Standardize the Samples (Compute the z-Score)

Question

Given a list of floating point numbers, standardize it.

Details

• 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.

Examples

[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.)

Answer 1

R, 51 45 38 37 bytes

Thanks to Giuseppe and J.Doe!

function(x)scale(x)/(1-1/sum(x|1))^.5

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Answer 2

CW for all trivial entries

---

Python 3 + scipy, 31 bytes

from scipy.stats import*
zscore

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Octave / MATLAB, 15 bytes

@(x)zscore(x,1)

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Answer 3

APL (Dyalog Classic), 21 20 19 bytes

(-÷.5*⍨⊢÷⌹×≢)+/-⊢×≢

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⊢÷⌹ is sum of squares

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

Answer 4

MATL, 10 bytes

tYm-t&1Zs/

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Explanation

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

Answer 5

APL+WIN, 41,32 30 bytes

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

x←v-(+/v)÷⍴v←⎕⋄x÷(+/x×x÷⍴v)*.5

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

Answer 6

R + pryr, 53 52 bytes

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

pryr::f((x-(y<-mean(x)))/(sum((x-y)^2)/sum(x|1))^.5)

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.

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Answer 7

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)}}

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Answer 8

Jelly, 10 bytes

_ÆmµL½÷ÆḊ×

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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

Answer 9

Mathematica, 25 bytes

Mean[(a=#-Mean@#)a]^-.5a&

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.

Answer 10

J, 22 bytes

-1 byte thanks to Cows quack!

(-%[:%:1#.-*-%#@[)+/%#

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J, 31 23 bytes

(-%[:%:#@[%~1#.-*-)+/%#

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                   +/%# - 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

Answer 11

APL (Dyalog Unicode), 33 29 bytes

{d÷.5*⍨l÷⍨+/×⍨d←⍵-(+/⍵)÷l←≢⍵}

-4 bytes thanks to @ngn

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Answer 12

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.

Expanded:

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

    -- 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)

Answer 13

MathGolf, 7 bytes

▓-_²▓√/

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Explanation

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

Answer 14

JavaScript (ES7),  80  79 bytes

a=>a.map(x=>(x-g(a))/g(a.map(x=>(x-m)**2))**.5,g=a=>m=eval(a.join`+`)/a.length)

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Commented

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()

Answer 15

Python 3 + numpy, 46 bytes

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

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Answer 16

Haskell, 59 bytes

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

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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

(n*y-s)/sqrt(n*v-s^2)

(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.

Answer 17

K (oK), 33 23 bytes

-10 bytes thanks to ngn!

{t%%(+/t*t:x-/x%#x)%#x}

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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...)

Answer 18

MATLAB, 26 bytes

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

f=@(x)(x-mean(x))/std(x,1)

Answer 19

TI-Basic (83 series), 14 11 bytes

Ans-mean(Ans
Ans/√(mean(Ans²

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.

Answer 20

05AB1E, 9 bytes

ÅA-DnÅAt/

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

Try it online or verify all test cases.

Explanation:

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]

Answer 21

Jelly, 10 bytes

_Æm÷²Æm½Ɗ$

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Answer 22

Pyth, 21 19 bytes

mc-dJ.OQ@.Om^-Jk2Q2

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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

Answer 23

SNOBOL4 (CSNOBOL4), 229 bytes

	DEFINE('Z(A)')
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)
	Z =A	:(RETURN)

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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.

Answer 24

Julia 0.7, 37 bytes

a->(a-mean(a))/std(a,corrected=false)

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Answer 25

Charcoal, 25 19 bytes

≧⁻∕ΣθＬθθＩ∕θ₂∕ΣＸθ²Ｌθ

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.