Muruganandham Subramanian el 26 de Mayo de 2012
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 comentario
Muruganandham Subramanian
Muruganandham Subramanian el 26 de Mayo de 2012
for lm=1:100
if (~isempty(find(A == lm)))
%calcultion
end
end

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Wayne King
Wayne King el 26 de Mayo de 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.

Más respuestas (1)

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

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