# Find nearest number in a given array

This is inspired by a real world problem I had. I'm curious to see if there is any clever way to go about this.

You are given two unsorted arrays, A and B, each containing an arbitrary number of floats. A and B don't necessarily have the same lengths. Write a function that takes the elements of A sequentially and finds the nearest value in array B. The result has to be contained in a new array.

Win condition

Shortest code wins (as usual).

• Round to the nearest integer? – n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ Jun 11 '14 at 14:10
• @n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ I read that as "round each element of A to the nearest element of B" – John Dvorak Jun 11 '14 at 14:14
• @JanDvorak: Well, I understand the part about rounding direction, but the problem didn't specify to how many digits. – n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ Jun 11 '14 at 14:15
• @n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ Round to nearest float. The answer has to output floats from array/list B. – Orhym Jun 11 '14 at 14:16
• Will arrays A and B be sorted? – Level River St Jun 11 '14 at 14:21

# APL, 13 17

(21 byte in UTF-8)

B[{↑⍋|⍵-B}¨A]


If you want true lambda (A as left argument and B as right):

{⍵[⍺{↑⍋|⍺-⍵}¨⊂⍵]}


## How it works:

{...}¨A invokes lambda function {...} with every A value (instead of invoking with A as array), gathering results to array of same shape

|⍵-B computes absolute values of difference between argument ⍵ and all in B (- is subtraction, | is abs).

↑⍋ takes index of least element (⍋ sorts array returning indices, ↑ get first element)

B[...] is just fetching element(s) by index(es).

The solution is quite strightforward, altough it uses wonderful feature of APL's sorting function returning permutation vector (sorted element's indices in original array) rather than sorted array itself.

• How does this work? – John Dvorak Jun 11 '14 at 14:30
• Explained in answer – Vovanium Jun 11 '14 at 14:43
• How on earth do you know how to write this? – Martijn Jun 12 '14 at 12:43
• This is like writing chinese. For me, there's no great difference writing either alien words or alien characters... – Vovanium Jun 18 '14 at 20:42

## Mathematica - 17

#&@@@Nearest@A/@B


How does it work? Yes, I admit that there's a bit of cheating here because Mathematica has built-in nearest functionality. The rest is straightforward and is concerned with arranging the result in a 1D array. It looks ugly only because of the extra effort to make it short.

• Ha! Welcome! :) – Dr. belisarius Jun 12 '14 at 0:52

# C# - 10397 87 Bytes

I'm not quite sure if I understood this question correctly but here is my solution anyway. I used Lists instead of arrays, because it allows me to write shorter code.

A integer array is shorter than a integer list.

Input:

t(new int[] { 0, 25, 10, 38 }, new int[] { 3, 22, 15, 49, 2 });


Method:

void t(int[]a,int[]b){var e=a.Select(c=>b.OrderBy(i=>Math.Abs(c-i)).First()).ToArray();


Output:

2, 22, 15, 49


EDIT: AS @grax pointed out, the question is now about floats. Therefore I'd like to include his answer too.

float[]t(float[]a,float[]b){return a.Select(d=>b.OrderBy(e=>Math.Abs(e-d)).First()).ToArray();}

• Lists are fine too. – Orhym Jun 11 '14 at 14:15
• Rename item to i and you will safe 6 additional characters ;) – Aschratt Jun 11 '14 at 14:31
• @Aschratt thank you very much! – tsavinho Jun 11 '14 at 14:38
• 1. The function doesn't specifically say to return the new value, but I think you should. 2. Since the question called for float, I think you should use float float[] t(float[] a, float[] b) {return a.Select(d=>b.OrderBy(e=>Math.Abs(e-d)).First()).ToArray();} – Grax Jun 11 '14 at 14:51
• @Grax As I wrote my first answer, the question wasn't about floats. Since the question has been updated, I included your answer too. Thank you very much. – tsavinho Jun 12 '14 at 9:16

## R, 41 chars

B[apply(abs(outer(A,B,-)),1,which.min)]


Explanation:

outer(A,B,-) computes for each element x of A the difference x-B and outputs the result as a matrix (of dimension length(A) x length(B)).
which.min picks the index of the minimal number.
apply(x, 1, f) applies function f on each row of matrix x.
So apply(abs(outer(A,B,-)),1,which.min) returns the indices of the minimal absolute difference between each element of A and the elements of vector B.

Usage:

> A <- runif(10,0,50)
> B <- runif(10,0,50)
> A
[1] 10.0394987 23.4564467 19.6667152 36.7101256 47.4567670 49.8315028  2.1321263 19.2866901  0.7668489 22.5539178
> B
[1] 44.010174 32.743469  1.908891 48.222695 16.966245 23.092239 24.762485 30.793543 48.703640  6.935354
> B[apply(abs(outer(A,B,-)),1,which.min)]
[1]  6.935354 23.092239 16.966245 32.743469 48.222695 48.703640  1.908891 16.966245  1.908891 23.092239


# CJam - 14

q~
f{{1$-z}$0=\;}
p


The main code is on the second line, the rest is for using the standard input and pretty output.

Explanation:

q~ reads and evaluates the input
f{...} executes the block for each element of the first array and the next object (which is the second array), collecting the results in an array
{...}$ sorts the second array using the block to calculate a key for each item 1$ copies the current item from the first array
-z subtracts then takes the absolute value
0= takes the first value of the sorted array (the one with the minimum key)
\; discards the item from the first array
p prints the string representation of the result

Input: [10.1 11.2 12.3 13.4 9.5] [10 12 14]
Output: [10 12 12 14 10]

Input: [0 25 10 38] [3 22 15 49 2]
Output: [2 22 15 49]

# Javascript (E6) 54 56 59

Minimize distance. Using square instead of abs just so save chars.
Edit algebra ...
Edit fix useless assignment (a remainder of a test w/o the function definition)

F=(A,B)=>A.map(a=>B.sort((x,y)=>x*x-y*y+2*a*(y-x))[0])


Was F=(A,B)=>D=A.map(a=>B.sort((x,y)=>((x-=a,y-=a,x*x-y*y))[0])

Test

F([10.1, 11.2, 12.3, 13.4, 9.5],[10, 12, 14])


Result: [10, 12, 12, 14, 10]

• D= isn't needed, as map returns a new array. Alternative (same length) sort function: (x,y)=>(x-=a)*x-(y-=a)*y – nderscore Jun 11 '14 at 16:42

# Python 3.x - 55 chars

f=lambda a,b:[min((abs(x-n),x)for x in b)[1]for n in a]


a and b are the input arrays, and the desired array is the expression's result.

• I edited the answer to make it a function since the question requires a function. – user80551 Jun 12 '14 at 12:37

c a b=[[y|y<-b,(y-x)^2==minimum[(z-x)^2|z<-b]]!!0|x<-a]


At first, I thought to use minimumBy and comparing, but since those aren't in Prelude, it took a ton of characters to qualify them. Also stole the squaring idea from some other answers to shave off a character.

## PowerShell - 44

$a|%{$n=$_;($b|sort{[math]::abs($n-$_)})[0]}


Example

With $a and $b set to:

$a = @(36.3, 9, 50, 12, 18.7, 30)$b = @(30, 10, 40.5, 20)


Output is

40.5, 10, 40.5, 10, 20, 30

• you could use floats in the example to make it clear it handles floats too – bebe Jun 11 '14 at 21:02
• @bebe - Thanks, updated to make that clear. – Rynant Jun 12 '14 at 13:03
• -3 bytes: $a|%{$n=$_;($b|sort{($n-$_)*($n-$_)})[0]} – mazzy Dec 23 '18 at 8:09

# Ruby, 40

f=->a,b{a.map{|x|b.min_by{|y|(x-y)**2}}}


Same as the Python answer, but squaring is a little terser than any way I could think of to take absolute value.

# Pyth - 12 11 bytes

## Note: Pyth is much younger than this challenge, so this answer is not eligible to win.

Simple method, uses o order function to get minimal distance and maps it over list a.

mho.a-dNQvz

m    vz    Map over evaled first input and implicitly print
ho Q      Minimal mapped over evaled second input
.a-      Absolute difference
d       Lambda param 1
b       Lambda param 2

• @Jakube oh yeah, sorry. – Maltysen Jun 7 '15 at 1:53

# TI-BASIC, 24

∟A+seq(min(∟B+i²∟A(N)),N,1,dim(∟A


Doesn't come close to APL, but uses less powerful functions-- this uses no "sorted by" or "index of least" function. The disadvantage of TI-BASIC here is its lack of those functions and multidimensional arrays.

Ungolfed:

seq(       ,N,1,dim(∟A           #Sequence depending on the Nth element of list A
∟A(N)+min(   +0i)            #Number with minimum absolute value, add to ∟A(N)
∟B-∟A(N)           #Subtracts Nth element of ∟A from all elements of B


The min( function has two behaviors: when used with real numbers or lists, it gives the smallest value; however, when used with complex numbers or lists, it gives the value with the smallest absolute value. Adding 0i or multiplying by i^2 causes the interpreter to use the second behavior, so min(1,-2) returns -2 whereas min(1+0i,-2+0i) returns 1.

# Fortran 90: 88

function f();integer::f(size(a));f(:)=[(b(minloc(abs(a(i)-b))),i=1,size(a))];endfunction


This requires it to be contained within a full program:

program main
real :: a(5), b(3)
integer :: i(size(a))
a = [10.1, 11.2, 12.3, 13.4, 9.5]
b = [10, 12, 14]
i = f()
print*,i
contains
function f()
integer :: f(size(a))
f(:)=[(b(minloc(abs(a(i)-b))),i=1,size(a))]
end function
end program main


The square braces declare an array while (...,i=) represents an implied do loop; I then return the value of b for which element a(i)-b is minimized.

# Matlab: 48

f=@(a)B(abs(B-a)==min(abs(B-a)));C=arrayfun(f,A)


Assumes that A and B are 1D matrices in the workspace, Final result is C in the workspace. This would likely also work in Octave as well. Conditional indexing makes doing this fairly trivial.

# C 144 163

#define f float
f T, *C, m;
f *q(f *A, f *B, int S, f s)
{
if(m)
return abs(T - *A) - abs(T - *B);
for (
C = malloc(S * 4);
m = S--;
C[S] = *B
)
T = A[S],
qsort(B, s, 4, q);
return C;
}


Okay... I think this little code needs explanation.

At first i tried to do the job with two level of for loop finding the min difference and set the current value to min of B's value. That's very basic.

The same thing can be reached with qsort and a comparator function. I make it sort B by the difference instead of B's elements. Too many functions for such a little algorithm. So the function q now serves two purposes. At first, it's the algorithm itself, secondly (when qsort calls it) a comparator. For communication between the two states, I had to declare globals.

m stands for whether it's in comparator state or the main one.

example:

float A[] = {1.5, 5.6, 8.9, -33.1};
float B[] = {-20.1, 2.2, 10.3};
float *C;

C = q(A, B, sizeof(A)/sizeof(*A), sizeof(B)/sizeof(*B));
// C holds 2.2,2.2,10.3,-20.1

• Does the 166/163 count the whitespace or not? – Kyle Kanos Jun 11 '14 at 16:31
• Of course not. Spaces and newlines are for ease of understanding. – bebe Jun 11 '14 at 16:48

# GolfScript, 49 bytes

Note: this is a partial solution. I'm working on making it a complete solution

{{\.@\.[.,,\]zip@{[\~@-abs]}+%{~\;}\$0=0==}%\;}:f;


Yes. GolfScript does support floating point. Try it out here. Example:

# B is [-20.1 2.2 10.3]
[-201 10 -1?*
22 10 -1?*
103 10 -1?*]

# A. No floating point numbers allowed here.
# This is because 1.5{}+ (where the 1.5 is a
# single floating point number, not 1.5,
# which would be 1 1 5) results in the block
# {1.5 }, which leads to 1 1 5 when executed
[1 5 9 -30]


Output:

[2.2 2.2 10.3 -20.1]


## C# 262

Program finds min differences and saves closest value from Array B. I'll work on the golfing shortly.

List<float> F(List<float> a, List<float> b)
{
List<float> c = new List<float>();
float diff,min;
int k;
for(int i=0; i<a.Count;i++)
{
diff=0;
min=1e6F;
k = 0;
for(int j=0; j<b.Count;j++)
{
diff = Math.Abs(a[i] - b[j]);
if (diff < min)
{
min = diff;
k = j;
}
}
}
return c;
}


Full program with test code

using System;
using System.Collections.Generic;
public class JGolf
{
static List<float> NearestValues(List<float> a, List<float> b)
{
List<float> c = new List<float>();
float diff,min;
int k;
for(int i=0; i<a.Count;i++)
{
diff=0;
min=1e6F;
k = 0;
for(int j=0; j<b.Count;j++)
{
diff = Math.Abs(a[i] - b[j]);
if (diff < min)
{
min = diff;
k = j;
}
}
}
return c;
}

public static void Main(string[] args)
{
List<float> A = RandF(8413);
Console.WriteLine("A");
Print(A);
List<float> B = RandF(9448);
Console.WriteLine("B");
Print(B);

List<float> d = JGolf.NearestValues(A, B);
Console.WriteLine("d");
Print(d);
}

private static List<float> RandF(int seed)
{
Random r = new Random(seed);
int n = r.Next(9) + 1;
List<float> c = new List<float>();
while (n-- > 0)
{
}
return c;
}

private static void Print(List<float> d)
{
foreach(float f in d)
{
Console.Write(f.ToString()+", ");
}
}
}


# C#: 120

Linq is awesome:

float[] t(float[] A, float[] B){return A.Select(a => B.First(b => Math.Abs(b-a) == B.Min(c=>Math.Abs(c-a)))).ToArray();}