Nekomata, 8 bytes

,o;↔ᶻÐlµ

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A port of @ბიმო's Husk answer.

,           Join two copies of the input
 o          Sort
  ;         Split into two parts
   ↔        Reverse the second part
    ᶻÐ      Zip the two parts into a list of pairs
      l     Take the last pair
       µ    Take the mean

Nekomata, 8 bytes

oʷ{tiN}µ

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oʷ{tiN}µ
o           Sort
 ʷ{   }     Loop until failure:
   t            Remove the first element
    i           Remove the last element
     N          Check if the list is non-empty
       µ    Take the mean

Python 3.8 (pre-release), 49 44 bytes

-5 Inspired by Luis Mendos Octave answer.

lambda x:sum(sorted(x+x)[(n:=~len(x)):-n])/2

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Octave, 38 bytes

@(x)mean(([1;1]*sort(x))(end/2+[0 1]))

This defines an anonymous function. Input is a row vector.

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Explanation

            sort(x)                 % Sort input x, of length k
      [1;1]*                        % Matrix-multiply by column vector of two ones
                                    % This vertically concatenates sort(x) with 
                                    % itself. In column-major order, this effectively 
                                    % repeats each entry of sort(x)
     (             )(end/2+[0 1])   % Select the entry at position end/2 and the next.
                                    % Entries are indexed in column-major order. Since
                                    % the array has 2*k elements, this picks the k-th 
                                    % and (k+1)-th (1-based indexing). Because entries
                                    % were repeated, for odd k this takes the original
                                    % (k+1)/2-th entry twice. For even k this takes
                                    % the original (k/2)-th and (k/2+1)-th entries
mean(                            )  % Mean of the two selected entries

Thunno 2, 1 byte

ṁ

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Built-in.

Thunno 2, 11 bytes

l⁻½çṃN$Ṡsim

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Non built-in.

Explanation

l⁻½çṃN$Ṡsim  # Implicit input
l⁻           # Length of input, decremented
  ½          # Halve this number
   ç         # Parallelly apply:
    ṃ        #  Ceiling
     N       #  Floor
      $Ṡ     # Input list sorted
        si   # Index into this list
          m  # Mean of this list
             # Implicit output

JavaScript, 63 bytes

a=>(l=a.sort((k,v)=>k-v).length/2,l*2%2?a[l|0]:(a[l]+a[l-1])/2)

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Thunno, \$ 2 \log_{256}(96) \approx \$ 1.65 bytes

Aa

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Built-ins FTW!

Pyt, 1 byte

Ṁ

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Gotta love built-ins

Factor, 6 bytes

median

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Pip, 16 bytes

SN:g(CHg+Rgoi)/2

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Explanation

SN:g(CHg+Rgoi)/2
   g              List of command-line arguments
SN:               Sort it (using numeric comparison) in-place
       g          Take that sorted list
        +         Add it (element-wise) with
         Rg       Its reverse
     CH           Chop into two halves (if odd length, second half is longer)
    (      oi)    Select the item at index 0 of the sublist at index 1 (o=1, i=0)
              /2  Divide it by 2

Fig, \$11\log_{256}(96)\approx\$ 9.054 bytes

KY
HSt2s{HL

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-3 chars because bugfixes

KY
HSt2s{HL
--------
KY       # First helper function
 Y       # Interleave with self
K        # Sort
-------- # Send the result to the next function:
HSt2s{HL # Main function
     {HL # Compute len(input)/2-1
    s    # Drop that many items from the input
  t2     # Take 2
 S       # Sum
H        # Halve

Vyxal, 6 bytes

s∆ṁwfṁ

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sort the input, get the ∆ṁiddle item(s), wrap in a list, flatten and get the ṁean of that. (wf is needed to handle cases where the list is of odd length)

Or, without a trivial built-in

Vyxal, 11 bytes

L‹½₍⌈⌊$s$İṁ

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Explained

L‹½₍⌈⌊$s$İṁ
L‹         # Length of the input - 1 (this accounts for 0 indexing)
  ½        # halved
   ₍⌈⌊      # a list of the ceiling and floor of that number
     $s    # the input list sorted
       $İ  # indexed at the positions in the ceil,floor list
         ṁ # take the average of that

Arn, 13 12 bytes

Version: 1.1.5

Iõå)n├┼U■¨Mõ

Explained

Unpacked: z&:}+\$.:<:_&:-

z&               Zip array with itself, bind to the following
  :_             Flatten
    :<           Sort ascending
      $.         Split in half
        :}+\     Fold with `:}+`.
            &:-  Bind the half symbol to the expression

How :}+\ works: Takes the last element of the first array and adds to the second array, which is casted to the first element of that array automatically (finally, that feature actually is useful!).

Where a symbol needs to take a value but none is provided, _ is inserted, the implied variable. Ungolfed this looks like

( _ z _ ) & ( :}+\ ( $. ( :< ( _ :_ ) ) ) ) & ( :- _ )

Arn, the boring way, 3 bytes

med

Built in function, can't be compressed.

BQN, 20 bytes

{2÷˜(≠𝕩)⊑»+˝˘2↕2/∧𝕩}

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Tcl, 100 bytes

proc M L {expr ([lindex [set S [lsort -r $L]] [set h [expr [llength $L]/2]]]+[lindex $S end-$h])/2.}

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proc M L {expr ([lindex [set S [lsort $L]] [set h [expr [llength $L]/2]]]+[lindex $S end-$h])/2.}

proc M L {set I [lindex [set S [lsort $L]] [expr [set n [llength $L]]/2]]
expr {$n%2?$I:($I+[lindex $S [expr $n/2-1]])/2.}}

proc M L {set n [llength [set S [lsort $L]]]
set I [lindex $S [expr $n/2]]
expr {$n%2?$I:($I+[lindex $S [expr $n/2-1]])/2.}}

proc M L {proc G L\ i {lindex $L [expr $i]}
expr {[set n [llength [set S [lsort $L]]]]%2?[G $S $n/2]:([G $S $n/2]+[G $S $n/2-1])/2.}}

proc M L {expr {[set n [llength [set S [lsort $L]]]]%2?[lindex $S [expr $n/2]]:([lindex $S [expr $n/2]]+[lindex $S [expr $n/2-1]])/2.}}

Still very ungolfed, my first minimum viable product!

Lua, 64 63 bytes

function f(t)table.sort(t)h=#t//2return(t[h+1]+t[h+#t%2])/2 end

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Sort table, get the position halfway through the table by integer-dividing table length by two, return average of element at half position plus one and element at half position if table length is even, else at half position plus one.

This solution is only valid in Lua 5.3 and onwards where there is integer division, // (and where integers can be squished right next to the keyword return). In Lua 5.1, the equivalent is math.floor(a/b), which would add several bytes.

Python 3, 59 bytes

f=lambda l:l.sort()or(len(l)<3)*(l[0]+l[-1])/2or f(l[1:-1])

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This is a recursive version:

• the list is sorted
• if there are 1 or 2 elements left, we output the median since 0 and -1 are both first and last with a single or atwo element list
• if not, we remove first and last elements and call f.

APL (Dyalog Unicode), 14 bytes

≢⊃2+/2/⊂∘⍋⌷÷∘2

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This is a train. The original dfn was {(2+/2/⍵[⍋⍵])[≢⍵]÷2}.

The train is structured as follows

┌─┼───┐
≢ ⊃ ┌─┼───┐
    2 / ┌─┼───┐
    ┌─┘ 2 / ┌─┼─┐
    +       ∘ ⌷ ∘
           ┌┴┐ ┌┴┐
           ⊂ ⍋ ÷ 2

⊢ denotes the right argument.

⌷ index

• ⊂∘⍋ the indices which indexed into ⊢ results in ⊢ being sorted

• ÷∘2 into ⊢ divided by 2

2/ replicate this twice, so 1 5 7 8 becomes 1 1 5 5 7 7 8 8

2+/ take the pairwise sum, this becomes (1+1)(1+5)(5+5)(5+7)(7+7)(7+8)(8+8)

⊃ from this pick

• ≢ element with index equal to the length of ⊢

Previous solutions

{.5×+/(⍵[⍋⍵])[(⌈,⌊).5×1+≢⍵]}
{+/(2/⍵[⍋⍵]÷2)[0 1+≢⍵]}
{+/¯2↑(1-≢⍵)↓2/⍵[⍋⍵]÷2}
{(2+/2/⍵[⍋⍵])[≢⍵]÷2}
{(≢⍵)⊃2+/2/⍵[⍋⍵]÷2}
≢⊃2+/2/2÷⍨⊂∘⍋⌷⊢
≢⊃2+/2/⊂∘⍋⌷÷∘2

J, ₁₆ 14 bytes

2%~#{2#/:~+\:~

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In addition to BMO's array duplication trick, I found that we can add the whole array sorted in two directions. Then I realized that the two steps can be reversed, i.e. add the two arrays, then duplicate them and take the nth element.

How it works

2%~#{2#/:~+\:~
                Input: array of length n
       /:~      Sort ascending
           \:~  Sort descending
          +     Add the two element-wise
     2#         Duplicate each element
   #{           Take n-th element
2%~             Halve

Previous answers

J with stats addon, 18 bytes

load'stats'
median

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Library function FTW.

median's implementation looks like this:

J, 31 bytes

-:@(+/)@((<.,>.)@(-:@<:@#){/:~)

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How it works

-:@(+/)@((<.,>.)@(-:@<:@#){/:~)
         (<.,>.)@(-:@<:@#)       Find center indices:
                  -:@<:@#          Compute half of given array's length - 1
          <.,>.                    Form 2-element array of its floor and ceiling
                          {/:~   Extract elements at those indices from sorted array
-:@(+/)                          Sum and half

A bit of golfing gives this:

J, 28 bytes

2%~[:+/(<.,>.)@(-:@<:@#){/:~

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APL(NARS), 31 chars, 62 bytes

{2÷⍨x[⌊k]+(⌈k←2÷⍨1+≢⍵)⌷x←⍵[⍋⍵]}

test

  t←{2÷⍨x[⌊k]+(⌈k←2÷⍨1+≢⍵)⌷x←⍵[⍋⍵]}
  t 5 4 3 2 1     
3
  t 4 3 2 1     
2.5
  t 5 40 30 2 1     
5
  t 5 40 30 2     
17.5
  35÷2
17.5
   t ,80
80
  t 9 3 4 8 7 6
6.5

GolfScript, 27 25 20 17 bytes

~..+$\,(>2<~+"/2"

Takes input as an array of integers on stdin. Outputs as an unreduced fraction. Try it online!

Explanation

The median of the array, as BMO's Husk answer explains, is equal to the median of an array twice as long where each element is repeated twice. So we concatenate the array to itself, sort, and take the mean of the middle two elements. If the length of the original array is \$l\$, the middle two elements of the doubled array are at indices \$l-1\$ and \$l\$.

~                  Evaluate input (converting string -> array)
 ..                Duplicate twice
   +               Concatenate two of the copies
    $              Sort the doubled array
     \,            Swap with the non-doubled array and get its length: l
       (           Decrement: l-1
        >          Array slice: all elements at index (l-1) and greater
         2<        Array slice: first two elements (originally at indices l-1 and l)
           ~       Dump array elements to stack
            +      Add
             "/2"  Push that string
                   Output all items on stack without separator

The output will be something like 10/2.

R without using the median builtin, 51 bytes

function(x,n=sum(x|1)+1)mean(sort(x)[n/2+0:1*n%%2])

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05AB1E, 2 bytes

Åm

Try it online or verify all test cases.

No need for an explanation, since Åm is a builtin which will:

Median. Sorts the list, then returns either the middle element or the average of the middle elements depending on the parity of the length of the list.

Haskell 1.2 (Gofer), 44 bytes

f[x]=x
f[x,y]=(x+y)/2.0
f(x:y)=f.init$sort y

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Husk, 10 bytes

½ΣF~e→←½OD

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Explanation

This function uses that the median of \$[a_1 \dots a_N]\$ is the same as the median of \$[a_1 \; a_1 \dots a_N \; a_N]\$ which avoids the ugly distinction of odd-/even-length lists.

½ΣF~e→←½OD  -- example input: [2,3,4,1]
         D  -- duplicate: [2,3,4,1,2,3,4,1]
        O   -- sort: [1,1,2,2,3,3,4,4]
       ½    -- halve: [[1,1,2,2],[3,3,4,4]]
  F         -- fold the following
   ~        -- | compose the arguments ..
     →      -- | | last element: 2
      ←     -- | | first element: 3
    e       -- | .. and create list: [2,3]
            -- : [2,3]
 Σ          -- sum: 5
½           -- halve: 5/2

_{Unfortunately ½ for lists has the type [a] -> [[a]] and not [a] -> ([a],[a]) which doesn't allow F~+→← since foldl1 needs a function of type a -> a -> a as first argument, forcing me to use e.}

PHP, 70 77 bytes

Not exactly optimal but works.

Requires that the values are passed over GET.

<?sort($_GET);die(($C=count($G=$_GET))&1?$G[~-$C/2]:($G[$C/2]+$G[$C/2-1])/2);

The result will be displayed in the browser and as the return code.

Thanks to Titus for fixing it, at the cost of 7 bytes.

SmileBASIC, 45 bytes

DEF M A
L=LEN(A)/2SORT A?(A[L-.5]+A[L])/2
END

Gets the average of the elements at floor(length/2) and floor(length/2-0.5) Very simple, but I was able to save 1 byte by moving things around:

DEF M A
SORT A    <- extra line break
L=LEN(A)/2?(A[L-.5]+A[L])/2
END

Jelly, 2 bytes

Æṁ

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k, 23 bytes

Basically a slightly golfed version of q's canonical med in k.

{avg x(<x)@_.5*-1 0+#x}

dc, 120 122 bytes

9kzsa0si[li:sli1+dsila>A]dsAx[1scddd;sr1-;sli:sr1-:s]sR[lidd1-;sr;s<R1+dsila>S]sS[1si0sclSxlc1=M]dsMxla2/dd1%-;sr.5-;s+2/p

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My original code worked for all the provided test cases, but was actually faulty, so +2 bytes for the fix. Dang!

Lots of bytes since dc doesn't have any inbuilt sorting mechanism, and very little in the way of stack manipulation.

9k sets the precision to 9 places since we need the possibility of digits past the decimal point. dc doesn't float, so hopefully this is satisfactory.

zsa0si[li:sli1+dsila>A]dsAx dumps the entirety of the stack into array s, and preserves the number of items in register a.

Macros M, S, and R all make up a bubble sort. M is our 'main' macro, so to speak, so I'll cover that one first.

[1si0sclSxlc1=M]dsMx We reset increment register i to 1, and check register c to 0. We run macro S, which is one pass through the array. If S (actually, R, but S by proxy) made any changes, it would have set register c to one, so if this is the case we loop through M again.

[lidd1-;sr;s<R1+dsila>S]sS One pass through the array. We load the increment counter i, duplicate it twice, and decrement the top version of it by one. Essentially i is always high, so we compare i and i-1. Load the two values from array s, compare them, and if they're going the wrong way we run our swapping macro, R. Then we keep on incrementing i, comparing it to a, and running S until that comparison tells us we've hit the end of our array.

[1scddd;sr1-;sli:sr1-:s]sR An individual instance of swappery in array a. First we set our check register c to 1 so that M knows we made changes to the array. i should still be on the stack from earlier, so we duplicate it three times. We retrieve i indexed item from a, swap our top-of-stack so that i is present again, subtract one from it, and then retrieve that item from a. Here we run into a stack manipulation limitation, so we have to load i again and we store our previous i-1 value into that index in a. Now we just have our old i-indexed a value on the stack and i itself, so we swap these, subtract 1 from i, and replace the value in a.

Eventually M will stop running when it sees no changes have been made, and now that things are sorted we can do the actual median operation.

la2/dd1%-;sr.5-;s+2/p Since a already has the length of array s, we load it and divide by two. Testing for evenness would be costly, so we rely on the fact that dc uses the floor of a non-whole value for its index. We divide a by two and duplicate the value. We then get from s the values indexed by (a/2-.5) and (a/2-((a/2)mod 1)). This gives us the middle value twice for an odd number of values, or the middle two values for an even number. +2/p averages them and prints the result.

Java 8, 71 bytes

Parity is fun! Here's a lambda from double[] to Double.

l->{java.util.Arrays.sort(l);int s=l.length;return(l[s/2]+l[--s/2])/2;}

Nothing too complex going on here. The array gets sorted, and then I take the mean of two numbers from the array. There are two cases:

• If the length is even, then the first number is taken from just ahead of the middle of the array, and the second number is taken from the position before that by integer division. The mean of these numbers is the median of the input.
• If the length is odd, s and s-1 both divide to the index of the middle element. The number is added to itself and the result divided by two, yielding the original value.

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GolfScript, 44 bytes

~$.,:l;~l 2%{l 2/$}{{l 2/$}2*+'/2'+}if{\;}l*

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Explanation

~$.,:l;~l 2%{l.2/-}{{l 2/$}2*+'/2'+}if{\;}l*   |
~                                              Create array from input string
 $                                             Sort array
  .                                            Duplicate array
   ,                                           Pop and count the top array
    :l                                         Assign variable l
      ;                                        Pop
       ~                                       Convert array into individual integers
        l                                      Push variable l onto stack
          2%                                   Push 2 and perform mod
            {l 2/$}                            If block
            {l 2/                              push variable l and divide by 2
                 $}                            Copy/push value at index (push(stack[pop()]))
                   {{l 2/$}2*+'/2'+}           Else block
                    {l 2/                      Push l/2
                         $}                    Copy
                           2*                  Perform block {} twice
                             +                 Add top two of stack (result of copies)
                              '/2'+}if         Push and add '/2'. End if
                                      {\;}     New block. Swap top two elements then pop
                                          l*   Perform previous block l times

APL, 26 bytes

{3>≢⍵:(+/÷≢)⍵⋄∇1↓¯1↓⍵[⍋⍵]}

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How?

• 3>≢⍵:(+/÷≢)⍵ - if the length of the array is less then 3, return the average
• ⋄ - otherwise
• ∇1↓¯1↓⍵[⍋⍵] - return the sorted array with the first and last elements removed.

Julia 0.6.0 (9 bytes) (6 bytes)

median(a)

median

where a is an array. It's not a very exciting answer but it's cool that Julia has a built in function for the median.

edit: I didn't know I could just write median!

Perl 5, 58 + 2 (-ap) = 60 bytes

$_=(@F=sort{$a<=>$b}@F)%2?@F[@F/2]:@F[@F/2]/2+@F[@F/2-1]/2

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Input is split into the @F array by the '-a' flag. @F gets sorted. Then, its length is checked to see if it is odd or even. If odd, result is the middle element. If even, result is half of the element to the left of middle plus half of the element to the right of the middle.

8th, 108 105 93 bytes

: m ' n:cmp a:sort a:len 2 n:/mod swap not if n:1- 2 a:slice a:open n:+ 2 n:/ else a:@ then ;

SED (Stack Effect Diagram) is a -- a n

Test

[1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4] m .

Output

1.50000

Ungolfed version (with comments)

\ Median
: m \ a -- a n
    ' n:cmp a:sort \ Sort array
    a:len          \ Get array length
    2 n:/mod       \ Remainder and quotient
    swap           \ Remainder on TOS
    not if         
        \ Array contains an even number of items
        \ Get arithmetic mean of the two values closest to the center of the sorted list
        n:1- 2 a:slice a:open n:+ 2 n:/
    else
        \ Array contains an odd number of items
        \ Get the central value           
        a:@        
    then ;

C++ 112 Bytes

Thanks to @original.legin for helping me save bytes.

#include<vector>
#include<algorithm>
float a(float*b,int s){std::sort(b,b+s);return(b[s/2-(s&1^1)]+b[s/2])/2;}

Usage:

    int main()
    {
        int n = 4;
        float e[4] = {1,4,3,2};
        std::cout<<a(e,n); /// Prints 2.5

        n = 9;
        float e1[9] = {1,2,3,4,5,6,7,8,9};
        std::cout<<a(e1,n); /// Prints 5

        n = 13;
        float e2[13] = {1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4};
        std::cout<<a(e2,n); /// Prints 1.5

        return 0;
    }

Brain-Flak, 914 + 1 = 915 bytes

([]){({}[()]<(([])<{({}[()]<([([({}<(({})<>)<>>)<><({}<>)>]{}<(())>)](<>)){({}())<>}{}({}<><{}{}>){{}<>(<({}<({}<>)<>>)<>({}<>)>)}{}({}<>)<>>)}{}<>{}>[()]){({}[()]<({}<>)<>>)}{}<>>)}{}([]<(()())>(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<(())>){((<{}{}([[]]()){({}()()<{}>)}{}(({}){}<([]){{}{}([])}{}>)>))}{}{(<{}([[]]()()){({}()()<{}>)}{}({}{}<([]){{}{}([])}{}>)>)}{}([(({}<((((((()()()){}){}){}()){})[()()()])>)<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({}<>)<{(<{}([{}])>)}{}{(({})<((()()()()()){})>)({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})({}<({}<>)<>>((((()()()){}){}){}){})((()()()()()){})<>({}<>)(()()){({}[()]<([([({})](<()>))](<>())){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({})<>)<>{(<{}([{}])>)}{}({}<>)<>({}<><({}<>)>)>)}{}({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)<>(({}{}[(())])){{}{}(((<{}>)))}{}{}{(<{}<>([{}])><>)}{}<>}{}>){(<{}(((((()()()()())){}{})){}{})>)}{}

Requires the -A flag to run.

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Explanation

The backbone of this algorithm is a bubble sort I wrote a while ago.

([]){({}[()]<(([])<{({}[()]<([([({}<(({})<>)<>>)<><({}<>)>]{}<(())>)](<>)){({}())<>}{}({}<><{}{}>){{}<>(<({}<({}<>)<>>)<>({}<>)>)}{}({}<>)<>>)}{}<>{}>[()]){({}[()]<({}<>)<>>)}{}<>>)}{}

I don't remember how this works so don't ask me. But I do know it sorts the stack and even works for negatives

After everything has been sorted I find 2 times the median with the following chunk

([]<(()())>(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<(())>)  #Stack height modulo 2
{((<{}{}          #If odd
 ([[]]())         #Push negative stack height +1
 {                #Until zero 
  ({}()()<{}>)    #add 2 to the stack height and pop one
 }{}              #Get rid of garbage
 (({}){}<         #Pickup and double the top value
 ([]){{}{}([])}{} #Remove everything on the stack
 >)               #Put it back down
>))}{}            #End if
{(<{}                     #If even
  ([[]]()())              #Push -sh + 2
  {({}()()<{}>)}{}        #Remove one value for every 2 in that value
  ({}{}<([]){{}{}([])}{}>)#Add the top two and remove everything under them
>)}{}                     #End if

Now all that is left is to make convert to ASCII

([(({}<((((((()()()){}){}){}()){})[()()()])>)<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({}<>)<
{(<{}([{}])>)}{}  #Absolute value (put "/2" beneath everything)

{                 #Until the residue is zero 
(({})<            #|Convert to base 10
((()()()()()){})  #|
>)                #|...
({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})
({}<({}<>)<>>((((()()()){}){}){}){})((()()()()()){})<>({}<>)
                  #|
(()()){({}[()]<([([({})](<()>))](<>())){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({})<>)<>{(<{}([{}])>)}{}({}<>)<>({}<><({}<>)>)>)}{}({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)<>(({}{}[(())])){{}{}(((<{}>)))}{}{}{(<{}<>([{}])><>)}{}<>
}{}               #|
>)
{(<{}(((((()()()()())){}{})){}{})>)}{}  #If it was negative put a minus sign

Racket, 95 bytes

Using the trusty old match syntax. The pattern (list _ m ... _) matches the middle of a list (that is, it omits the first and last element).

(λ(l)(let f([l(sort l <)])(match l[(list x)x][(list x y)(/(+ x y)2)][(list _ m ... _)(f m)])))

Ungolfed

(λ (l)
  (let f ([l (sort l <)])
    (match l
      [(list x) x]
      [(list x y) (/ (+ x y) 2)]
      [(list _ m ... _) (f m)])))

Japt, 20 bytes

n gV=0|½*Ul)+Ug~V)/2

Test it online! Japt really lacks any built-ins necessary to create a really short answer for this challenge...

Explanation

n gV=0|½*Ul)+Ug~V)/2  // Implicit: U = input list
n                     // Sort U.
   V=0|½*Ul)          // Set variable V to floor(U.length / 2).
  g                   // Get the item at index V in U.
            +Ug~V     // Add to that the item at index -V - 1 in U.
                 )/2  // Divide by 2 to give the median.
                      // Implicit: output result of last expression

C#, 126 bytes

using System.Linq;float m(float[] a){var x=a.Length;return a.OrderBy(g=>g).Skip(x/2-(x%2==0?1:0)).Take(x%2==0?2:1).Average();}

Pretty straightforward, here with LINQ to order the values, skip half the list, take one or two values depending on even/odd and average them.

Ruby 50 48 Bytes

-2 bytes thanks to @Conor O'Brien

->(l){l.sort!;e=l.length;(l[~-e/2]+l[e/2])/2.0}

Python 2, 48 bytes

An unnamed function which returns the result. -1 byte thanks to xnor.

lambda l:l.sort()or(l[len(l)/2]+l[~len(l)/2])/2.

The first step is obviously to sort the array, using l.sort(). However, we can only have one statement in a lambda, so we utilise the fact that the sort function returns None by adding an or - as None is falsy in Python, this tells it to evaluate and return the next part of the statement.

Now we have the sorted list, we need to find either the middle, or the middle two, values.

Using a conditional to check the length's parity would be too verbose, so instead we get the indexes len(l)/2 and ~len(l)/2:

• The first is floor(length / 2), which gets the middle element if the length is odd, or the left item in the central pair if the length is even.
• The second is the binary inversion of the list length's, evaluating to -1 - floor(length / 2). Due Python's negative indexing, this essentially does the same as the first index, but backwards from the end of the array.

If the list is of odd length, these indexes will point to the same value. If it is of even length, then they will point to the central two items.

Now that we have these two indexes, we find these values in the list, sum them, and divide them by 2. The trailing decimal place in /2. makes sure that it is float division rather than integer division.

The result is implicitly returned, as this is a lambda function.

Try it online!

Pari/GP - 37 39 Bytes

Let a be a rowvector containing the values.

b=vecsort(a);n=#b+1;(b[n\2]+b[n-n\2])/2  \\ 39 byte              

n=1+#b=vecsort(a);(b[n\2]+b[n-n\2])/2    \\ obfuscated but only 37 byte

Since Pari/GP is interactive, no additional command is needed to display the result.

a=vector(8,r,random(999))           
n=1+#b=vecsort(a);w=(b[n\2]+b[n-n\2])/2      
print(a);print(b);print(w)       

Try it online!

CJAM - 21

q~]$__,2/=\_,(2/=+2d/

• q~] reads input to array
• $__ sorts it and makes 2 copies
• , gets length of array
• 2/ divides that by 2 rounded down
• = finds the number at that index
• /_ puts original array at top of stack and copies it
• ,( gets length of array - 1
• 2/ divides that by 2 rounded down
• = finds the number at that index
• + adds the two array elements extracted 2d/ divides them by 2 as a double (so no rounding)

If the number of array elements N is odd, floor(N/2) = floor((N-1)/2). If N is even the two center elements are selected and the mean is found.

Longer but working alternative strategies:

q~]$__,2/)<_,@W%<&_:+\,d/
q~]$:A,2/_(A,2%$A=@A=+2d/\;
q~]$_Vf*_,2/.5t_W%.+.*:+

C#, 75 bytes

a=>{Array.Sort(a);int m=a.Length;return m%2>0?a[m/2]:(a[m/2-1]+a[m/2])/2;};

An anonymous function which computes the median.

Full program with ungolfed method and test cases:

using System;

public class Program
{
    public static void Main()
    {
        Func<double[],double> f =
        a =>
        {
            Array.Sort(a);  // built-in sort function for arrays
            int m = a.Length;   // stores the number of elements from the array
            return m % 2 > 0 ? a[m/2] : ( a[m/2-1] + a[m/2] ) / 2;
            // if the array has an odd number of elements, the central number will be returned
            // otherwise, the average of the two central elements
        };

        // test cases:
        Console.WriteLine(f(new double[]{1,2,3,4,5,6,7,8,9}));  // 5
        Console.WriteLine(f(new double[]{1,4,3,2}));    // 2.5
        Console.WriteLine(f(new double[]{1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4}));  // 1.5
    }
}

T-SQL, 101 67

DECLARE @ table(i real)
INSERT @ values(1),(3),(20),(4)

SELECT top 1PERCENTILE_CONT(.5)WITHIN GROUP(ORDER BY i)OVER()FROM @

Try it out

Swift, 93 bytes

let m:([Double])->Double={{c,s in c%2==0 ?(s[c/2-1]+s[c/2])/2:s[c/2]}($0.count,$0.sorted())}

This takes about 10 seconds to compile on my machine but it works. It declares the constant m of type [Double] -> Double.

Haxe, 104 bytes

Not amazing, what with the function keywords and a mandatory sorting function …

function f(l,?a)return(l[(a={l[0]+=.0;l.sort(function(x,y)return x>y?1:-1);l.length;})>>1]+l[a-1>>1])/2;

With some whitespace:

function f(l, ?a)
  return (
      l[(a = {
          l[0] += .0;
          l.sort(
               function(x, y) return x > y ? 1 : -1
             );
          l.length;
        }) >> 1]
      + l[a - 1 >> 1]
    ) / 2;

I used l[0]+=.0; to let Haxe know the type of l. The alternative would be l:Array<Float> in the arguments. Then l is sorted, its length is stored in a, and then we basically do (l[a / 2] + l[(a - 1) / 2]) / 2.

Python3 - 31 30 bytes

Saved a byte thanks to @Dennis!

I wasn't planning on a builtin answer, but I found this module and thought it was really cool cuz I had no idea it existed.

from statistics import*;median

Try it online here.

Matlab/Octave, 6 bytes

A boring built-in:

median

Try it online!

TI-Basic, 2 bytes

median(Ans

Very straightforward.

Vim, 62 bytes

I originally did this in V using only text manipulation until the end, but got frustrated with handling [X] and [X,Y], so here's the easy version. They're about the same length.

c$:let m=sort(")[(len(")-1)/2:len(")/2]
=(m[0]+m[-1])/2.0

Try it online!

Unprintables:

c$^O:let m=sort(^R")[(len(^R")-1)/2:len(^R")/2]
^R=(m[0]+m[-1])/2.0

Honorable mention:

• ^O takes you out of insert mode for one command (the let command).
• ^R" inserts the text that was yanked (in this case the list)

Java 7, 99 bytes

Golfed:

float m(Float[]a){java.util.Arrays.sort(a);int l=a.length;return l%2>0?a[l/2]:(a[l/2-1]+a[l/2])/2;}

Ungolfed:

float m(Float[] a)
{
    java.util.Arrays.sort(a);
    int l = a.length;
    return l % 2 > 0 ? a[l / 2] : (a[l / 2 - 1] + a[l / 2]) / 2;
}

Try it online

Clojure, 65 bytes

#(/(apply +(take 2(drop(-(count %)1)(sort(for[c % i[0 1]]c)))))2)

An other approach I tried:

#(apply +(map *(for[i(range)](get{-2 0.5 -1 1 0 0.5}(-(* i 2)(count %))0))(sort %)))

Common Lisp, 89

(lambda(s &aux(m(1-(length s)))(s(sort s'<)))(/(+(nth(floor m 2)s)(nth(ceiling m 2)s))2))

I compute the mean of elements at position (floor middle) and (ceiling middle), where middle is the zero-based index for the middle element of the sorted list. It is possible for middle to be a whole number, like 1 for an input list of size 3 such as (10 20 30), or a fraction for lists with an even numbers of elements, like 3/2 for (10 20 30 40). In both cases, we compute the expected median value.

(lambda (list &aux
             (m (1-(length list)))
             (list (sort list #'<)))
  (/ (+ (nth (floor m 2) list)
        (nth (ceiling m 2) list))
     2))

R, 6 bytes

median

Not surprising that R, a statistical programming language, has this built-in.

Jelly, 9 bytes

L‘HịṢµ÷LS

Try it online!

Explanation

I'm still getting the hang of Jelly... I wasn't able to find built-ins for either the median or the mean of a list, but it's very convenient for this challenge that Jelly allows non-integer indices into lists, in which case it will return a pair of the two closest values. That means we can work with half the input length as an index, and get a pair of values when we need to average it.

L          Get the length of the input.
 ‘         Increment it.
  H        Halve it. This gives us the index of the median for an odd-length list
           (Jelly uses 1-based indexing), and a half-integer between the two indices
           we need to average for even-length lists.
   ịṢ      Use this as an index into the sorted input. As explained above this will
           either give us the median (in case of an odd-length list) or a pair of
           values we'll now need to average.
     µ     Starts a monadic chain which is then applied to this median or pair...
      ÷L     Divide by the length. L treats atomic values like singleton lists.
        S    Sum. This also treats atomic values like singleton lists. Hence this
             monadic chain leaves a single value unchanged but will return the
             mean of a pair.

JavaScript, 57 52 bytes

v=>(v.sort((a,b)=>a-b)[(x=v.length)>>1]+v[--x>>1])/2

Sort the array numerically. If the array is an even length, find the 2 middle numbers and average them. If the array is odd, find the middle number twice and divide by 2.

Racket 113 bytes

(let*((L(sort L >))(n(length L))(r list-ref))(if(odd? n)(r L(floor(/ n 2)))(/(+(r L(-(/ n 2)1))(r L(/ n 2)))2)))

Ungolfed:

(define (median L)
  (let* ((L (sort L >))
         (n (length L))
         (lr list-ref))
    (if (odd? n)
        (lr L (floor (/ n 2)))
        (/(+ (lr L (sub1(/ n 2)))
             (lr L (/ n 2)))
          2))))

Testing:

(median '(1 2 3))
(median '(1 2 3 4))

Output:

2
2 1/2

Perl 6, 31 bytes

*.sort[{($/=$_/2),$/-.5}].sum/2

Try it

Expanded:

*\     # WhateverCode lambda ( this is the parameter )

.sort\ # sort it

[{     # index into the sorted list using a code ref to calculate the positions

  (
    $/ = $_ / 2 # the count of elements divided by 2 stored in ｢$/｣
  ),            # that was the first index

  $/ - .5       # subtract 1/2 to get the second index

                # indexing operations round down to nearest Int
                # so both are effectively the same index if given
                # an odd length array

}]\

.sum / 2        # get the average of the two values

JavaScript, 273 Bytes

function m(l){a=(function(){i=l;o=[];while(i.length){p1=i[0];p2=0;for(a=0;a<i.length;a++)if(i[a]<p1){p1=i[a];p2=a}o.push(p1);i[p2]=i[i.length-1];i.pop()}return o})();return a.length%2==1?l[Math.round(l.length/2)-1]:(l[Math.round(l.length/2)-1]+l[Math.round(l.length/2)])/2}

Actually, 1 byte

║

Try it online!

J, 19 bytes

<.@-:@#{(/:-:@+\:)~

Explanation:

        (        )~   apply monadic argument twice to dyadic function 
         /:           /:~ = sort the list upwards
               \:     \:~ = sort the list downwards
           -:@+       half of sum of both lists, element-wise
<.@-:@#               floor of half of length of list
       {              get that element from the list of sums

Pyth - 11 bytes

Finds the average of the middle item taken both backwards and forwards.

.O@R/lQ2_BS

Test Suite.

MATL, 4 bytes

.5Xq

This finds the 0.5-quantile, which is the median.

Try it online!

Mathematica, 6 bytes

Median

As soon as I figure out Mthmtca, I'm posting a solution in it.