Check an array of values if within an upper and lower limit

38 vues (au cours des 30 derniers jours)
Ayman Fathy
Ayman Fathy le 15 Oct 2018
Commenté : Ayman Fathy le 15 Oct 2018
I am trying to do the same thing as the following example but checking an array of numbers and not just x. However it is not working. can someone help?
x = 10;
minVal = 2;
maxVal = 6;
if (x >= minVal) && (x <= maxVal)
disp('Value within specified range.')
elseif (x > maxVal)
disp('Value exceeds maximum value.')
else
disp('Value is below minimum value.')
end

Réponse acceptée

Dennis
Dennis le 15 Oct 2018
If you want to operate on arrays you could do it like this:
x = 1:10;
minVal = 2;
maxVal = 6;
x(x(x<=maxVal)>=minVal)
disp('Values within specified range.')
  3 commentaires
Dennis
Dennis le 15 Oct 2018
x = [randi(10,19,1),randi(100,19,1)];
minVal = 2;
maxVal = 6;
x(:,3)=NaN;
idx=find(x(:,1)>=minVal&x(:,1)<=maxVal);
x(idx,3)=x(idx,2)
Ayman Fathy
Ayman Fathy le 15 Oct 2018
Thanks a lot. It worked!!

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Plus de réponses (1)

KSSV
KSSV le 15 Oct 2018
minVal = 2;
maxVal = 6;
if any(x >= minVal) && any(x <= maxVal)
disp('Value within specified range.')
elseif any(x > maxVal)
disp('Value exceeds maximum value.')
else
disp('Value is below minimum value.')
end
  3 commentaires
Ayman Fathy
Ayman Fathy le 15 Oct 2018
Lowerlimit = 1000; Upperlimit = 2000;
if any(s(:,1) >= Lowerlimit) && any(s(:,1) <= Upperlimit) s(:,3) = s(:,2); else s(:,3) = NaN; end
%Itried this code but didnt work
Ayman Fathy
Ayman Fathy le 15 Oct 2018
this was my code previously and again didnt work:
% for x = size(s, 1) % % if (s(x, 1) >=Lowerlimit) (s(x,1) <= Upperlimit) % s(x, 3) = s(x, 2); % else % s(x, 2) = NaN; % end % end

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