Given a list of floating point numbers, standardize it.


  • A list \$x_1,x_2,\ldots,x_n\$ is standardized if the mean of all values is 0, and the standard deviation is 1. One way to compute this is by first computing the mean \$\mu\$ and the standard deviation \$\sigma\$ as $$ \mu = \frac1n\sum_{i=1}^n x_i \qquad \sigma = \sqrt{\frac{1}{n}\sum_{i=1}^n (x_i -\mu)^2} ,$$ and then computing the standardization by replacing every \$x_i\$ with \$\frac{x_i-\mu}{\sigma}\$.
  • You can assume that the input contains at least two distinct entries (which implies \$\sigma \neq 0\$).
  • Note that some implementations use the sample standard deviation, which is not equal to the population standard deviation \$\sigma\$ we are using here.
  • There is a CW answer for all trivial solutions.


[1,2,3] -> [-1.224744871391589,0.0,1.224744871391589]
[1,2] -> [-1,1]
[-3,1,4,1,5] -> [-1.6428571428571428,-0.21428571428571433,0.8571428571428572,-0.21428571428571433,1.2142857142857144]

(These examples have been generated with this script.)


25 Answers 25


R, 51 45 38 37 bytes

Thanks to Giuseppe and J.Doe!


Try it online!

  • \$\begingroup\$ Beat me by 2 bytes and 1 minute \$\endgroup\$ – Sumner18 Dec 14 '18 at 19:38

CW for all trivial entries

Python 3 + scipy, 31 bytes

from scipy.stats import*

Try it online!

Octave / MATLAB, 15 bytes


Try it online!


APL (Dyalog Classic), 21 20 19 bytes


Try it online!

⊢÷⌹ is sum of squares

⊢÷⌹×≢ is sum of squares divided by length

  • \$\begingroup\$ Wow. I shouldn't be surprised anymore, but I am every time \$\endgroup\$ – Quintec Dec 16 '18 at 16:19

MATL, 10 bytes


Try it online!


t       % Implicit input
        % Duplicate
Ym      % Mean
-       % Subtract, element-wise
t       % Duplicate
&1Zs    % Standard deviation using normalization by n
/       % Divide, element-wise
        % Implicit display

APL+WIN, 41,32 30 bytes

9 bytes saved thanks to Erik + 2 more thanks to ngn


Prompts for vector of numbers and calculates mean standard deviation and standardised elements of input vector

  • \$\begingroup\$ Can't you assign x←v-(+/v)÷⍴v←⎕ and then do x÷((+/x*2)÷⍴v)*.5? \$\endgroup\$ – Erik the Outgolfer Dec 14 '18 at 19:11
  • \$\begingroup\$ I can indeed. Thanks. \$\endgroup\$ – Graham Dec 14 '18 at 19:24
  • \$\begingroup\$ does apl+win do singleton extension (1 2 3+,4 ←→ 1 2 3+4)? if yes, you could rewrite (+/x*2)÷⍴v as +/x×x÷⍴v \$\endgroup\$ – ngn Dec 16 '18 at 10:24
  • \$\begingroup\$ @ngn That works for another 2 bytes. Thanks. \$\endgroup\$ – Graham Dec 16 '18 at 12:47

R + pryr, 53 52 bytes

-1 byte using sum(x|1) instead of length(x) as seen in @Robert S.'s solution


For being a language built for statisticians, I'm amazed that this doesn't have a built-in function. At least not one that I could find. Even the function mosaic::zscore doesn't yield the expected results. This is likely due to using the population standard deviation instead of sample standard deviation.

Try it online!

  • 2
    \$\begingroup\$ You can change the <- into a = to save 1 byte. \$\endgroup\$ – Robert S. Dec 14 '18 at 19:52
  • \$\begingroup\$ @J.Doe nope, I used the method I commented on Robert S.'s solution. scale is neat! \$\endgroup\$ – Giuseppe Dec 14 '18 at 19:57
  • 2
    \$\begingroup\$ @J.Doe since you only use n once you can use it directly for 38 bytes \$\endgroup\$ – Giuseppe Dec 14 '18 at 20:00
  • 2
    \$\begingroup\$ @RobertS. here on PPCG we tend to encourage allowing flexible input and output, including outputting more than is required, with the exception of challenges where the precise layout of the output is the whole point of the challenge. \$\endgroup\$ – ngm Dec 14 '18 at 21:09
  • 6
    \$\begingroup\$ Of course R built-ins wouldn't use "population variance". Only confused engineers would use such a thing (hencethe Python and Matlab answers ;)) \$\endgroup\$ – ngm Dec 14 '18 at 21:12

Tcl, 126 bytes

proc S L {lmap c $L {expr ($c-[set m ([join $L +])/[set n [llength $L]].])/sqrt(([join [lmap c $L {expr ($c-$m)**2}] +])/$n)}}

Try it online!


Jelly, 10 bytes


Try it online!

It's not any shorter, but Jelly's determinant function ÆḊ also calculates vector norm.

_Æm             x - mean(x)
   µ            then:
    L½          Square root of the Length
      ÷ÆḊ       divided by the norm
         ×      Multiply by that value
  • \$\begingroup\$ Hey, nice alternative! Unfortunately, I can't see a way to shorten it. \$\endgroup\$ – Erik the Outgolfer Dec 14 '18 at 22:34

Mathematica, 25 bytes


Pure function. Takes a list of numbers as input and returns a list of machine-precision numbers as output. Note that the built-in Standardize function uses the sample variance by default.


J, 22 bytes

-1 byte thanks to Cows quack!


Try it online!

J, 31 23 bytes


Try it online!

                   +/%# - mean (sum (+/) divided (%) by the number of samples (#)) 
(                 )     - the list is a left argument here (we have a hook)
                 -      - the difference between each sample and the mean
                *       - multiplied by 
               -        - the difference between each sample and the mean
            1#.         - sum by base-1 conversion
          %~            - divided by
       #@[              - the length of the samples list
     %:                 - square root
   [:                   - convert to a fork (function composition) 
 -                      - subtract the mean from each sample
  %                     - and divide it by sigma
  • 1
    \$\begingroup\$ Rearranging it gives 22 [:(%[:%:1#.*:%#)]-+/%# tio.run/##y/qfVmyrp2CgYKVg8D/…, I think one of those caps could be removed, but haven't had any luck so far, EDIT: a more direct byteshaving is (-%[:%:1#.-*-%#@[)+/%# also at 22 \$\endgroup\$ – Kritixi Lithos Dec 15 '18 at 13:12
  • \$\begingroup\$ @Cows quack Thanks! \$\endgroup\$ – Galen Ivanov Dec 15 '18 at 14:30

APL (Dyalog Unicode), 33 29 bytes


-4 bytes thanks to @ngn

Try it online!

  • \$\begingroup\$ you could assign ⍵-m to a variable and remove m← like this: {d÷.5*⍨l÷⍨+/×⍨d←⍵-(+/⍵)÷l←≢⍵} \$\endgroup\$ – ngn Dec 16 '18 at 10:39
  • \$\begingroup\$ @ngn Ah, nice, thanks, I didn't see that duplication somehow \$\endgroup\$ – Quintec Dec 16 '18 at 16:15

Haskell, 80 75 68 bytes

t x=k(/sqrt(f$sum$k(^2)))where k g=g.(-f(sum x)+)<$>x;f=(/sum(1<$x))

Thanks to @flawr for the suggestions to use sum(1<$x) instead of sum[1|_<-x] and to inline the mean, @xnor for inlining the standard deviation and other reductions.


-- Standardize a list of values of any floating-point type.
standardize :: Floating a => [a] -> [a]
standardize input = eachLessMean (/ sqrt (overLength (sum (eachLessMean (^2)))))

    -- Map a function over each element of the input, less the mean.
    eachLessMean f = map (f . subtract (overLength (sum input))) input

    -- Divide a value by the length of the input.
    overLength n = n / sum (map (const 1) input)
  • 1
    \$\begingroup\$ You can replace [1|_<-x] with (1<$x) to save a few bytes. That is a great trick for avoiding the fromIntegral, that I haven't seen so far! \$\endgroup\$ – flawr Dec 16 '18 at 10:54
  • \$\begingroup\$ By the way: I like using tryitonline, you can run your code there and then copy the preformatted aswer for posting here! \$\endgroup\$ – flawr Dec 16 '18 at 10:57
  • \$\begingroup\$ And you do not have to define m. \$\endgroup\$ – flawr Dec 16 '18 at 11:02
  • \$\begingroup\$ You can write (-x+) for (+(-x)) to avoid parens. Also it looks like f can be pointfree: f=(/sum(1<$x)), and s can be replaced with its definition. \$\endgroup\$ – xnor Dec 16 '18 at 20:00
  • \$\begingroup\$ @xnor Ooh, (-x+) is handy, I’m sure I’ll be using that in the future \$\endgroup\$ – Jon Purdy Dec 16 '18 at 21:15

MathGolf, 7 bytes


Try it online!


This is literally a byte-for-byte recreation of Kevin Cruijssen's 05AB1E answer, but I save some bytes from MathGolf having 1-byters for everything needed for this challenge. Also the answer looks quite good in my opinion!

▓         get average of list
 -        pop a, b : push(a-b)
  _       duplicate TOS
   ²      pop a : push(a*a)
    ▓     get average of list
     √    pop a : push(sqrt(a)), split string to list
      /   pop a, b : push(a/b), split strings

JavaScript (ES7),  80  79 bytes


Try it online!


a =>                      // given the input array a[]
  a.map(x =>              // for each value x in a[]:
    (x - g(a)) /          //   compute (x - mean(a)) divided by
    g(                    //   the standard deviation:
      a.map(x =>          //     for each value x in a[]:
        (x - m) ** 2      //       compute (x - mean(a))²
      )                   //     compute the mean of this array
    ) ** .5,              //   and take the square root
    g = a =>              //   g = helper function taking an array a[],
      m = eval(a.join`+`) //     computing the mean
          / a.length      //     and storing the result in m
  )                       // end of outer map()

Python 3 + numpy, 46 bytes

lambda a:(a-mean(a))/std(a)
from numpy import*

Try it online!


Haskell, 59 bytes

f l=[(0%l*y-1%l)/sqrt(2%l*0%l-1%l^2)|y<-l]

Try it online!

Doesn't use libraries.

The helper function % computes the sum of ith powers of a list, which lets us get three useful values.

  • 0%l is the length of l (call this n)
  • 1%l is the sum of l (call this s)
  • 2%l is the sum of squares of l (call this m)

We can express the z-score of an element y as


(This is the expression (y-s/n)/sqrt(v/n-(s/n)^2) simplified a bit by multiplying the top and bottom by n.)

We can insert the expressions 0%l, 1%l, 2%l without parens because the % we define has higher precedence than the arithmetic operators.

(%)i=sum.map(^i) is the same length as i%l=sum.map(^i)l. Making it more point-free doesn't help. Defining it like g i=... loses bytes when we call it. Although % works for any list but we only call it with the problem input list, there's no byte loss in calling it with argument l every time because a two-argument call i%l is no longer than a one-argument one g i.

  • \$\begingroup\$ We do have \$\LaTeX\$ here:) \$\endgroup\$ – flawr Dec 16 '18 at 9:59
  • \$\begingroup\$ I really like the % idea! It looks just like the discrete version of the statistical moments. \$\endgroup\$ – flawr Dec 16 '18 at 10:02

K (oK), 33 23 bytes

-10 bytes thanks to ngn!


Try it online!

First attempt at coding (I don't dare to name it "golfing") in K. I'm pretty sure it can be done much better (too many variable names here...)

  • 1
    \$\begingroup\$ nice! you can replace the initial (x-m) with t (tio) \$\endgroup\$ – ngn Dec 16 '18 at 9:53
  • 1
    \$\begingroup\$ the inner { } is unnecessary - its implicit parameter name is x and it has been passed an x as argument (tio) \$\endgroup\$ – ngn Dec 16 '18 at 9:56
  • 1
    \$\begingroup\$ another -1 byte by replacing x-+/x with x-/x. the left argument to -/ serves as initial value for the reduction (tio) \$\endgroup\$ – ngn Dec 16 '18 at 10:08
  • \$\begingroup\$ @ngn Thank you! Now I see that the first 2 golfs are obvious; the last one is beyond my current level :) \$\endgroup\$ – Galen Ivanov Dec 16 '18 at 10:14

MATLAB, 26 bytes

Trivial-ish, std(,1) for using population standard deviation


TI-Basic (83 series), 14 11 bytes


Takes input in Ans. For example, if you type the above into prgmSTANDARD, then {1,2,3}:prgmSTANDARD will return {-1.224744871,0.0,1.224744871}.

Previously, I tried using the 1-Var Stats command, which stores the population standard deviation in σx, but it's less trouble to compute it manually.


05AB1E, 9 bytes


Port of @Arnauld's JavaScript answer, so make sure to upvote him!

Try it online or verify all test cases.


ÅA          # Calculate the mean of the (implicit) input
            #  i.e. [-3,1,4,1,5] → 1.6
  -         # Subtract it from each value in the (implicit) input
            #  i.e. [-3,1,4,1,5] and 1.6 → [-4.6,-0.6,2.4,-0.6,3.4]
   D        # Duplicate that list
    n       # Take the square of each
            #  i.e. [-4.6,-0.6,2.4,-0.6,3.4] → [21.16,0.36,5.76,0.36,11.56]
     ÅA     # Pop and calculate the mean of that list
            #  i.e. [21.16,0.36,5.76,0.36,11.56] → 7.84
       t    # Take the square-root of that
            #  i.e. 7.84 → 2.8
        /   # And divide each value in the duplicated list with it (and output implicitly)
            #  i.e. [-4.6,-0.6,2.4,-0.6,3.4] and 2.8 → [-1.6428571428571428,
            #   -0.21428571428571433,0.8571428571428572,-0.21428571428571433,1.2142857142857144]

Jelly, 10 bytes


Try it online!


Pyth, 21 19 bytes


Try it online here.

mc-dJ.OQ@.Om^-Jk2Q2Q   Implicit: Q=eval(input())
                       Trailing Q inferred
    J.OQ               Take the average of Q, store the result in J
           m     Q     Map the elements of Q, as k, using:
             -Jk         Difference between J and k
            ^   2        Square it
         .O            Find the average of the result of the map
        @         2    Square root it
                       - this is the standard deviation of Q
m                  Q   Map elements of Q, as d, using:
  -dJ                    d - J
 c                       Float division by the standard deviation
                       Implicit print result of map

Edit: after seeing Kevin's answer, changed to use the average builtin for the inner results. Previous answer: mc-dJ.OQ@csm^-Jk2QlQ2


SNOBOL4 (CSNOBOL4), 229 bytes

Z	X =X + 1
	M =M + A<X>	:S(Z)
	N =X - 1.
	M =M / N
D	X =GT(X) X - 1	:F(S)
	A<X> =A<X> - M	:(D)
S	X =LT(X,N) X + 1	:F(Y)
	S =S + A<X> ^ 2 / N	:(S)
Y	S =S ^ 0.5
N	A<X> =A<X> / S
	X =GT(X) X - 1	:S(N)

Try it online!

Link is to a functional version of the code which constructs an array from STDIN given its length and then its elements, then runs the function Z on that, and finally prints out the values.

Defines a function Z which returns an array.

The 1. on line 4 is necessary to do the floating point arithmetic properly.


Julia 0.7, 37 bytes


Try it online!


Charcoal, 25 19 bytes


Try it online! Link is to verbose version of code. Explanation:

       θ    Input array
≧           Update each element
 ⁻          Subtract
   Σ        Sum of
    θ       Input array
  ∕         Divided by
     L      Length of
      θ     Input array

Calculate \$\mu\$ and vectorised subtract it from each \$x_i\$.

  θ         Updated array
 ∕          Vectorised divided by
   ₂        Square root of
     Σ      Sum of
       θ    Updated array
      X     Vectorised to power
        ²   Literal 2
    ∕       Divided by
         L  Length of
          θ Array
I           Cast to string
            Implicitly print each element on its own line.

Calculate \$\sigma\$, vectorised divide each \$x_i\$ by it, and output the result.

Edit: Saved 6 bytes thanks to @ASCII-only for a) using SquareRoot() instead of Power(0.5) b) fixing vectorised Divide() (it was doing IntDivide() instead) c) making Power() vectorise.

  • \$\begingroup\$ crossed out 25 = no bytes? :P (Also, you haven't updated the TIO link yet) \$\endgroup\$ – ASCII-only Dec 25 '18 at 10:59
  • \$\begingroup\$ @ASCII-only Oops, thanks! \$\endgroup\$ – Neil Dec 25 '18 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.