in the below mentioned code, i get an error in finding some elements (not in all) : 0×1 empty double column vector
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ravi shukla
le 15 Août 2023
Modifié(e) : Torsten
le 15 Août 2023
m=41;
n=51;
dx=0.0500;
dy=0.0500;
for i=1:m
for j=1:n
x(i,1)=(i-1)*dx;
y(j,1)=(j-1)*dy;
end
end
k=find(x==0.3)
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Réponse acceptée
Steven Lord
le 15 Août 2023
This behavior is a consequence of floating point arithmetic. See this Answers post and the "Avoiding Common Problems with Floating-Point Arithmetic" section of this documentation page for more information.
If you are using the == operator to attempt to locate a floating-point number in an array, instead subtract the number you're trying to find from the numbers in the array and locate those positions where the difference is smaller than some tolerance or use the ismembertol function.
x = 0:0.1:1
It appears that x contains the value 0.3, but it does not contain exactly 0.3.
checkWithExactEquality = x == 0.3
It does contain a value that is extremely close to 0.3, however.
tolerance = 1e-15;
checkWithTolerance = abs(x-0.3) < tolerance
whichValueTolerance = x(checkWithTolerance)
How far away from 0.3 is the value we found using a tolerance?
howDifferent = whichValueTolerance - 0.3
To do the same with ismembertol:
checkWithIsmembertol = ismembertol(x, 0.3, tolerance)
whichValueIsmembertol = x(checkWithIsmembertol)
The ismembertol function found the same value that the check with a tolerance did.
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Torsten
le 15 Août 2023
Déplacé(e) : Torsten
le 15 Août 2023
k = find(abs(x-0.3)==min(abs(x-0.3)))
2 commentaires
Torsten
le 15 Août 2023
Modifié(e) : Torsten
le 15 Août 2023
Asking for strict equality of expressions is dangerous because of floating point arithmetic. See @Steven Lord 's answer for a more detailed explanation.
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