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

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');

Answer 1

[Walter Roberson]

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

[Sarthak Jakar]

Thanks @Walter Roberson. The first code MIEIdx = find(abs(MIE_2 - MIE) <= tol); matches the nearest posssible value as per the tolerance.

[Walter Roberson]

No, that first code finds all of the values that are within the tolerance, not just the nearest.

[Sarthak Jakar]

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.