How do i add tolerance to the code. PLS HELP.

2022 4 2
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更新时间：2022 4 3
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MIE is has 3501 values inside it and MIE_2 is a single value. I am trying to find the index of MIE_2 value inside the MIE with next closest possible value so as to plot it using indexing. since MIE_2 us calculated using expression, upto 4 decimals it does not match with values in MIE.
tol = 0.0001;
MIEIdx = find(MIE == (MIE_2),1,'first');

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Walter Roberson
Walter Roberson 2022-4-2

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MIEIdx = find(abs(MIE_2 - MIE) <= tol);
This might have 0 or more matches.
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[dist, MIEIdx] = min(abs(MIE_2 - MIE));
if dist > tol
%no match
else
%MIEIdx is the match
end

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Sarthak Jakar
Sarthak Jakar 2022-4-3
编辑：Sarthak Jakar 2022-4-3
Thanks @Walter Roberson. The first code MIEIdx = find(abs(MIE_2 - MIE) <= tol); matches the nearest posssible value as per the tolerance.
Walter Roberson
Walter Roberson 2022-4-3
No, that first code finds all of the values that are within the tolerance, not just the nearest.
Sarthak Jakar
Sarthak Jakar 2022-4-3
Oh yes, when i run it for different data it gives me 4-5 values with accuracy described as per the tolerance. But it still works for me. thanks.

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