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)
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採用された回答
Steven Lord
2023 年 8 月 15 日
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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その他の回答 (1 件)
Torsten
2023 年 8 月 15 日
移動済み: Torsten
2023 年 8 月 15 日
k = find(abs(x-0.3)==min(abs(x-0.3)))
2 件のコメント
Torsten
2023 年 8 月 15 日
編集済み: Torsten
2023 年 8 月 15 日
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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