Non-Unique/Duplicate Elements

Question

Write a program which finds the non-unique elements of an array of signed integers. The resulting array can be in any order.

Your answer may be a snippet which assumes the input to be stored in a variable (d, say) and evaluates to the correct result.

Test Cases

Each test case is a single line in the format input => output. Note that other permutations of the output are valid as well.

[]                        => []
[-1, 0, 1]                => []
[1, 1]                    => [1]
[3, 0, 0, 1, 1, 0, 5, 3]  => [3, 0, 1]
[-34, 0, 1, -34, 4, 8, 4] => [-34, 4]

Order of the elements doesn't matter.

This is code golf, so the shortest answer (in bytes) wins.

Answer 1

K5, 5 bytes

Assuming the input is already in a variable called d,

?d^?d

Take the distinct elements (?) of d except (d^) the distinct elements of d (?d). Nicely symmetrical, no? This works because the "except" operator removes only the first occurrence of the right argument from the left argument.

More generally,

nu: {?x^?x}

In action:

  nu'(();-1 0 1;1 1;3 0 0 1 1 0 5 3;-34 0 1 -34 4 8 4)
(()
 ()
 ,1
 0 1 3
 -34 4)

Edit:

If we wanted to preserve the order of the first occurrence of non-unique elements, we could reverse the source list before and after we remove the unique elements via except at the cost of 4 extra bytes:

  nu: {?|(|x)^?x}
  nu'(();-1 0 1;1 1;3 0 0 1 1 0 5 3;-34 0 1 -34 4 8 4)
(()
 ()
 ,1
 3 0 1
 -34 4)

Answer 2

CJam, 10

Assuming the array is already in variable D (based on this comment):

D{De=(},_&

Try it online

Explanation:

D{…},   filter items of D based on the block
  De=   count occurrences in D
  (     decrement (resulting in true/false for duplicate/unique)
_&      remove duplicates from the results

Note: append a p if you want pretty printing, otherwise the resulting array is just printed out with no delimiters by default. That is acceptable since the question specifies the snippet only needs to "evaluate to the correct result".

Standard input/output version, 13:

q~_{1$e=(},&p

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Explanation:

q~      read and evaluate the input array
_       duplicate the array
{…},    filter items based on the block
  1$    copy the array
  e=    count occurrences
  (     decrement (resulting in true/false for duplicate/unique)
&       set intersection with the initial array (removes duplicates)
p       pretty print

Answer 3

Haskell - 32

import Data.List;f l=nub$l\\nub l

Pretty short, even with the import. a \\ b removes the first occurrence of each element of b from a, and nub makes all elements of a list unique.

Answer 4

Pyth, 7 bytes

S{.-Q{Q

Try it online.

How it works

Pyth automatically stores the evaluated input in Q and prints all unused return values.

     {Q  Convert Q into a set. This removes duplicates.
  .-Q    Perform "bagwise" difference of Q and set(Q).
         This removes the first occurrence of all elements in Q.
 {       Convert to set to deduplicate.
S        Sort. Returns a list.

Answer 5

SQL, 44 42 bytes

SELECT*FROM D GROUP BY I HAVING COUNT(*)>1

I hope it is OK to assume the integers are stored in table D? This will work in both SQLServer, PostgreSQL and possibly others. Thanks to @manatwork from the 2 bytes.

Answer 6

Mathematica, 29 26 bytes

Assuming that input is stored in d:

Select[d⋃d,d~Count~#>1&]

Otherwise, it's 29 bytes as an unnamed function:

Cases[#⋃#,n_/;#~Count~n>1]&

Here, d⋃d (or #⋃#) is a golfing trick to remove duplicates - by taking the set union with itself, Mathematica interprets the list as a set, removing duplicates automatically, while the actual union doesn't do anything.

Afterwards, both methods simply filter those elements which appear in the original list at least twice.

Answer 7

JavaScript (ES6), 37 bytes

Run this in the JavaScript console:

e={};d.filter(x=>(e[x]=1+e[x]||0)==1)

Answer 8

Matlab / Octave, 40

I'm assuming input values are real (not complex). The input is in a variable d.

unique(d(sum(triu(bsxfun(@eq,d,d')))>1))

Try it online in Octave.

Answer 9

Python 3.5, 30

[x for x in{*d}if~-d.count(x)]

Uses Python 3.5's set unpacking. The ~- subtracts 1, which takes a count of 1 to 0 which is Falsy.

This gives a list. If giving a set is OK, then we use a set comprehension, saving 1 char and not needing version 3.5:

{x for x in d if~-d.count(x)}

Answer 10

PowerShell, 31 29 Bytes

($d|group|?{$_.Count-1}).Name

Assumes that $d is already populated (as given) -- e.g., $d=@(-34,0,1,-34,4,8,4).

Pipes the array into the Group-Object cmdlet, which groups like-items together and spits out an object that's essentially an array of arrays. We pipe that to a Where-Object (the ? operator) that has Count greater than one (i.e., there are duplicates), and output the .Name of those items. Has a side bonus of preserving the initial ordering, too.

_{Edit - saved two bytes thanks to Danko Durbić}

Answer 11

APL (Dyalog Unicode), 13 9 bytes^SBCS

Anonymous tacit prefix function.

∊(⊂1↓⊣¨)⌸

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(…)⌸ for each unique element (left argument) and the indices where it occurs (right argument), apply the following tacit function:

 ⊣¨ one of the left (the unique element) for each on the right (the indices)

 1↓ drop one

 ⊂ enclose (prevents padding with zeros to create a non-ragged matrix)

∊ ϵnlist (flatten)

Answer 12

Julia, 30 29 bytes

∪(d[find(sum(d.==d',1)-1)])

d.==d' creates a symmetric matrix with the value at i,j being true if d[i]==d[j] and false otherwise. summing in one dimension and then subtracting 1 will produce zero if there's only one of the element and nonzero if there's more than one. find will obtain the indexes of the non-zero elements, which are then used to index the array d itself. ∪ (union) acts like unique when used in this way, removing the repeats.

Old solution:

∪(filter(i->sum(d.==i)>1,d))

Simple - for each entry, it checks if there's more than one of it in the array. Those for which there are more than one are returned by "filter", and then ∪ (union) acts like unique when used in this way, removing the repeats.

Note: originally had it as a function, but question allows array to be stored in a variable, for which I've chosen d as suggested in the question.

Answer 13

Octave, 22 bytes

Based on Luis Mendo's answer.

d(sum(triu(d==d'))==2)

Answer 14

R, 31 24 bytes

Thanks to flodel for the 7 bytes.

Assuming the input is already in d.

code:

unique(d[duplicated(d)])

edit: now it outputs correctly if there are more than 2 duplicates as pointed by aditsu.

Answer 15

Mathematica, 31 29

Cases[{s_,t_/;t>1}:>s]@*Tally

Answer 16

Shell + GNU coreutils, 12

sort|uniq -d

Test output:

$ printf "%s\n" -34 0 1 -34 4 8 4 | ./nonuniq.sh 
-34
4
$ 

Answer 17

Pyth, 7 bytes

ft/QT{Q

Explanation:

ft/QT{Q
           Q = eval(input())
     {Q    set(Q) - deduplicate
f          filter - with T as the filter variable.
  /QT      count in Q of T
 t         minus 1.

The filter removes all elements that appear exactly once from the set of elements.

Answer 18

LINQ,62 54 bytes

Kinda new here, but here goes nothing.

d.GroupBy(c=>c).Where(g=>g.Count()>1).Select(g=>g.Key)

Answer 19

Python 2.7, 36 42

list(set(filter(lambda x:d.count(x)>1,d)))

edit : surrounded the expression with list(..) in order to comply with the format required in the question

Answer 20

K (not K5), 10 bytes

x@&1<#:'=x

Assumes input is in x. I thought it'd be fun to do a non-K5 answer!

Answer 21

Python 3 - 33 30 bytes

{_ for _ in d if d.count(_)>1}

Repl output, d as input.

Answer 22

K (oK), 7 bytes

Solution:

&1<#:'=

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Explanation:

&1<#:'= / the solution
      = / group, key => value (index)
   #:'  / count length of each group
 1<     / 1 less than 
&       / keys where true

Answer 23

Haskell + hgl, 11 10 bytes

(!!1)~<<bg

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Explanation

• bg sort a list into lists of like values.
• (!!1) get the second value or [] if there is none.

We use ~<< to glue these together, which basically acts as a concat map.

Reflection

This works pretty well, but the operator precedence is killing me. Both of the following work but requires the parens to prevent an error.

nb<(df^*nb)
nb<(fdf~<nb)

If the operator precedence was different both of these would be competitive with the first one actually saving bytes.

I don't know if that means the precedence is wrong, but it's worth noticing.

(!!) should have a prefix function, and a bunch of versions with the index prefilled. I noticed this earlier but still didn't fix it, and now it's coming back to haunt me.

Answer 24

Mathematica, 23 bytes

With input stored in d:

Pick[#,#2>1]&@@@Tally@d

As a function, 24 bytes:

Pick[#,#2>1]&@@@Tally@#&

for example, with

d = {3, 0, 0, 1, 1, 0, 5, 3}
Tally@d

returns this:

   {{3, 2},
    {0, 3},
    {1, 2},
    {5, 1}}

(first element of each sublist is the element, second one is frequency of occurrence). Applying to this list Pick[#,#2>1]&@@@ transforms it to

{Pick[3,2>1], Pick[0,3>1], Pick[1,2>1], Pick[5,1>1]}

And where the second argument of Pick evaluates to True the first argument is returned.

Answer 25

Perl 6, 16 bytes

Assuming the list is stored in $_ you could use any of the following snippets.
_{^{( which was specifically allowed )}}

(--«.BagHash).Set.keys # 23 bytes

keys .Bag (-) .Set # 18 bytes

# U+2216 SET MINUS
keys .Bag∖.Set # 16 bytes in utf8

---

If you don't care that you get a Bag you could leave off keys .

$_ = [3, 0, 0, 1, 1, 0, 5, 3];
.Bag∖.Set ∋ 3 # True
.Bag∖.Set ∋ 5 # False

---

None of these have the limitation of only working on signed integers, or even just numbers for that matter.

say keys .Bag∖.Set given |(<a b c d a a c>), 1/3, 2/3 - 1/3;
# (a c 0.333333)

Answer 26

Ruby, 30 28 bytes

In the Interactive Ruby Shell:

d.select{|x|d.count(x)>1}|[]

^{Saved 2 bytes thanks to Kirill L.}

Answer 27

Husk, 4 bytes

u-u¹

Try it online!

u       # unique element of
 -      # set difference between
  u¹    # unique elements of input  
        # and input

Answer 28

Thunno D, \$7 \log_{256}(96) \approx\$ 5.76 bytes

Note: this only works on Thunno 1.1.3+

gc1>kZU

(No ATO link since that's on v1.1.0 at the moment)

Explanation

gc1>kZU   # Implicit input. The D flag duplicates it implicitly.
g   k     # Filter the input for elements where the following is truthy:
 c        #   The count of the element in the input
  1>      #   Is greater than 1
     ZU   # Deduplicate and output implicitly

Answer 29

JavaScript, 35 28 34 bytes

a=>a.filter(o=x=>(o[x]=-~o[x])==2)

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Answer 30

Common Lisp, 57 bytes

(remove-duplicates(remove-if(lambda(x)(<(count x d)2))d))