11
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Inspired by this question.

Given a list containing numbers, print:

  • The sum and product of the numbers in the list
  • The average and median
  • The differences between each term in the list (e.g. [1,2,3] -> [1,1]: 1+1=2, 2+1=3)
  • The list, sorted ascending
  • The minimum and maximum of the list
  • The standard deviation of the list

For reference:

Standard Deviation

$$\sigma=\sqrt{\frac1N\sum^N_{i=1}(x_i-\mu)^2}$$

Where \$\mu\$ is the mean average, \$x_i\$ is the \$i\$th term in the list, and \$N\$ is the length of the list.

Shortest code wins. Good luck!

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  • \$\begingroup\$ Do we have to print them in that order? \$\endgroup\$ – Titus Oct 2 '18 at 15:15

14 Answers 14

14
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Q, 41

{(+/;*/;avg;med;-':;asc;min;max;dev)@\:x}
| improve this answer | |
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  • \$\begingroup\$ 'med'...facepalm \$\endgroup\$ – skeevey Jun 8 '12 at 11:14
  • \$\begingroup\$ I was wondering what you were up to! \$\endgroup\$ – tmartin Jun 8 '12 at 12:04
5
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J, 73 70 characters

((+/;*/;a;(<.@-:@#{/:~);2&-~/\;/:~;<./;>./;%:@:(a@:*:@:(-a)))[a=.+/%#)

Usage:

   ((+/;*/;a;(<.@-:@#{/:~);2&-~/\;/:~;<./;>./;%:@:(a@:*:@:(-a)))[a=.+/%#)1 2 3 4
+--+--+---+-+-------+-------+-+-+-------+
|10|24|2.5|3|1 1 1 1|1 2 3 4|1|4|1.11803|
+--+--+---+-+-------+-------+-+-+-------+
| improve this answer | |
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  • \$\begingroup\$ It has to be 1 1 1 not 1 1 1 1 as difference itself next \$\endgroup\$ – user58988 Jan 6 '18 at 10:37
5
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TI-BASIC, 41 bytes

1-Var Stats is one byte, and Σx, , etc. are two bytes each.

Ans→L₁
1-Var Stats
SortA(L₁
Disp Σx,prod(Ans),x̄,Med,ΔList(Ans),L₁,minX,maxX,σx

If changing the output order is allowed, a close-paren can be saved, bringing the score to 40 bytes.

| improve this answer | |
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4
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Q (87 chars)

(sum;prd;avg;{.5*(sum/)x[((<)x)(neg(_)t;(_)neg t:.5*1-(#)x)]};(-':);asc;min;max;dev)@\:

eg.

q) (sum;prd;avg;{.5*(sum/)x[((<)x)(neg(_)t;(_)neg t:.5*1-(#)x)]};(-':);asc;min;max;dev)@\: 10 9 8 7 6 5 4 3 2 1
55
3628800
5.5
5.5
10 -1 -1 -1 -1 -1 -1 -1 -1 -1
`s#1 2 3 4 5 6 7 8 9 10
1
10
2.872281
| improve this answer | |
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4
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Ruby 187

O=->l{g=l.size
r=l.sort
s=l.inject(:+)+0.0
m=s/g
p s,l.inject(:*),m,g%2>0?r[g/2]:(r[g/2]+r[g/2-1])/2.0,l.each_cons(2).map{|l|l[1]-l[0]},r,r[0],r[-1],(l.inject(0){|e,i|e+(i-m)**2}/g)**0.5}

Usage syntax: O[<array>] (for example, O[[1,2,3]])

Outputs all the required values to the console, in the order specified in the question.

IdeOne examples:

| improve this answer | |
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2
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Scala 208 202 188:

val w=l.size
val a=l.sum/w
val s=l.sortWith(_<_)
Seq(l.sum,l.product,a,s((w+1)/2),(0 to w-2).map(i=>l(i+1)-l(i)),s,l.min,l.max,(math.sqrt((l.map(x=>(a-x)*(a-x))).sum*1.0/w))).map(println)

Test:

scala> val l = util.Random.shuffle((1 to 6).map(p=>math.pow(2, p).toInt))
l: scala.collection.immutable.IndexedSeq[Int] = Vector(64, 8, 4, 32, 16, 2)

scala> val a=l.sum/l.size
a: Int = 21

scala> val s=l.sortWith(_<_)
s: scala.collection.immutable.IndexedSeq[Int] = Vector(2, 4, 8, 16, 32, 64)

scala> Seq(l.sum,l.product,a,s((s.size+1)/2),(0 to l.size-2).map(i=>l(i+1)-l(i)),l.sortWith(_<_),l.min,l.max,(math.sqrt((l.map(x=>(a-x)*(a-x))).sum*1.0/l.size))).map(println)
126
2097152
21
16
Vector(-56, -4, 28, -16, -14)
Vector(2, 4, 8, 16, 32, 64)
2
64
21.656407827707714
| improve this answer | |
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  • \$\begingroup\$ For me "Vector(-56, -4, 28, -16, -14)" is wrong \$\endgroup\$ – user58988 Jan 6 '18 at 10:38
  • \$\begingroup\$ @RosLuP: Why is it wrong? \$\endgroup\$ – user unknown Jan 7 '18 at 0:31
  • \$\begingroup\$ Yes you are right if input is "Vector(64, 8, 4, 32, 16, 2)" ( i confuse the input) \$\endgroup\$ – user58988 Jan 7 '18 at 12:50
2
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Julia 0.6, 66 bytes

x->map(f->f(x),[sum,prod,mean,median,diff,sort,extrema,std])|>show

Try it online!

Julia 0.6, 88 bytes (uncorrected std dev, as in op)

x->map(f->f(x),[sum,prod,mean,median,diff,sort,extrema,x->std(x,corrected=false)])|>show

Try it online!

| improve this answer | |
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  • \$\begingroup\$ this isn't right, because Julia is using the sample standard deviation calculation (dividing by n-1) rather than the population std (dividing by n) as required in the problem. Multiplying by (n-1)/n wouldn't fix it either, because when dividing by n-1, NaN is produced. I ran into the same problems when trying to do this in R and haven't given it thought since. \$\endgroup\$ – Giuseppe Jan 5 '18 at 17:30
  • \$\begingroup\$ That didn't even occur to me. I added an alternate solution with the correct std deviation. \$\endgroup\$ – gggg Jan 5 '18 at 20:35
1
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C++14, 340 383 bytes

As generic unnamed lambda. First parameter L is the list as std::list of floating point type and second parameter is the desired output stream, like std::cout.

#import<cmath>
#define F(x);O<<x<<'\n';
#define Y l=k;++l!=L.end();
#define A auto
[](A L,A&O){A S=L;A l=L.begin(),k=l;A n=L.size();A s=*l,p=s,d=s*s,h=n/2.;for(S.sort(),Y s+=*l,p*=*l,d+=*l**l);for(l=S.begin();--h>0;++l)F(s)F(p)F(s/n)F(*l)for(Y)O<<*l-*k++<<","F(' ')for(A x:S)O<<x<<","F(' ')F(S.front())F(S.back())F(sqrt((d-s*s/n)/(n-1)))}

Compiles with a warning, C++ does not allow " directly followed by literals like F. Program still running.

  • -1 & -2 bytes thanks to Zacharý

Ungolfed:

#include<iostream>
#include<list>

#import<cmath>
#define F(x);O<<x<<'\n';
#define Y l=k;++l!=L.end();
#define A auto

auto f=
[](A L, A&O){
  A S=L;                  //copy the list for later sorting
  A l=L.begin(),          //main iterator
    k=l;                  //sidekick iterator
  A n=L.size();
  A s=*l,                 //sum, init with head of list
    p=s,                  //product, same
    d=s*s,                //standard deviation, formula see https://en.wikipedia.org/wiki/Algebraic_formula_for_the_variance
    h=n/2.;               //for the median later   
  for(
    S.sort(),             //now min/med/max is at known positions in S
    Y //l=k;++l!=L.end(); //skip the headitem-loop
    s += *l,              //l points the next element which is fine
    p *= *l,              //since the head given at definiten
    d += *l * *l          //needs the sum of the squares
  );
  for(
    l=S.begin();          //std::list has no random access
    --h>0;                //that's why single increment loop
    ++l                   //until median is crossed
  )
  F(s)  //;O<<s<<'\n';    //sum
  F(p)                    //product
  F(s/n)                  //average
  F(*l)                   //median (in S)
  for(Y) //l=k;++l!=L.end(); //set l back to L
    O<<*l-*k++<<","       //calc difference on the fly
  F(' ')
  for(A x:S)              //output sorted list
    O<<x<<"," 
  F(' ')
  F(S.front())            //minimum
  F(S.back())             //maximum
  F(sqrt((d-s*s/n)/(n-1))) //standard deviation
}

;


using namespace std;

int main() {
 list<double> l = {10,3,1,2,4};
 f(l, cout);
}
| improve this answer | |
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  • \$\begingroup\$ I think you can save a few bytes by changing F to ;F(x)O<<x<<'\n'; and the last line to: [](A L,A&O){A S=L;A l=L.begin(),k=l;A n=L.size();A s=*l,p=s,d=s*s,h=n/2.;for(S.sort(),Y s+=*l,p*=*l,d+=*l**l);for(l=S.begin();--h>0;++l)F(s)F(p)F(s/n)F(*l)for(Y)O<<*l-*k++<<","F(' ')for(A x:S)O<<x<<","F(' ')F(S.front())F(S.back())F(sqrt((d-s*s/n)/(n-1)));} \$\endgroup\$ – Zacharý Aug 11 '17 at 22:38
  • \$\begingroup\$ @Zacharý There was indeed an unnecessary ; quite at the end. That could be removed, but the compiler does not like " "F: warning: invalid suffix on literal; C++11 requires a space between literal and string macro it compiles though... \$\endgroup\$ – Karl Napf Aug 11 '17 at 22:48
  • \$\begingroup\$ Does it work though?! \$\endgroup\$ – Zacharý Aug 11 '17 at 22:54
  • \$\begingroup\$ @Zacharý yes it does work. \$\endgroup\$ – Karl Napf Aug 11 '17 at 23:01
  • \$\begingroup\$ 327 bytes \$\endgroup\$ – ceilingcat Mar 17 '19 at 23:13
1
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Perl 5, 204 + 1 = 205 bytes

@L=sort{$a<=>$b}@F;$p=1;$s+=$_,$p*=$_,$a+=$_/@F for@L;for(0..$#F){$o=($F[$_]-$a)**2/@F;push@d,$F[$_]-$F[$_-1]if$_}$o=sqrt$o;$m=@F%2?$F[@F/2]:$F[@F/2]/2+$F[@F/2-1]/2;say"$s $p$/$a $m$/@d$/@L$/@L[0,-1]$/$o"

Try it online!

| improve this answer | |
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0
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Pyt, 39 bytes

←ĐĐĐĐĐĐĐŞ⇹Ʃ3ȘΠ4Șµ5Ș₋⇹6Ș↕⇹7ȘṀ↔Đе-²Ʃ⇹Ł/√

This outputs, in order, the median, the product, the differences, the list reversed, the sum, the maximum and minimum, the mean, and the standard deviation.q

Try it online!

Explanation:

←ĐĐĐĐĐĐĐ                                              Push the array onto the stack 8 times
        ş                                             Sort in ascending order
         ⇹                                            Stack management
          Ʃ                                           Sum
           3Ș                                         Stack management
             Π                                        Product
              4Ș                                      Stack management
                µ                                     Mean (as a float)
                 5Ș                                   Stack management
                   ₋                                  Differences
                    ⇹6Ș                               Stack management
                       ↕                              Minimum and maximum
                        ⇹7Ș                           Stack management
                           Ṁ                          Median
                            ↔                         Stack management
                             Đе-²Ʃ⇹Ł/√               Standard Deviation
| improve this answer | |
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0
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APL NARS, 119 chars, 182 bytes

{m←(s←+/w)÷n←⍴w←,⍵⋄s,(×/w),m,(n{j←⌊⍺÷2⋄2|⍺:⍵[1+j]⋄2÷⍨⍵[j]+⍵[j+1]}t),(⊂¯1↓(1⌽w)-w),(⊂t←w[⍋w]),(⌊/w),(⌈/w),√n÷⍨+/(w-m)*2}
  

test

  h←{m←(s←+/w)÷n←⍴w←,⍵⋄s,(×/w),m,(n{j←⌊⍺÷2⋄2|⍺:⍵[1+j]⋄2÷⍨⍵[j]+⍵[j+1]}t),(⊂¯1↓(1⌽w)-w),(⊂t←w[⍋w]),(⌊/w),(⌈/w),√n÷⍨+/(w-m)*2}
  ⎕fmt h 0 
┌9──────────────────────┐
│        ┌0─┐ ┌1─┐      │
│0 0 0 0 │ 0│ │ 0│ 0 0 0│
│~ ~ ~ ~ └~─┘ └~─┘ ~ ~ ~2
└∊──────────────────────┘
  ⎕fmt h 3
┌9──────────────────────┐
│        ┌0─┐ ┌1─┐      │
│3 3 3 3 │ 0│ │ 3│ 3 3 0│
│~ ~ ~ ~ └~─┘ └~─┘ ~ ~ ~2
└∊──────────────────────┘
  ⎕fmt h 1 2 3
┌9───────────────────────────────────────┐
│        ┌2───┐ ┌3─────┐                 │
│6 6 2 2 │ 1 1│ │ 1 2 3│ 1 3 0.8164965809│
│~ ~ ~ ~ └~───┘ └~─────┘ ~ ~ ~~~~~~~~~~~~2
└∊───────────────────────────────────────┘
  ⎕fmt h 1 2 3 4
┌9────────────────────────────────────────────────┐
│              ┌3─────┐ ┌4───────┐                │
│10 24 2.5 2.5 │ 1 1 1│ │ 1 2 3 4│ 1 4 1.118033989│
│~~ ~~ ~~~ ~~~ └~─────┘ └~───────┘ ~ ~ ~~~~~~~~~~~2
└∊────────────────────────────────────────────────┘
  ⎕fmt h 1 2 7 3 4 5 
┌9──────────────────────────────────────────────────────────────────┐
│                       ┌5──────────┐ ┌6───────────┐                │
│22 840 3.666666667 3.5 │ 1 5 ¯4 1 1│ │ 1 2 3 4 5 7│ 1 7 1.972026594│
│~~ ~~~ ~~~~~~~~~~~ ~~~ └~──────────┘ └~───────────┘ ~ ~ ~~~~~~~~~~~2
└∊──────────────────────────────────────────────────────────────────┘
| improve this answer | |
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0
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Ocaml - 288 bytes

Assuming the given list is a non-empty list of floats (to avoid conversions), and that the returned median is the weak definition of the median :

median l = n such that half the elements of l are smaller or equal to n and half the elements of l are greater or equal to n

open List
let f=fold_left
let z=length
let s l=f(+.)0. l
let a l=(s l)/.(float_of_int(z l))let rec i=function|a::[]->[]|a::b->(hd b -. a)::(i b)let r l=let t=sort compare l in(s,f( *.)1. l,a t,nth t((z t)/2+(z t)mod 2-1),t,i l,nth t 0,nth t((z t)-1),sqrt(a(map(fun n->(n-.(a l))**2.)l)))

The readable version is

open List

let sum l = fold_left (+.) 0. l
let prod l = fold_left ( *. ) 1. l
let avg l = (sum l) /. (float_of_int (length l))
let med l =
        let center = (length l) / 2 + (length l) mod 2 -1 in
        nth l center
let max l = nth l 0
let min l = nth l ((length l) - 1)
let dev l =
let mean = avg l in
        sqrt (avg (map (fun n -> (n -. mean)**2.) l))

let rec dif =
        function
        | a::[] -> []
        | a::b -> ((hd b) - a) :: (dif b)

let result l =
        let sorted = sort compare l in
        (
                sum sorted,
                prod sorted,
                avg sorted,
                med sorted,
                sorted,
                dif l,
                max sorted,
                min sorted,
                dev sorted
        )
| improve this answer | |
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0
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PHP, 213 bytes

function($a){echo$s=array_sum($a),_,array_product($a),_,$v=$s/$c=count($a);foreach($a as$i=>$x){$d+=($x-$v)**2;$i&&$f[]=$x-$a[$i-1];}sort($a);var_dump(($a[$c/2]+$a[$c/2+~$c%2])/2,$f,$a,$a[0],max($a),sqrt($d/$c));}

Try it online.

| improve this answer | |
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0
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Jelly, 27 bytes

_ÆmÆḊ÷L½$
Wẋ9SPÆmÆṁIṢṂṀÇ9ƭ€

Try it online!

How it works

_ÆmÆḊ÷L½$ - Helper link. Takes a list l on the left
 Æm       - Yield the mean of l
_         - Subtract the mean from each element of l
   ÆḊ     - Calculate the norm
        $ - Run the previous two commands over l:
      L   -   Length of l
       ½  -   Square root
     ÷    - Divide the norm by the square root of l's length

Wẋ9SPÆmÆṁIṢṂṀÇ9ƭ€ - Main link. Takes a list l on the left
W                 - Yield [l]
 ẋ9               - Repeat 9 times
              9ƭ€ - Generate a 9 element list by running the previous 9 commands over l
                    and collecting the results together:
   S              -   Sum
    P             -   Product
     Æm           -   Mean
       Æṁ         -   Median
         I        -   Forward increments
          Ṣ       -   Sort
           Ṃ      -   Minimum
            Ṁ     -   Maximum
             Ç    -   Helper link (standard deviation)
| improve this answer | |
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