# Operations with Lists

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.x. [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 Where μ is the mean average, xi is the ith term in the list, and N is the length of the list.

Shortest code wins. Good luck!

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

# Q, 41

{(+/;*/;avg;med;-':;asc;min;max;dev)@\:x}

• 'med'...facepalm – skeevey Jun 8 '12 at 11:14
• I was wondering what you were up to! – tmartin Jun 8 '12 at 12:04

## 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|
+--+--+---+-+-------+-------+-+-+-------+

• It has to be 1 1 1 not 1 1 1 1 as difference itself next – RosLuP Jan 6 '18 at 10:37

# TI-BASIC, 41 bytes

1-Var Stats is one byte, and Σx, x̄, 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.

# 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


### 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-l},r,r,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:

### Scala 208202 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

• For me "Vector(-56, -4, 28, -16, -14)" is wrong – RosLuP Jan 6 '18 at 10:38
• @RosLuP: Why is it wrong? – user unknown Jan 7 '18 at 0:31
• Yes you are right if input is "Vector(64, 8, 4, 32, 16, 2)" ( i confuse the input) – RosLuP Jan 7 '18 at 12:50

# 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!

• 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. – Giuseppe Jan 5 '18 at 17:30
• That didn't even occur to me. I added an alternate solution with the correct std deviation. – gggg Jan 5 '18 at 20:35

# 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
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);
}

• 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)));} – Zacharý Aug 11 '17 at 22:38
• @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... – Karl Napf Aug 11 '17 at 22:48
• Does it work though?! – Zacharý Aug 11 '17 at 22:54
• @Zacharý yes it does work. – Karl Napf Aug 11 '17 at 23:01
• 327 bytes – ceilingcat Mar 17 at 23:13

@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! # Pyt, 39 bytes ←ĐĐĐĐĐĐĐŞ⇹Ʃ3ȘΠ4Șµ5Ș₋⇹6Ș↕⇹7ȘṀ↔ĐĐµ-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 ĐĐµ-²Ʃ⇹Ł/√ Standard Deviation  # 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 └∊──────────────────────────────────────────────────────────────────┘  # 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 )  # 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,$a,max($a),sqrt($d/$c));}
`