Normalize a Vector

Question

To normalize a vector is to scale it to a length of 1 (a unit vector), whilst keeping the direction consistent.

For example, if we wanted to normalize a vector with 3 components, u, we would first find its length:

|u| = sqrt(u_x² + u_y² + u_z²)

...and then scale each component by this value to get a length 1 vector.

û = u ÷ |u|

The Challenge

Your task is to write a program or function which, given a non-empty list of signed integers, interprets it as a vector, and normalizes it. This should work for any number of dimensions, for example (test cases rounded to two decimal places):

[20]           -> [1]
[-5]           -> [-1]
[-3, 0]        -> [-1, 0]
[5.5, 6, -3.5] -> [0.62, 0.68, -0.40]
[3, 4, -5, -6] -> [0.32, 0.43, -0.54, -0.65]
[0, 0, 5, 0]   -> [0, 0, 1, 0]

Rules:

You can assume the input list will:
- Have at least one non-zero element
- Only contain numbers within your language's standard floating point range
Your output should be accurate to at least two decimal places. Returning "infinite precision" fractions / symbolic values is also allowed, if this is how your language internally stores the data.
Submissions should be either a full program which performs I/O, or a function. Function submissions can either return a new list, or modify the given list in place.
Builtin vector functions/classes are allowed. Additionally, if your language has a vector type which supports an arbitrary number of dimensions, you can take one of these as input.

This is a code-golf contest, so you should aim to achieve the shortest solution possible (in bytes).

Does it have to have at least two decimal places for every possible input (which is not possible for any standard type of floating point values) or only for the examples you provide? E.g. Steadybox's answer provides 2 decimal places of precision for all your test but he uses ints for the sum of squares which of course fails for almost all inputs (e.g. [0.1, 0.1]). — Christoph, Nov 27, 2017 at 11:07
... now we just wait for a lang with built-in norm function mapped to one char... — user18967, Nov 27, 2017 at 13:55
It should be to at least 2dp for every possible input @Christoph — FlipTack, Nov 27, 2017 at 16:09
@FlipTack but that rules out basically all languages because floatings points have bigger exponents than mantissa which means they do not always have enough precision to have any decimal places. — Christoph, Nov 27, 2017 at 19:34
Why don't the 6 in the 4th example and the -6 in the 5th respectively normalize to 1 and -1? — Mast, Nov 28, 2017 at 13:03

Adnan · Accepted Answer · 2017-11-26 22:07:41Z

15

05AB1E, 4 bytes

Code:

nOt/

Try it online!

Explanation

n     # Square each element of the input
 O    # Sum all elements
  t   # Take the square root of the sum
   /  # Divide each element by the square root of the sum

edited Nov 26, 2017 at 22:07

answered Nov 26, 2017 at 21:42

Adnan

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12

\$\begingroup\$ n0t what I expected / \$\endgroup\$
– YSC
Nov 27, 2017 at 10:19

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Arnauld · Accepted Answer · 2017-11-26 21:42:07Z

JavaScript (ES6), 31 bytes

a=>a.map(n=>n/Math.hypot(...a))

Test cases

let f =

a=>a.map(n=>n/Math.hypot(...a))

console.log(JSON.stringify(f([20]          ))) // -> [1]
console.log(JSON.stringify(f([-5]          ))) // -> [-1]
console.log(JSON.stringify(f([-3, 0]       ))) // -> [-1, 0]
console.log(JSON.stringify(f([5.5, 6, -3.5]))) // -> [0.62, 0.68, -0.40]
console.log(JSON.stringify(f([3, 4, -5, -6]))) // -> [0.32, 0.43, -0.54, -0.65]
console.log(JSON.stringify(f([0, 0, 5, 0]  ))) // -> [0, 0, 1, 0]

Community · Accepted Answer · 2020-06-17 09:04:33Z

10

Mathematica, 9 bytes

Normalize

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2017 at 21:45

ZaMoC

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13

\$\begingroup\$ Or #/Norm@#& for the same byte count. \$\endgroup\$
– Martin Ender
Nov 26, 2017 at 22:03

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Community · Accepted Answer · 2020-06-17 09:04:33Z

9

J, 8 bytes

%+/&.:*:

Try it online!

6 bytes %|@j./ works if the vector is at least 2-dimensional.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2017 at 21:53

FrownyFrog

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\$\begingroup\$ Love the way of getting the magnitude. \$\endgroup\$
– cole
Nov 26, 2017 at 22:54
1

\$\begingroup\$ @cole 1 byte longer: %1%:@#.*: \$\endgroup\$
– FrownyFrog
Nov 27, 2017 at 4:10
6

\$\begingroup\$ Could you please add an explanation for the uninitiated in J? \$\endgroup\$
– MechMK1
Nov 27, 2017 at 7:49
\$\begingroup\$ % (divide by) +/ (sum) &.: (under) *: (square). + sums two things. +/ sums a list of things. &.: modifies the preceding operation by applying the following operation first and its inverse afterwards. % normally takes two arguments, but (% f) is a function from x to x % (f x). Most operators automagically work on lists. \$\endgroup\$
– Roman Odaisky
Nov 28, 2017 at 21:20
\$\begingroup\$ And by the same principles, the function that “normalizes” a vector by adding such a number to each component that they sum to zero is “- +/ % #”. \$\endgroup\$
– Roman Odaisky
Nov 28, 2017 at 21:26

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caird coinheringaahin g · Accepted Answer · 2017-11-27 00:16:54Z

8

Jelly, 5 3 bytes

÷ÆḊ

Try it online!, or see the test suite

Saved 2 bytes thanks to miles!

edited Nov 27, 2017 at 0:16

answered Nov 26, 2017 at 22:02

caird coinheringaahin g♦

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\$\begingroup\$ 3 bytes with ÷ÆḊ \$\endgroup\$
– miles
Nov 26, 2017 at 23:46
\$\begingroup\$ @miles Huh, never knew about that builtin. Thanks \$\endgroup\$
– caird coinheringaahin g ♦
Nov 27, 2017 at 0:12
\$\begingroup\$ unfortunately that built-in gives +ve mod for scalars per TIO examplelike absolute value .. the problem requested keeping the sign \$\endgroup\$
– jayprich
May 1, 2018 at 15:34

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Stewie Griffin · Accepted Answer · 2017-11-26 21:50:40Z

6

Octave, 13 bytes

@(x)x/norm(x)

Try it online!

answered Nov 26, 2017 at 21:50

Stewie Griffin

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Steadybox · Accepted Answer · 2017-11-27 11:27:55Z

6

C, 73 70 bytes

Thanks to @Christoph for saving a byte!

s,i;f(v,n)float*v;{for(s=0;i++<n;)s+=*v**v++;for(;--i;)*--v/=sqrt(s);}

Try it online!

edited Nov 27, 2017 at 11:27

answered Nov 26, 2017 at 22:03

Steadybox

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\$\begingroup\$ +1. s=0,i=0 instead of s=i=0 saves one \$\endgroup\$
– caylee
Nov 27, 2017 at 6:10
\$\begingroup\$ I love the use of s[-i]but sadly *--v/=sqrt(s); is 1 byte shorter. \$\endgroup\$
– Christoph
Nov 27, 2017 at 11:08
1

\$\begingroup\$ @xanoetux Thanks, but I need to initialize the variables inside the function, because functions need to be reusable. Besides, as global variables, s and i are automatically initialized to 0. (Turns out I don't need to initialize i in the function, because the function always leaves it at value 0) \$\endgroup\$
– Steadybox
Nov 27, 2017 at 11:30
1

\$\begingroup\$ @Christoph Thanks! I was initially printing the values from the function, so I needed v[-i] to get the values in correct order. \$\endgroup\$
– Steadybox
Nov 27, 2017 at 11:31

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Community · Accepted Answer · 2020-06-17 09:04:33Z

4

Python, 47 46 bytes

lambda v:[e/sum(e*e for e in v)**.5for e in v]

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2017 at 21:50

PattuX

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Uriel · Accepted Answer · 2017-11-26 21:57:43Z

4

Julia, 9 bytes

normalize

Try it online!

answered Nov 26, 2017 at 21:57

Uriel

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Martin Ender · Accepted Answer · 2017-11-26 22:07:23Z

3

CJam, 9 bytes

{_:mhzf/}

Try it online!

Explanation

_    e# Duplicate input.
:mh  e# Fold hypothenuse-length over the vector. This gives the norm, unless the vector
     e# has only one component, in which case it just gives that component.
z    e# Abs. For the case of a single negative vector component.
f/   e# Divide each vector component by the norm.

answered Nov 26, 2017 at 22:07

Martin Ender

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

TI-Basic, 6 bytes

Ans/√(sum(Ans²

Run with {1,2,3}:prgmNAME, where {1,2,3} is the vector to be normalized.

Divides each element in the vector by the square root of the sum of the squares of its elements.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2017 at 22:54

pizzapants184

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\$\begingroup\$ We got the same answer! \$\endgroup\$
– kamoroso94
Nov 27, 2017 at 3:24
\$\begingroup\$ @kamoroso94 Whoops! Didn't see yours when I posted this. If you want to add the explanation from this to your answer I'll delete this. \$\endgroup\$
– pizzapants184
Nov 27, 2017 at 3:30
\$\begingroup\$ Nah I'll just remove mine. You put more effort into your answer :P \$\endgroup\$
– kamoroso94
Nov 27, 2017 at 5:48

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pajonk · Accepted Answer · 2017-11-27 06:44:01Z

3

R, 23 bytes

function(v)v/(v%*%v)^.5

Try it online!

v%*%v computes the dot product of v with itself.
The function will issue a warning for length 2 or greater vectors.

edited Nov 27, 2017 at 6:44

pajonk

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answered Nov 26, 2017 at 23:00

Giuseppe

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Olivier Grégoire · Accepted Answer · 2017-11-28 16:24:03Z

3

Java (OpenJDK 8), 57 bytes

v->v.stream().map(x->x/v.stream().reduce(0d,Math::hypot))

Try it online!

edited Nov 28, 2017 at 16:24

answered Nov 27, 2017 at 18:10

Olivier Grégoire

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Stewie Griffin · Accepted Answer · 2017-11-26 22:12:47Z

2

MATL, 5 bytes

t2&|/

Try it online!

I'm not entirely sure this is the shortest way to do this. First, we duplicate the input, then select the second output type of | (which is either abs, norm or determinant). Finally, we divide the input by the norm.

Alternative for 7 bytes:

t2^sX^/

edited Nov 26, 2017 at 22:12

answered Nov 26, 2017 at 21:57

Stewie Griffin

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ბიმო · Accepted Answer · 2017-11-26 22:30:25Z

2

Haskell, 29 bytes

f x=map(/sqrt(sum$(^2)<$>x))x

Try it online!

Or for 1 byte more pointfree: map=<<flip(/).sqrt.sum.map(^2)

answered Nov 26, 2017 at 22:30

ბიმო

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ATaco · Accepted Answer · 2017-11-26 22:54:56Z

2

Funky, 42 bytes

a=>(d=a::map)(c=>c/d(b=>b^2)::reduce@+^.5)

Try it online!

answered Nov 26, 2017 at 22:54

ATaco

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Cinaski · Accepted Answer · 2017-11-27 09:37:05Z

2

Ohm v2, 5 bytes

D²Σ¬/

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answered Nov 27, 2017 at 9:37

Cinaski

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Colera Su · Accepted Answer · 2017-11-27 12:29:05Z

2

C++ (gcc), 70 bytes

Input by std::valarray<float>. Overwrites the original vector.

#import<valarray>
int f(std::valarray<float>&a){a/=sqrt((a*a).sum());}

Try it online!

edited Nov 27, 2017 at 12:29

answered Nov 27, 2017 at 1:47

Colera Su

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\$\begingroup\$ I am just lurking codegolf every now and then, but isn't this invalid C++, given "#import", which is a Microsoft specific extension? \$\endgroup\$
– phresnel
Nov 27, 2017 at 11:01
\$\begingroup\$ @phresnel #import works at least with GCC, Clang and MinGW, too. But, yeah, it's not standard C++. \$\endgroup\$
– Steadybox
Nov 27, 2017 at 11:36
\$\begingroup\$ @phresnel I forgot to specify gcc. Fixed. \$\endgroup\$
– Colera Su
Nov 27, 2017 at 12:30

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Common Lisp, 69 bytes

(lambda(v)(mapcar(lambda(x)(/ x(sqrt(loop as y in v sum(* y y)))))v))

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 27, 2017 at 20:07

Renzo

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Uriel · Accepted Answer · 2017-11-28 13:10:06Z

2

APL (Dyalog), 13 12 10 bytes

1 byte saved thanks to @Adám

2 bytes saved thanks to @ngn

⊢÷.5*⍨+.×⍨

Try it online!

How?

⊢  ÷  .5*⍨  +.  ×⍨
u  ÷    √   Σ   u²

edited Nov 28, 2017 at 13:10

answered Nov 26, 2017 at 21:46

Uriel

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\$\begingroup\$ Train for less: ⊢÷.5*⍨(+/×⍨) \$\endgroup\$
– Adám
Nov 26, 2017 at 22:15
\$\begingroup\$ @Adám thanks alot! I've been trying for hours, couldn't get any train to work \$\endgroup\$
– Uriel
Nov 26, 2017 at 22:27
\$\begingroup\$ We should do something about that, as it is really not so hard. When you have a monadic function (other than the rightmost one), begin a parenthesis to its left (or use a ∘ if it isn't derived). Other than that, just swap ⍺ and ⍵ for ⊣ and ⊢: {⍵÷.5*⍨+/×⍨⍵} → {⍵÷.5*⍨(+/(×⍨⍵))} → ⊢÷.5*⍨(+/(×⍨⊢)) → ⊢÷.5*⍨(+/(×⍨)) → ⊢÷.5*⍨(+/×⍨) \$\endgroup\$
– Adám
Nov 27, 2017 at 10:13
\$\begingroup\$ (+/×⍨) -> +.×⍨ \$\endgroup\$
– ngn
Nov 28, 2017 at 7:15

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emanresu A · Accepted Answer · 2023-02-28 21:30:50Z

2

Vyxal, 4 bytes

²∑√/

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²    # Square each
 ∑   # Sum
  √  # Square root
   / # Divide by that

answered Feb 28 at 21:30

emanresu A

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ბიმო · Accepted Answer · 2017-11-26 22:06:56Z

1

Husk, 8 7 bytes

´(Ṁ/√ṁ□

Try it online!

answered Nov 26, 2017 at 22:06

ბიმო

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Community · Accepted Answer · 2020-06-17 09:04:33Z

C# (.NET Core), 51+64=115 bytes

v=>v.Select(d=>d/Math.Sqrt(v.Select(x=>x*x).Sum()))

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+64 bytes for the using System;using System.Collections.Generic;using System.Linq;

C# (.NET Core), 94+13=107 bytes

v=>{var m=0d;foreach(var x in v)m+=x*x;for(int i=0;i<v.Length;)v[i++]/=Math.Sqrt(m);return v;}

Try it online!

+13 bytes for using System;

The non-Linq approach

DeGolfed

v=>{
    var m=0d;
    foreach (var x in v)
        m+=x*x;

    for (int i=0; i < v.Length;)
        v[i++] /= Math.Sqrt(m);

    return v;
}

Xcali · Accepted Answer · 2017-11-27 16:54:47Z

1

Perl 5, 45 + 1 (`-a`) = 46 bytes

say$_/sqrt eval join'+',map"($_)**2",@F for@F

Try it online!

answered Nov 27, 2017 at 16:54

Xcali

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DLosc · Accepted Answer · 2017-11-27 19:33:55Z

1

Pip, 10 bytes

9 bytes of code, +1 for -p flag.

g/RT$+g*g

Takes the vector as separate command-line arguments. Try it online!

How it works

      g*g  Arglist, multiplied by itself itemwise
    $+     Sum
  RT       Square root
g/         Divide arglist itemwise by that scalar
           Result is autoprinted (-p flag to format as list)

answered Nov 27, 2017 at 19:33

DLosc

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyth, 5 bytes

cR.aQ

Try it online: Test Suite

Explanation:

cR.aQQ   implicit Q at the end
c        divide
 R   Q   each element of the input
  .aQ    by the L2 norm of the input vector

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 27, 2017 at 20:14

Jakube

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Sean · Accepted Answer · 2017-11-28 01:42:02Z

1

Perl 6, 25 bytes

{$_ »/»sqrt sum $_»²}

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$_, the list argument to the function, is divided elementwise (»/») by the square root of the sum of the squares of the elements (»²).

answered Nov 28, 2017 at 1:42

Sean

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m-chrzan · Accepted Answer · 2017-11-28 08:52:33Z

1

Ruby, 39 35 bytes

->v{v.map{|x|x/v.sum{|x|x*x}**0.5}}

-4 bytes thanks to G B.

edited Nov 28, 2017 at 8:52

answered Nov 26, 2017 at 22:11

m-chrzan

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1

\$\begingroup\$ Save some bytes by using sum{...} instead of map{...}.sum \$\endgroup\$
– G B
Nov 27, 2017 at 11:34

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Dennis · Accepted Answer · 2017-12-13 02:21:20Z

1

Coconut, 40 bytes

x->map((/)$(?,sum(map(t->t*t,x))**.5),x)

I have no idea what I'm doing...

Try it online!

answered Dec 13, 2017 at 2:21

Dennis

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The Thonnu · Accepted Answer · 2023-03-02 10:57:05Z

1

Thunno `D`, \$ 5 \log_{256}(96) \approx \$ 4.12 bytes

2^St/

Attempt This Online!

Explanation

2^St/  # D flag duplicates the input
2^     # Square each
  S    # Sum this list
   t   # Push the square root
    /  # Divide the input by this number

answered Mar 2 at 10:57

The Thonnu

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Normalize a Vector

The Challenge

38 Answers 38

05AB1E, 4 bytes

Code:

Explanation

JavaScript (ES6), 31 bytes

Test cases

Mathematica, 9 bytes

J, 8 bytes

Jelly, 5 3 bytes

Octave, 13 bytes

C, 73 70 bytes

Python, 47 46 bytes

Julia, 9 bytes

CJam, 9 bytes

Explanation

TI-Basic, 6 bytes

R, 23 bytes

Java (OpenJDK 8), 57 bytes

MATL, 5 bytes

Haskell, 29 bytes

Funky, 42 bytes

Ohm v2, 5 bytes

C++ (gcc), 70 bytes

Common Lisp, 69 bytes

APL (Dyalog), 13 12 10 bytes

Vyxal, 4 bytes

Husk, 8 7 bytes

C# (.NET Core), 51+64=115 bytes

C# (.NET Core), 94+13=107 bytes

DeGolfed

Perl 5, 45 + 1 (-a) = 46 bytes

Pip, 10 bytes

How it works

Pyth, 5 bytes

Explanation:

Perl 6, 25 bytes

Ruby, 39 35 bytes

Coconut, 40 bytes

Thunno D, \$ 5 \log_{256}(96) \approx \$ 4.12 bytes

Explanation

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Perl 5, 45 + 1 (`-a`) = 46 bytes

Thunno `D`, \$ 5 \log_{256}(96) \approx \$ 4.12 bytes

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