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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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)

Accepted Answer

Steven Lord
Steven Lord on 15 Aug 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
x = 1×11
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000
It appears that x contains the value 0.3, but it does not contain exactly 0.3.
checkWithExactEquality = x == 0.3
checkWithExactEquality = 1×11 logical array
0 0 0 0 0 0 0 0 0 0 0
It does contain a value that is extremely close to 0.3, however.
tolerance = 1e-15;
checkWithTolerance = abs(x-0.3) < tolerance
checkWithTolerance = 1×11 logical array
0 0 0 1 0 0 0 0 0 0 0
whichValueTolerance = x(checkWithTolerance)
whichValueTolerance = 0.3000
How far away from 0.3 is the value we found using a tolerance?
howDifferent = whichValueTolerance - 0.3
howDifferent = 5.5511e-17
To do the same with ismembertol:
checkWithIsmembertol = ismembertol(x, 0.3, tolerance)
checkWithIsmembertol = 1×11 logical array
0 0 0 1 0 0 0 0 0 0 0
whichValueIsmembertol = x(checkWithIsmembertol)
whichValueIsmembertol = 0.3000
The ismembertol function found the same value that the check with a tolerance did.

More Answers (1)

Torsten
Torsten on 15 Aug 2023
Moved: Torsten on 15 Aug 2023
k = find(abs(x-0.3)==min(abs(x-0.3)))

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