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 Commenti
Jose
il 15 Feb 2023
Modificato: Jose
il 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
il 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?
Risposta accettata
per isakson
il 27 Mar 2015
Modificato: per isakson
il 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
2 Commenti
Peter Kövesdi
il 2 Apr 2019
Take care: This routine fails with uint data types. Transform them to double first:
double(x);
or
double(val);
as needed.
Più risposte (3)
Peter Kövesdi
il 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 Commento
Andoni Medina Murua
il 18 Ago 2022
Modificato: Andoni Medina Murua
il 18 Ago 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 ) );
Revant Adlakha
il 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));
1 Commento
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