Regsrding for loop with array

2012 5 26
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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

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Wayne King
Wayne King 2012-5-26

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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.

更多回答（1 个）

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Hi wayne, This also works!!! for lm=1:100 if (~isempty(find(A == lm))) %calcultion end end

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