How to find the index of the closest value to some number in 1D array ?

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How to find the index in 1D array that has closest value to some number ?
val =-1.03
val1 = 1.04
x = -10:0.009:10
ind1 = find (A==val) % will work if the val is exact match
  2 Comments
Jose
Jose on 15 Feb 2023
Edited: Jose on 15 Feb 2023
Index=find(min(abs(Array-target))==abs(Array-target))
that should work even if the # of sig digits change in your array. It finds the location of value in the array, that when substracted from your target value has the smallest difference (i.e. closest match).
Din N
Din N on 17 Mar 2023
This does give the closest value, but if you want the closest value to be smaller than your target value? For example if my target value is 7300, how can I specify here that I only want the index for the closest value that is smaller than 7300?

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Accepted Answer

per isakson
per isakson on 27 Mar 2015
Edited: per isakson on 2 Apr 2019
Hint:
>> [ d, ix ] = min( abs( x-val ) );
>> x(ix-1:ix+1)
ans =
-1.0360 -1.0270 -1.0180
ix is the "index in 1D array that has closest value to" val
"if the val is exact match"   that's tricky with floating point numbers

More Answers (3)

Peter Kövesdi
Peter Kövesdi on 1 Apr 2019
ind = interp1(x,1:length(x),val,'nearest');
also does it.
But a short comparison shows disadvantages in timing:
f1=@()interp1(x,1:length(x),val,'nearest');
f2=@()min( abs( x-val ) );
timeit(f1)>timeit(f2)
  1 Comment
Andoni Medina Murua
Andoni Medina Murua on 18 Aug 2022
Edited: Andoni Medina Murua on 18 Aug 2022
However
interp1(x,1:length(x),val,'nearest');
works in case val is an array, which doesn't happen with
min( abs( x-val ) );

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Revant Adlakha
Revant Adlakha on 26 Feb 2021
You could also use something like this, where f(x) is the function and x is the value of interest.
ind = find(min(abs(f(x) - x)) == abs(f(x) - x));

Ernest Nachaki
Ernest Nachaki on 27 May 2022
Try this
index = fix(interp1(array,find(array),value));

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