Finding closest match in collection of numbers [closed]

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So I got asked today what was the best way to find the closes match within a collection.

For example, you've got an array like this:

1, 3, 8, 10, 13, ...

What number is closest to 4?

Collection is numerical, unordered and can be anything. Same with the number to match.

Lets see what we can come up with, from the various languages of choice.

Answer 1

11 bytes in J:

C=:0{]/:|@-

Examples:

>> a =: 1 3 8 10 13
>> 4 C a
3
>> 11 C a
10
>> 12 C a
13

my breakdown for the layman:

0{         First element of
]          the right argument
/:         sorted by
|          absolute value 
@          of
-          subtraction

Answer 2

Shorter Python: 41 chars

f=lambda a,l:min(l,key=lambda x:abs(x-a))

Answer 3

My attempt in python:

def closest(target, collection) :
    return min((abs(target - i), i) for i in collection)[1]

Answer 4

Groovy 28B

f={a,n->a.min{(it-n).abs()}}

Answer 5

Some C# Linq ones... too many ways to do this!

decimal[] nums = { 1, 3, 8, 12 };
decimal target = 4;

var close1 = (from n in nums orderby Math.Abs(n-target) select n).First();
var close2 = nums.OrderBy(n => Math.Abs(n - target)).First();

Console.WriteLine("{0} and {1}", close1, close2);

Even more ways if you use a list instead, since plain ol arrays have no .Sort()

Answer 6

Assuming that the values start in a table called T with a column called N, and we are looking for the value 4 then in Oracle SQL it takes 59 characters:

select*from(select*from t order by abs(n-4))where rownum=1

I've used select * to reduce the whitespace requirements.

Answer 7

Because I actually needed to do this, here is my PHP

$match = 33;

$set = array(1,2,3,5,8,13,21,34,55,89,144,233,377,610);

foreach ($set as $fib)
    {
        $diff[$fib] = (int) abs($match - $fib);
    }
$fibs = array_flip($diff);
$closest = $fibs[min($diff)];

echo $closest;

Answer 8

PostgreSQL:

select n from tbl order by abs(4 - n) limit 1

In the case where two records share the same value for "abs(4 - id)" the output would be in-determinant and perhaps not a constant. To fix that I suggest something like the untested guess:

select n from tbl order by abs(4 - n) + 0.5 * 4 > n limit 1;

This solution provides performance on the order of O(N log N), where O(log N) is possible for example: https://stackoverflow.com/a/8900318/1153319

Answer 9

Ruby like Python has a min method for Enumerable so you don't need to do a sort.

def c(value, t_array)
  t_array.min{|a,b|  (value-a).abs <=> (value-b).abs }
end

ar = [1, 3, 8, 10, 13]
t = 4
c(t, ar) = 3

Answer 10

Language: C, Char count: 79

c(int v,int*a,int A){int n=*a;for(;--A;++a)n=abs(v-*a)<abs(v-n)?*a:n;return n;}

Signature:

int closest(int value, int *array, int array_size);

Usage:

main()
{
    int a[5] = {1, 3, 8, 10, 13};
    printf("%d\n", c(4, a, 5));
}

Answer 11

Scala (62 chars), based on the idea of the J and Ruby solutions:

def c(l:List[Int],n:Int)=l.sort((a,b)=>(a-n).abs<(b-n).abs)(0)

Usage:

println(c(List(1,3,8,10,13),4))

Answer 12

PostgreSQL:

This was pointed out by RhodiumToad on FreeNode and has performance on the order of O(log N)., much better then the other PostgreSQL answer here.

select * from ((select * from tbl where id <= 4
order by id desc limit 1) union
(select * from tbl where id >= 4
order by id limit 1)) s order by abs(4 - id) limit 1;

Both of the conditionals should be "or equal to" for much better handling of the id exists case. This also has handling in the case where two records share the same value for "abs(4 - id)" then that other PostgreSQL answer here.

Answer 13

The above code doesn't works for floating numbers.
So here's my revised php code for that.

function find_closest($match, $set=array()) {
    foreach ($set as $fib) {  
        $diff[$fib] = abs($match - $fib);  
    }  
    return array_search(min($diff), $diff);  
}

$set = array('2.3', '3.4', '3.56', '4.05', '5.5', '5.67');  
echo find_closest(3.85, $set); //return 4.05  

Answer 14

Python by me and https://stackoverflow.com/users/29253/igorgue based on some of the other answers here. Only 34 characters:

min([(abs(t-x), x) for x in a])[1]

Answer 15

Haskell entry (tested):

import Data.List

near4 = head . sortBy (\n1 n2 -> abs (n1-4) `compare` abs (n2-4))

Sorts the list by putting numbers closer to 4 near the the front. head takes the first element (closest to 4).

Answer 16

Ruby

def c(r,t)
r.sort{|a,b|(a-t).abs<=>(b-t).abs}[0]
end

Not the most efficient method, but pretty short.

Answer 17

returns only one number:

var arr = new int[] { 1, 3, 8, 10, 13 };
int numToMatch = 4;

Console.WriteLine("{0}", 
   arr.OrderBy(n => Math.Abs(numToMatch - n)).ElementAt(0));

Answer 18

returns only one number:

var arr = new int[] { 1, 3, 8, 10, 13 };
int numToMatch = 4;
Console.WriteLine("{0}", 
     arr.Select(n => new{n, diff = Math.Abs(numToMatch - n) }).OrderBy(x => x.diff).ElementAt(0).n);

Answer 19

Perl -- 66 chars:

perl -e 'for(qw/1 3 8 10 13/){$d=($_-4)**2; $c=$_ if not $x or $d<$x;$x=$d;}print $c;'

Answer 20

EDITED = in the for loop

int Closest(int val, int[] arr)
{
    int index = 0;
    for (int i = 0; i < arr.Length; i++)
        if (Math.Abs(arr[i] - val) < Math.Abs(arr[index] - val))
            index = i;
    return arr[index];
}

Answer 21

Here's another Haskell answer:

import Control.Arrow
near4 = snd . minimum . map (abs . subtract 4 &&& id)

Answer 22

Haskell, 60 characters -

f a=head.Data.List.sortBy(compare`Data.Function.on`abs.(a-))

Answer 23

Kdb+, 23B:

C:{x first iasc abs x-}

Usage:

q)a:10?20
q)a
12 8 10 1 9 11 5 6 1 5

q)C[a]4
5

Answer 24

Python, not sure how to format code, and not sure if code will run as is, but it's logic should work, and there maybe builtins that do it anyways...

list = [1,4,10,20]
num = 7
for lower in list:
         if lower <= num:
           lowest = lower #closest lowest number

for higher in list:
     if higher >= num:
           highest = higher #closest highest number

if highest - num > num - lowest: # compares the differences
    closer_num = highest
else:
    closer_num = lowest

Answer 25

In Java Use a Navigable Map

NavigableMap <Integer, Integer>navMap = new ConcurrentSkipListMap<Integer, Integer>();  

navMap.put(15000, 3);  
navMap.put(8000, 1);  
navMap.put(12000, 2);  

System.out.println("Entry <= 12500:"+navMap.floorEntry(12500).getKey());  
System.out.println("Entry <= 12000:"+navMap.floorEntry(12000).getKey());  
System.out.println("Entry > 12000:"+navMap.higherEntry(12000).getKey()); 

Answer 26

int numberToMatch = 4;

var closestMatches = new List<int>();
closestMatches.Add(arr[0]); // closest tentatively

int closestDifference = Math.Abs(numberToMatch - arr[0]);


for(int i = 1; i < arr.Length; i++)
{
    int difference = Math.Abs(numberToMatch - arr[i]);
    if (difference < closestDifference)
    {
        closestMatches.Clear();
        closestMatches.Add(arr[i]);
        closestDifference = difference;
    }
    else if (difference == closestDifference)
    {       
        closestMatches.Add(arr[i]);
    }
}


Console.WriteLine("Closest Matches");
foreach(int x in closestMatches) Console.WriteLine("{0}", x);

Answer 27

Some of you don't seem to be reading that the list is unordered (although with the example as it is I can understand your confusion). In Java:

public int closest(int needle, int haystack[]) { // yes i've been doing PHP lately
  assert haystack != null;
  assert haystack.length; > 0;
  int ret = haystack[0];
  int diff = Math.abs(ret - needle);
  for (int i=1; i<haystack.length; i++) {
    if (ret != haystack[i]) {
      int newdiff = Math.abs(haystack[i] - needle);
      if (newdiff < diff) {
        ret = haystack[i];
        diff = newdiff;
      }
    }
  }
  return ret;
}

Not exactly terse but hey its Java.

Answer 28

Common Lisp using iterate library.

(defun closest-match (list n)
     (iter (for i in list)
            (finding i minimizing (abs (- i n)))

Answer 29

41 characters in F#:

let C x = Seq.min_by (fun n -> abs(n-x))

as in

#light

let l = [1;3;8;10;13]

let C x = Seq.min_by (fun n -> abs(n-x))

printfn "%d" (C 4 l)   // 3 
printfn "%d" (C 11 l)  // 10
printfn "%d" (C 12 l)  // 13

Answer 30

Ruby. One pass-through. Handles negative numbers nicely. Perhaps not very short, but certainly pretty.

class Array
  def closest int
    diff = int-self[0]; best = self[0]
    each {|i|
      if (int-i).abs < diff.abs
        best = i; diff = int-i
      end
    }
    best
  end
end

puts [1,3,8,10,13].closest 4