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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body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

• Mutador: You mean the output must be a floating point OR integer?
• Mutador: I think most build-in standard deviation functions calculates the the sample standard deviation.
• user58988: What about if input list is void?175656.78441352615 result to me 175656.78441352614
• Beta Decay: @RosLuP You don't have to worry about that
• Beta Decay: @a13a22 As per PPCG's standard rules, you are fine to take input via function arguments

Answer 1

Clip, 3

.sk

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

• flawr: What formula is used for the standard deviation? I could not find it int he reference.
• Conor O'Brien: @flawr It's this chart, towards the bottom.
• flawr: @CᴏɴᴏʀO'Bʀɪᴇɴ I saw that, but there is no formula given.
• Conor O'Brien: @flawr Oh, I see. Perhaps then it is up to the interpreter, if such a thing exists.
• flawr: @CᴏɴᴏʀO'Bʀɪᴇɴ I found it here on line 493, it seems to be ok!

Answer 2

Mathematica, (削除) 24 (削除ここまで) 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.

• Alex A.: You can save two bytes by removing g= since the function handle doesn't need a name to be a valid submission.

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

R, (削除) 41 (削除ここまで) (削除) 40 (削除ここまで) (削除) 39 (削除ここまで) (削除) 36 (削除ここまで) (削除) 30 (削除ここまで) 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.

• beaker: I don't know R, but would it be possible to augment the input array with the mean of the array? It seems that might be shorter.
• Alex A.: On this site we typically can't assume a REPL environment unless explicitly allowed by the question. Thus in this case you'll need to use cat to print to the console.
• Alex A.: Also, R uses ^ for exponentiation, which is a byte shorter than **.
• Alex A.: You don't need to sum the mean since mean returns a scalar; sum has no effect. 36 bytes: x=scan();cat(mean((x-mean(x))^2)^.5)
• MickyT: @AndréMuta apologies, when I tested it I had an X hanging around.

Answer 6

Dyalog APL, (削除) 24 (削除ここまで) (削除) 23 (削除ここまで) (削除) 21 (削除ここまで) (削除) 20 (削除ここまで) (削除) 19 (削除ここまで) 17 bytes

×ばつM←+/÷≢

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

{.5*⍨M(×ばつ⍨⍵)×ばつ(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.

×ばつM

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

×ばつ⍨-(×ばつM)

This is yet another fork that calculates the square of the arithmetic mean as explained before, applies ×ばつ⍨ (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)(×ばつ⍨-(×ばつ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.

formula

Answer 7

Julia 0.6, (削除) 26 (削除ここまで) 19 bytes

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

Try it online!

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 8

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

• flawr: I agree with AndréMuta, this does not produce the required result, see here.
• lirtosiast: @AndréMuta @flawr TI's builtin stdDev( calculates the sample SD; stdDev(augment(Ans,{mean(Ans computes the population SD. That's on the page you linked to.

Answer 9

Pyth, (削除) 20 (削除ここまで) (削除) 19 (削除ここまで) (削除) 17 (削除ここまで) 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.

• Conor O'Brien: I like how the decomposition of a Pyth program looks like a skewed parabola.

Answer 10

CJam, (削除) 24 (削除ここまで) (削除) 22 (削除ここまで) 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.

• aditsu quit because SE is EVIL: I think you can convert just the length to double
• Dennis: @aditsu Of course. Thanks!
• aditsu quit because SE is EVIL: :mh is genius btw :)
• lirtosiast: Reduce by hypotenuse. isn't something you see every day.

Answer 11

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

• Adám: Not every APL implementation supports {dfns}, ⍨, or ≢. However, every version supports R←F Y R←(+/((Y-+/Y÷⍴Y)*2)÷⍴Y)*.5

Answer 12

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.

• Laikoni: You can use sum(1<$n) and <$> for map.
• Laikoni: It just occurred to me that those functions might not be present because of an older GHC version at the time of this answer, but according to this tip they were introduced to prelude in March 2015, and the site policy has changed anyway to allow newer language features.

Answer 13

Python 3.4+, 30 bytes

from statistics import*;pstdev

Imports the builtin function pstdev, e.g.

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

• FryAmTheEggman: I think just pstdev after the first line is ok? I believe xnor did that a while ago with sum. It sort of makes sense wrt how anonymous lambdas would be used i.e. p=pstdev or map(pstdev, [...])
• xnor: I was going to say the same thing. Meta posts seem to support just putting a function literal.
• xnor: I think you still need to write the literal pstdev though, like from statistics import*;pstdev. Otherwise, this could be any function from that library.
• Sp3000: @xnor Edited. tbh I'm not really sure about the ruling on these situations...
• Beta Decay: Maybe a meta question would be helpful? :)

Answer 14

Haskell, 45 bytes

(?)i=sum.map(^i)
f l=sqrt2ドル?l/0?l-(1?l/0?l)^2

Try it online!

The value i?l is the sum of the i'th powers of elements in l, so that 0?l is the length and 1?l is the sum.

Answer 15

Arn, (削除) 19 (削除ここまで) (削除) 18 (削除ここまで) 14 bytes

 ̄‡iʠØbmÅQƥaÈaÆ

Try it!

Explained

Unpacked: :/mean(n{:*n-mean:s}\

:/ Square root
 mean Mean function
 ( Begin expression
 n{ Block with key of n
 :* Square
 n
 - Subtraction
 mean
 _ Variable initialized to STDIN; implied
 :s Split on spaces
 } End of block
 \ Mapped over
 _ Implied
 ) End of expression; Implied

• Dominic van Essen: I'm very intrigued by Arn as a cool new different language. But am I getting it wrong or is the compression ('packing') quite inefficient? The 'unpacked' code is only 21 characters, including 2x the word 'mean' (which surely must be a single-byte token (?)) and 1x parenthesis (which could be completely avoided by rearranging to RPN), so I would have guessed that the 'packed' code would be at most 14 characters long. What am I missing?
• ZippyMagician: Packing does not use that method of compression (mainly because most of the time you are NOT using a single word, but rather punctuation 90% of the time). Because of this, I essentially had two options for compression; pack the 7-bit characters into 8-bit characters (11% compression) or do something a bit stranger. The latter involves converting the ascii into a massive number (base 95) and then reading it off into a codepage, which results in approximately a 17% compression rate. Naturally, I chose option B here. That is why the compression doesn’t seem nearly as efficient as you thought.
• Dominic van Essen: Thanks for the explanation! It's particularly interesting to hear other languages' strategies now that there's a new 'golf a golfing language' challenge! Could you (or do you) at least reduce the text of 'mean' to 'M' (or 'ME' once you run-out of one-character function names) before doing the base95->256 compression, though?
• ZippyMagician: I’ve considered shortening function names, but so far I haven’t. If a user wanted to shorten it manually, they could do this: m(_):=.mean
• ZippyMagician: However, what you suggested is a good idea, and so I’ll test it out a bit

Answer 16

Jelly

11 bytes

S÷L
Ç2_2ÇN1⁄2

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.
Ç2_2ÇN1⁄2 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.
 1⁄2 Take the square root of the result.

Answer 17

☾, 20 characters (58 bytes)

√○しろまる󷺹ᴍ⬤2○しろまる⬤-←⟞⨁÷🃌⟞→⨁÷🃌⟞

Try it: WASMoon online REPL

highlighted explanation

• Ganer: bro this taught me ⬤² and i made the lang what

Answer 18

JavaScript (ES6), 73 bytes

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

• Mwr247: @BetaDecay Regarding precision of the output? My original actually didn't have that correct, and I fixed it right after, only to discover floating point was ok hehe... So is it good now as it is?
• Beta Decay: Yeah that's fine :)
• George Reith: Psst... you can shave off 5 bytes by using this summing method eval(a.join`+`) instead of a.reduce((e,f)=>e+f)
• Mwr247: @GeorgeReith Nice trick! I'm going to have to remember that one for later...
• The Fifth Marshal: @StevenPenny It seems to work for me.

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

J, 18 bytes

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

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

Try it online!

• Conor O'Brien: I had no idea M was a predefined built in.
• Dennis: It isn't. M=:+/%# is an inline function definition.
• Conor O'Brien: But it's predefined, right? Perhaps builtin is the wrong term
• Dennis: No, it's not predefined. M=:+/%# saves the verb +/%# in M, then calls it.
• Conor O'Brien: I am sorry XD I didn't see the last part

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

05AB1E, (削除) 10 (削除ここまで) 7 bytes

ÅA-nÅAt

-3 bytes thanks to @ovs.

Try it online or verify all test cases.

Explanation:

ÅA # Get the arithmetic mean of the (implicit) input-list
 - # Subtract it from each value in the (implicit) input-list
 n # Square each of those
 ÅA # Take the arithmetic mean of that
 t # And take the square-root of that
 # (after which it is output implicitly as result)

• ovs: Just found this challenge, ÅA-nÅAt is a more literal implementation of the given formula at 7 bytes. (This scores about 404 points in the other challenge)

Answer 23

Brachylog, 16 bytes

⟨∋-⟨+/l⟩⟩f^2m↰2√

Try it online! (all cases at once)

I feel like maybe there's some shorter way to compute the squared deviations than 13 bytes.

Answer 24

Uiua, 15 Characters

√-n2∩(/+÷⊸⧻)⤙n2

This takes advantage of the fact that $$\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2 = \frac{1}{n}\sum_{i=1}^n x_i^2 - 2\mu\frac{1}{n}\sum_{i=1}^nx_i+\mu^2 = \frac{1}{n}\sum_{i=1}^nx_i^2 - \mu^2.$$ Explanation:

√ # take a square root
 - # take the difference between top two numbers on stack
 n2 # square first entry of stack
 ∩ # apply the next function on top two items on stack
 (/+÷⊸⧻) # take average of a list of numbers
 ⤙n2 # square the top item on stack, but place result beneath the original list

Answer 25

Tcl, 90 bytes

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

Try it online!

---

(削除) # [Tcl], 98 bytes

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

Try it online!

(削除ここまで)

• sergiol: Failed outgolf, 102: proc S L {expr sqrt((([join $L -([join $L +])/[set n [llength $L]].)**2+(]-([join $L +])/$n.)**2)/$n)} — tio.run/…

Answer 26

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 27

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 28

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

Answer 29

Perl5, (削除) 39 (削除ここまで) 38

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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.

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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 30

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$@L I=INSTR(O,L,ドル",")IF I>-1THEN PUSH L,VAL(MID$(L,ドルO,I-O))O=I+1GOTO@L ELSE PUSH L,VAL(MID$(L,ドルO,LEN(L$)-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,ドル",") 'find next comma
IF I>-1 THEN 'we have a comma
 PUSH L,VAL(MID$(L,ドルO,I-O)) 'get substring of number, parse & store
 O=I+1 'set next search location
 GOTO @L 'go again
ELSE 'we don't have a comma
 PUSH L,VAL(MID$(L,ドル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)