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

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

[Jose]

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]

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?

Answer 1

[per isakson]

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

[Richard Barrett-Jolley]

Bingo. That does it. Thanks

[Peter Kövesdi]

Take care: This routine fails with uint data types. Transform them to double first:

double(x);

or

double(val);

as needed.

Answer 2

[Peter Kövesdi]

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)

[Andoni Medina Murua]

However

interp1(x,1:length(x),val,'nearest');

works in case val is an array, which doesn't happen with

min( abs( x-val ) );

Answer 3

[Revant Adlakha]

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]

Try this

index = fix(interp1(array,find(array),value));

Answer 4

[Ernest Nachaki]

Try this

index = fix(interp1(array,find(array),value));