Get duplicates from an array [duplicate]

Question

So, I have an array

['a', 42, 'b', 42, 'a']

I want to alert users to the fact that 42 and 'a' appear in it twice.

some_function(['a', 42, 'b', 42, 'a'])

should return

['a', 42]

---

Write some code to extract the duplicates from an array.

• It should accept an array as input.
• The array can have multiple types of object.
• Duplication should be defined as per the language's standard equality function (examples are from ruby).
• It outputs an array of the duplicated items.
• The order of the returned items is irrelevant.
• Duplicates should only appear in the returned array once.
• Golf rules, do it in the smallest number of bytes.

Test cases

[1,2,3] => []

[1,2,3,2,1] => [1,2]

[1.0, 2, 2.0, 1.0] => [1, 2] or [1.0, 2.0] or [1, 2.0] or [1.0, 2]

[1, '1', '1.0'] => []

[nil, nil, false, false, true, true] => [nil, false, true]

Answer 1

Brachylog, 9 bytes

ọ{tℕ₂&h}ˢ

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Explanation

ọ           Occurrences
 {     }ˢ   Select:
     &h       The elements…
  tℕ₂         …which occur at least 2 times

Answer 2

05AB1E, 8 bytes

{γʒg≠}€н

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Explanation

{γ       # Sort input and split into chunks of identical elements
  ʒg≠}   # Keeps chunks of length != 1
      €н # Map with the first element of each chunk

Answer 3

Pyth, 5 bytes

{.-Q{

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How it works?

{.-Q{ – Full program.
    { – Deduplicate.
 .-Q  – And apply bagwise subtraction between the input Q and the above.
{     – Deduplicate the result.

Answer 4

Python 2, 39 38 bytes

lambda l:{v for v in l if~-l.count(v)}

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---

Python 3, 37 bytes

lambda l:[*map(l.remove,{*l})]and{*l}

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Answer 5

Javascript ES6, 48 bytes

a=>[...new Set(a.filter((b,c)=>a.indexOf(b)<c))]

Test cases

f=
a=>[...new Set(a.filter((b,c)=>a.indexOf(b)<c))];

tests=[
    [1,2,3],
    [1,2,3,2,1],
    [1.0, 2, 2.0, 1.0],
    [1, '1', '1.0'],
    [undefined, undefined, false, false, true, true],
    [1,2,1,1,1,2,1,2,2,1,1]
]
console.log(tests.map(b=>f(b)))

Answer 6

[R], 58 bytes

function(x)x[duplicated(x)]

(My original code was nonsense.)

This works with an array input (in R an "atomic vector") but all the elements have to have the same type, or will be coerced to the most general type.

This also works with a list input (the only way in R an "array"-ish object can have elements of different types) which is what the challenge seems to want. The list elements can be other lists, atomic vectors, matrices, data frames - anything, really.

But then the "equality" function used by duplicated is fairly strict and won't give the same answers as the challenge test cases indicate, which is a flaw in the challenge specs and not this answer.

Answer 7

MATL, 13 12 bytes

"GX@&)m?6Mvu

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Thanks to Luis Mendo for saving a byte!

using m is the only way I could use mixed-type equality checking. Takes input as a row cell array.

		% (implicit input x)
"		% for loop:
GX@&)		% push input and index, then get x(i) and x(-i),
                % where x(-i) means "x excluding element at index i"
    m		% ismember -- true if x(i) is in x(-i)
     ?		% if true
      6M	% get x(i)
	v	% concatenate with prior results
	 u	% and take unique values
		% implicit end of if
		% implicit end of for loop
		% implicit end of program, output results as cell array

Answer 8

Jelly, 4 bytes

œ-QQ

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The last test case ([None, None, False, False, True, True]) isn't possible in Jelly.

Answer 9

PHP, 75 bytes

function a($b){return array_unique(array_diff_assoc($b,array_unique($b)));}

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Answer 10

Javascript (ES6), 51 bytes

a=>[...new Set(a.sort().filter((v,i)=>v===a[i+1]))]

run on tio.

Try it:

a=a=>[...new Set(a.sort().filter((v,i)=>v===a[i+1]))]


console.log(a([1,2,3]));
console.log(a([1,2,3,2,1]));
console.log(a([1.0, 2, 2.0, 1.0]));
console.log(a([1, '1', '1.0']));
console.log(a([null, null, false, false, true, true]));
console.log(a([1,1,1,3,2,3,2,3,4]));

Answer 11

Powershell, 38 32 bytes

($args|group|?{$_.count-1}).Name

Try it online, thanks AdmBorkBork

using:

1. save the code to a file get-duplicates.ps1
2. run .\get-duplicates.ps1 'a' 42 'b' 42 'a' 42

Known issues: Powershell supress $null value. Result of

 .\get-duplicates.ps1 $null $null $false $false $true $true

is False, True without null.

Answer 12

Java (JDK 10), 77 bytes

l->{java.util.Set.copyOf(l).forEach(l::remove);return l.stream().distinct();}

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Answer 13

Crystal, 42 bytes

def f(a)a.reject{|x|a.count(x)<2}.uniq
end

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Sometimes, when you can't get the desired results, the problem can be solved... by running your code in a different language.

Initially, I tried solving this in Ruby, but ironically, despite the OP's statement that all test cases correspond to Ruby comparison rules, the above code when run in Ruby actually fails for the third test case, as both uniq and array intersection operator use stricter equality checks. However, it passes all tests in Crystal (which shares most of its syntax with Ruby) at the expense of having to declare it as a defined function (-> proc would be longer in Crystal).

As a bonus, here is the golfier Ruby-only version of the code (doesn't pass test 3):

Ruby, 33 bytes

->a{a.reject{|x|a.count(x)<2}|[]}

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Answer 14

Japt -h, 7 bytes

â £kX â

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Explanation:

â £kX â
â         // Get all unique items from the input
  £       // Map X through the results
   kX     //   Remove X from the input
      â   //   Get unique items
-h        // Return the last item

Answer 15

Gaia, 4 bytes

:uDu

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: duplicates the input, u uniquifies it, D retrieves the multiset difference between the two and u uniquifies the result.

Answer 16

Husk, 4 bytes

uṠ-u

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Explanation

     -- input, e.g. [2,1,1,3,2,2]
   u -- remove duplicates: [2,1,3]
 Ṡ-  -- subtract this from original list: [2,1,1,3,2,2] - [2,1,3] = [1,2,2]
u    -- remove duplicates: [1,2]

Answer 17

Physica, 33 bytes

->a:Set@Filter[->e:a.count@e>1;a]

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Answer 18

Haskell, 33 bytes

nub.((\\)<*>nub)
import Data.List

Try it online! Port of my Husk answer.

The point-free version given above is equivalent to f x=nub$x\\nub x which is one byte longer.

---

Haskell, 50 bytes

([]#)
a#(x:r)=([x|any(==x)r,all(/=x)a]++a)#r
a#_=a

Try it online! Alternative without imports.

Answer 19

Proton, 32 bytes

l=>set(v for v:l if~-l.count(v))

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Port of TFeld's Python answer. Could be shorter but {...} apparently doesn't work for setcomps.

Answer 20

Red, 46 bytes

func[b][foreach c unique b[alter b c]unique b]

Doesn't work for numbers in different representation as float or string.

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Answer 21

Stax, 5 bytes

═H+U(

Run and debug it

Explanation:

cu|-uJ Full program, unpacked
cu     Copy and make unique
  |-   Remove each element once
    u  Make unique again
     J Join by space. 'm' would be equally long, but as F has a lower charcode, it can be packed better.

Answer 22

Retina, 23 22 bytes

Lmsl`(.+)(?=.+^\1$)
D`

Try it online! Explanation:

Lmsl`(.+)(?=.+^\1$)

List duplicated lines. The m modifier allows the ^ and $ to match the start and end of a line in the lookahead. The s modifier allows us to look ahead past newlines using just .+. The l modifier automatically anchors the match to whole lines. I can't use the p flag though as the match could be more than one line at once, however the next stage will deal with that.

D`

Deduplicate that list.