Calculate Standard Deviation

Question

Challenge

Given a list of numbers, calculate the population standard deviation of the list.

Use the following equation to calculate population standard deviation:

Input

The input will a list of integers in any format (list, string, etc.). Some examples:

56,54,89,87
67,54,86,67

The numbers will always be integers.

Input will be to STDIN or function arguments.

Output

The output must be a floating point number.

Rules

You may use built in functions to find the standard deviation.

Your answer can be either a full program or a function.

Examples

10035, 436844, 42463, 44774 => 175656.78441352615

45,67,32,98,11,3 => 32.530327730015607

1,1,1,1,1,1 => 0.0

Winning

The shortest program or function wins.

Leaderboard

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Answer 1

Clip, 3

.sk

.s is the standard deviation, k parses the input in the form {1,2,3}.

Answer 2

Mathematica, 24 22 bytes

Nice, Mathematica has a built-in StandardDevi... oh... that computes the sample standard deviation, not the population standard deviation.

But what if we use Variance... oh... same deal.

But there is yet another related built-in:

CentralMoment[#,2]^.5&

Yay. :)

This also works for 22 bytes:

Mean[(#-Mean@#)^2]^.5&

And this for 27:

N@RootMeanSquare[#-Mean@#]&

Answer 3

Octave, 14 bytes

g=@(a)std(a,1)

Try it on ideone.

Answer 4

kdb+, 3 bytes

dev

One of the APL derviates had to have this as a built-in.

Test run

q)dev 56, 54, 89, 87
16.53028
q)f:dev
q)f 10035, 436844, 42463, 44774
175656.8
q)f 45,67,32,98,11,3
32.53033

Answer 5

Dyalog APL, 24 23 21 20 19 17 bytes

*∘.5∘M×⍨-M×M←+/÷≢

This defines an unnamed, monadic function train, which is equivalent to the following function.

{.5*⍨M(×⍨⍵)-M⍵×(M←{(+/⍵)÷≢⍵})⍵}

Try them online on TryAPL.

How it works

The code consists of several trains.

M←+/÷≢

This defines a monadic 3-train (fork) M that executes +/ (sum of all elements) and ≢ (length) for the right argument, then applies ÷ (division) to the results, returning the arithmetic mean of the input.

M×M

This is another fork that applies M to the right argument, repeats this a second time, and applies × (product) to the results, returning μ².

×⍨-(M×M)

This is yet another fork that calculates the square of the arithmetic mean as explained before, applies ×⍨ (product with itself) to the right argument, and finally applies - (difference) to the results.

For input (x₁, …, x_N), this function returns (x₁ - μ², …, x_N - μ²).

*∘.5∘M

This composed function is applies M to its right argument, then *∘.5. The latter uses right argument currying to apply map input a to a*0.5 (square root of a).

(*∘.5∘M)(×⍨-(M×M))

Finally, we have this monadic 2-train (atop), which applies the right function first, then the left to its result, calculating the standard deviation as follows.

Answer 6

CJam, 24 22 21 bytes

q~_,_@_:+d@/f-:mh\mq/

Thanks to @aditsu for golfing off 1 byte!

Try it online in the CJam interpreter.

How it works

q~                    e# Read all input and evaluate it.
  _,                  e# Copy the array and push its length.
    _@                e# Copy the length and rotate the array on top.
      _:+d            e# Copy the array and compute its sum. Cast to Double.
          @/          e# Rotate the length on top and divide the sum by it.
            f-        e# Subtract the result (μ) from the array's elements.
              :mh     e# Reduce by hypotenuse.
                      e# a b mh -> sqrt(a^2 + b^2)
                      e# sqrt(a^2 + b^2) c mh -> sqrt(sqrt(a^2 + b^2)^2 + c^2)
                      e#                           = sqrt(a^2 + b^2 + c^2)
                      e# ⋮
                 \mq/ e# Divide the result by the square root of the length.

Answer 7

R, 41 40 39 36 30 28 bytes

code

Thanks to beaker, Alex A. and MickyT for much bytes.

cat(sd(c(v=scan(),mean(v))))   

old codes

v=scan();n=length(v);sd(v)/(n/(n-1))**0.5
m=scan();cat(sqrt(sum(mean((m-mean(m))^2))))
m=scan();cat(mean((m-mean(m))^2)^.5) 

This should yield the population standard deviation.

Answer 8

Pyth, 20 19 17 13 bytes

@.O^R2-R.OQQ2

Thanks to @FryAmTheEggman for golfing off 4 bytes!

Try it online.

How it works

        .OQ    Compute the arithmetic mean of the input (Q).
      -R   Q   Subtract the arithmetic mean of all elements of Q.
   ^R2         Square each resulting difference.
 .O            Compute the arithmetic mean of the squared differences.
@           2  Apply square root.

Answer 9

TI-BASIC, 7 bytes

stdDev(augment(Ans,{mean(Ans

I borrowed the algorithm to get population standard deviation from sample standard deviation from here.

The shortest solution I could find without augment( is 9 bytes:

stdDev(Ans√(1-1/dim(Ans

Answer 10

APL, 24 bytes

{.5*⍨+/(2*⍨⍵-+/⍵÷≢⍵)÷≢⍵}

A little different approach than Dennis' Dyalog APL solution. This should work with any APL implementation.

This creates an unnamed monadic function that computes the vector (x - µ)² as 2*⍨⍵-+/⍵÷≢⍵, divides this by N (÷≢⍵), takes the sum of this vector using +/, and then takes the square root (.5*⍨).

Try it online

Answer 11

Julia, 26 19 bytes

x->std([x;mean(x)])

This creates an unnamed function that accepts an array and returns a float.

Ungolfed, I guess:

function f(x::Array{Int,1})
    # Return the sample standard deviation (denominator N-1) of
    # the input with the mean of the input appended to the end.
    # This corrects the denominator to N without affecting the
    # mean.
    std([x; mean(x)])
end

Answer 12

Python 3.4+, 30 bytes

from statistics import*;pstdev

Imports the builtin function pstdev, e.g.

>>> pstdev([56,54,89,87])
16.53027525481654

Answer 13

Haskell, 61 bytes

d n=1/sum(n>>[1])
f a=sqrt$d a*sum(map((^2).(-)(d a*sum a))a)

Straightforward, except maybe my custom length function sum(n>>[1]) to trick Haskell's strict type system.

Answer 14

JavaScript (ES6), 73 bytes

a=>Math.sqrt(a.reduce((b,c)=>b+(d=c-eval(a.join`+`)/(l=a.length))*d,0)/l)

Answer 15

J, 18 bytes

[:%:@M*:-M*M=:+/%#

This is a direct translation of my APL answer to J.

Try it online!

Answer 16

Jelly, non-competing

11 bytes This answer is non-competing, since it uses a language that postdates the challenge.

S÷L
Dz_²ÇN½

This is a direct translation of my APL answer to Jelly. Try it online!

How it works

S÷L        Helper link. Argument: z (vector)

S          Compute the sum of z.
  L        Compute the length of z.
 ÷         Divide the former by the latter.
           This computes the mean of z.

Dz_²ÇN½    Main link. Argument: z (vector)

Ç          Apply the previous link, i.e., compute the mean of z.
 ²         Square the mean.
   ²       Square all number in z.
  _        Subtract each squared number from the squared mean.
    Ç      Take the mean of the resulting vector.
     N     Multiply it by -1.
      ½    Take the square root of the result.

Answer 17

Simplex v.0.5, 43 bytes

Just 'cuz. I really need to golf this one more byte.

t[@u@RvR]lR1RD@wA@T@{j@@SR2ERpR}u@vR@TR1UEo   
t[      ]                                     ~~ Applies inner function to entire strip (left-to-right)
  @                                           ~~ Copies current value to register
   u                                          ~~ Goes up a strip level
    @                                         ~~ Dumps the register on the current byte
     R                                        ~~ Proceeds right (s1)
      v                                       ~~ Goes back down
       R                                      ~~ Proceeds right (s0)
                                              ~~ Go right until an empty byte is found
         lR1RD                                ~~ Push length, 1, and divide.
              @                               ~~ Store result in register (1/N)
               wA                             ~~ Applies A (add) to each byte, (right-to-left)
                 @T@                          ~~ Puts 1/N down, multiplies it, and copies it to the register
                    {          }              ~~ Repeats until a zero-byte is met
                     j@@                      ~~ inserts a new byte and places register on it
                        SR                    ~~ Subtract it from the current byte and moves right
                          2E                  ~~ Squares result
                            RpR               ~~ Moves to the recently-created cell, deletes it, and continues
                                u@v           ~~ takes 1/N again into register
                                   R@T        ~~ multiplies it by the new sum
                                      R1UE    ~~ takes the square root of previous
                                          o   ~~ output as number

Answer 18

Perl5, 39 38

---

 16 for the script
+22 for the M switch
+ 1 for the E switch
=39

perl -MStatistics::Lite=:all -E"say stddevp@ARGV" .1 .2 300

Tested in Strawberry 5.20.2.

---

Oh, but then I realized that you said our answers can be functions instead of programs. In that case,

{use Statistics::Lite":all";stddevp@_}

has just 38. Tested in Strawberry 5.20.2 as

print sub{use Statistics::Lite":all";stddevp@_}->( .1, .2, 300)

Answer 19

Prolog (SWI), 119 bytes

Code:

q(U,X,A):-A is(X-U)^2.
p(L):-sumlist(L,S),length(L,I),U is S/I,maplist(q(U),L,A),sumlist(A,B),C is sqrt(B/I),write(C).

Explanation:

q(U,X,A):-A is(X-U)^2.   % calc squared difference of X and U
p(L):-sumlist(L,S),      % sum input list
      length(L,I),       % length of input list
      U is S/I,          % set U to the mean value of input list
      maplist(q(U),L,A), % set A to the list of squared differences of input and mean
      sumlist(A,B),      % sum squared differences list
      C is sqrt(B/I),    % divide sum of squares by length of list
      write(C).          % print answer

Example:

p([10035, 436844, 42463, 44774]).
175656.78441352615

Try it out online here

Answer 20

SmileBASIC, 105 bytes (as a function)

I just noticed it's allowed to be a function. Whoops, that reduces my answer dramatically. This defines a function S which takes an array and returns the population standard deviation. Go read the other one for an explanation, but skip the parsing part. I don't want to do it again.

DEF S(L)N=LEN(L)FOR I=0TO N-1U=U+L[I]NEXT
U=1/N*U FOR I=0TO N-1T=T+POW(L[I]-U,2)NEXT RETURN SQR(1/N*T)END

As a program, 212 bytes

Unfortunately, I have to take the input list as a string and parse it myself. This adds over 100 bytes to the answer, so if some input format other than a comma-separated list is allowed I'd be glad to hear it. Also note that because VAL is buggy, having a space before the comma or trailing the string breaks the program. After the comma or at the start of the string is fine.

DIM L[0]LINPUT L$@L I=INSTR(O,L$,",")IF I>-1THEN PUSH L,VAL(MID$(L$,O,I-O))O=I+1GOTO@L ELSE PUSH L,VAL(MID$(L$,O,LEN(L$)-O))
N=LEN(L)FOR I=0TO N-1U=U+L[I]NEXT
U=1/N*U FOR I=0TO N-1T=T+POW(L[I]-U,2)NEXT?SQR(1/N*T)

Ungolfed and explained:

DIM L[0]  'define our array
LINPUT L$ 'grab string from input

'parse list
'could've used something cleaner, like a REPEAT, but this was shorter
@L
I=INSTR(O,L$,",")                 'find next comma
IF I>-1 THEN                      'we have a comma
 PUSH L,VAL(MID$(L$,O,I-O))       'get substring of number, parse & store
 O=I+1                            'set next search location
 GOTO @L                          'go again
ELSE                              'we don't have a comma
 PUSH L,VAL(MID$(L$,O,LEN(L$)-O)) 'eat rest of string, parse & store
ENDIF                             'end

N=LEN(L) 'how many numbers we have

'find U
'sum all of the numbers, mult by 1/N
FOR I=0 TO N-1
 U=U+L[I]
NEXT
U=1/N*U

'calculate our popstdev
'sum(pow(x-u,2))
FOR I=0 TO N-1
 T=T+POW(L[I]-U,2)
NEXT
PRINT SQR(1/N*T) 'sqrt(1/n*sum)

Answer 21

Axiom, 137 bytes

m(a:List Float):Complex Float==(#a=0=>%i;reduce(+,a)/#a)
s(a:List Float):Complex Float==(#a=0=>%i;n:=m(a);sqrt(m([(x-n)^2 for x in a])))

The function m() would return the mean of the list in input. Both the functions on error return %i the imaginary constant sqrt(-1). Code for test and results. [but the result if it is ok, it is the real part of one complex number]

(6) -> s([45,67,32,98,11,3])
   (6)  32.5303277300 15604966

(7) -> s([10035,436844,42463,44774])
   (7)  175656.7844135261 4035

(8) -> s([1,1,1,1,1,1])
   (8)  0.0

Answer 22

Python 3, 49 bytes

lambda l,N:(sum((i-sum(l)/N)**2for i in l)/N)**.5

Try it online!

Takes in l, a list of integers, and N, the number of integers present.

Answer 23

Pyt, 13 bytes

←Đ↔Ł↔е-²Ʃ⇹/√

Implements the formula for standard deviation

Answer 24

Python, 57 bytes

lambda l:(sum((x-sum(l)/len(l))**2for x in l)/len(l))**.5

Takes input as a list

Thanks @xnor

Answer 25

PowerShell, 122

:\>type stddev.ps1
$y=0;$z=$args -split",";$a=($z|?{$_});$c=$a.Count;$a|%{$y+=$_};$b=$y/$c;$a|%{$x+
=(($_-$b)*($_-$b))/$c};[math]::pow($x,0.5)

explanation

<#
$y=0                            init
$z=$args -split","              split delim ,
$a=($z|? {$_})                  remove empty items
$c=$a.Count                     count items
$a|%{$y+=$_}                    sum
$b=$y/$c                        average
$a|%{$x+=(($_-$b)*($_-$b))/$c}  sum of squares/count
[math]::pow($x,0.5)             result
#>

result

:\>powershell -nologo -f stddev.ps1 45,67,32,98,11,3
32.5303277300156

:\>powershell -nologo -f stddev.ps1 45,  67,32,98,11,3
32.5303277300156

:\>powershell -nologo -f stddev.ps1 45,  67,32, 98 ,11,3
32.5303277300156

:\>powershell -nologo -f stddev.ps1 10035, 436844, 42463, 44774
175656.784413526

:\>powershell -nologo -f stddev.ps1 1,1,1,1,1,1
0

Answer 26

Fortran, 138 bytes

Just a straightforward implementation of the equation in Fortran:

double precision function std(x)
integer,dimension(:),intent(in) :: x
std = norm2(dble(x-sum(x)/size(x)))/sqrt(dble(size(x)))
end function