Count repetitions of an array

Question

You will receive an array and must return the number of integers that occur more than once.

[234, 2, 12, 234, 5, 10, 1000, 2, 99, 234]

This will return 2, since each of 234 and 2 appear more than once.

[234, 2, 12, 234] [2, 12, 234, 5, 10, 1000, 2]

The list will never be more than 100k integers long, and the integers inside the list will always be in between -100k and 100k.

Integers should be counted if they occur more than once, so if an integer occurs 3 times then it will still only count as one repeated integer.

Test cases

[1, 10, 16, 4, 8, 10, 9, 19, 2, 15, 18, 19, 10, 9, 17, 15, 19, 5, 13, 20] = 4 [11, 8, 6, 15, 9, 19, 2, 2, 4, 19, 14, 19, 13, 12, 16, 13, 0, 5, 0, 8] = 5 [9, 7, 8, 16, 3, 9, 20, 19, 15, 6, 8, 4, 18, 14, 19, 12, 12, 16, 11, 19] = 5 [10, 17, 17, 7, 2, 18, 7, 13, 3, 10, 1, 5, 15, 4, 6, 0, 19, 4, 17, 0] = 5 [12, 7, 17, 13, 5, 3, 4, 15, 20, 15, 5, 18, 18, 18, 4, 8, 15, 13, 11, 13] = 5 [0, 3, 6, 1, 5, 2, 16, 1, 6, 3, 12, 1, 16, 5, 4, 5, 6, 17, 4, 8] = 6 [11, 19, 2, 3, 11, 15, 19, 8, 2, 12, 12, 20, 13, 18, 1, 11, 19, 7, 11, 2] = 4 [6, 4, 11, 14, 17, 3, 17, 11, 2, 16, 14, 1, 2, 1, 15, 15, 12, 10, 11, 13] = 6 [0, 19, 2, 0, 10, 10, 16, 9, 19, 9, 15, 0, 10, 18, 0, 17, 18, 18, 0, 9] = 5 [1, 19, 17, 17, 0, 2, 14, 10, 10, 12, 5, 14, 16, 7, 15, 15, 18, 11, 17, 7] = 5

What do you mean by Once it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input? — Gymhgy
– Gymhgy, Commented Feb 24, 2019 at 18:19
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases. — Riker
– Riker, Commented Feb 24, 2019 at 18:21
I have added some answers to the test cases, sorry if I go them wrong — MickyT
– MickyT, Commented Feb 24, 2019 at 19:20
I've voted to close this question until you confirm this is what you intended. — Riker
– Riker, Commented Feb 24, 2019 at 21:01
Related (output the non-unique items, instead of the amount of non-unique items). — Kevin Cruijssen
– Kevin Cruijssen, Commented Feb 24, 2019 at 21:38

MickyT · Accepted Answer · 2019-02-24 18:34:02Z

16

R, 20 bytes

Is this what you are after? Uses table to count the occurrences of each of the scan input values. Tests if count is > 1 and sums the trues.

sum(table(scan())>1)

Try it online!

answered Feb 24, 2019 at 18:34

MickyT

12.3k2 gold badges27 silver badges49 bronze badges

\$\begingroup\$ my mind went to duplicated first -- the humble table is so useful for golfing! \$\endgroup\$

Giuseppe
– Giuseppe

2019-02-25 15:41:37 +00:00
Commented Feb 25, 2019 at 15:41
\$\begingroup\$ @giuseppe table is a favorite now :) \$\endgroup\$

MickyT
– MickyT

2019-02-25 17:17:23 +00:00
Commented Feb 25, 2019 at 17:17

Add a comment |

Digital Trauma · Accepted Answer · 2019-02-24 19:19:13Z

9

Bash + coreutils, 18

sort|uniq -d|wc -l

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answered Feb 24, 2019 at 19:19

Digital Trauma

73.8k10 gold badges117 silver badges268 bronze badges

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Laikoni · Accepted Answer · 2019-02-24 22:16:37Z

Haskell, 42 bytes

f s=sum[1|x<-[-9^6..9^6],filter(==x)s>[x]]

Try it online! Abuses the fact the the integers in the list are guaranteed to be within -100k and 100k.

Adám · Accepted Answer · 2019-03-27 14:10:49Z

APL (Dyalog Unicode), 9 8 bytes^SBCS

-1 thanks to ngn

Anonymous tacit prefix function.

+/1<⊢∘≢⌸

Try it online!

+/ sum of

1< whether 1 is less than

…⌸ for each unique element:

⊢∘ ignoring the actual unique element,

≢ the count of its occurrences

\$\begingroup\$ {1<≢⍵}⌸ -> 1<⊢∘≢⌸ \$\endgroup\$

ngn
– ngn

2019-03-22 16:13:26 +00:00
Commented Mar 22, 2019 at 16:13 — ngn
– ngn, Commented Mar 22, 2019 at 16:13
\$\begingroup\$ @ngn Thanks. Incorporated. \$\endgroup\$

Adám
– Adám

2019-03-27 14:11:05 +00:00
Commented Mar 27, 2019 at 14:11 — Adám
– Adám, Commented Mar 27, 2019 at 14:11

Gymhgy · Accepted Answer · 2019-02-25 16:20:13Z

6

C# (Visual C# Interactive Compiler), 40 bytes

n=>n.GroupBy(c=>c).Count(c=>c.Count()>1)

The first draft of the spec was unclear, and I thought it mean return all the elements that appear more than once. This is the updated version.

Somehow I didn't notice that my code returned the number of elements that appeared once. Thanks to Paul Karam for catching that!

Try it online!

edited Feb 25, 2019 at 16:20

answered Feb 24, 2019 at 19:03

Gymhgy

8,03813 silver badges35 bronze badges

1

\$\begingroup\$ Your output is wrong, it needs to count the elements with 2 or more occurences. It should be n=>n.GroupBy(c=>c).Count(c=>c.Count()>=2). The OP says the answer of this list is 2. Your code returns 5. The change I gave you returns 2. \$\endgroup\$

Paul Karam
– Paul Karam

2019-02-25 06:26:37 +00:00
Commented Feb 25, 2019 at 6:26
1

\$\begingroup\$ Or just >1 to keep the 40 bytes count \$\endgroup\$

Paul Karam
– Paul Karam

2019-02-25 07:13:07 +00:00
Commented Feb 25, 2019 at 7:13
\$\begingroup\$ @PaulKaram I didn't notice that, thanks! \$\endgroup\$

Gymhgy
– Gymhgy

2019-02-25 16:20:33 +00:00
Commented Feb 25, 2019 at 16:20

Add a comment |

Collin Phillips · Accepted Answer · 2019-02-25 01:27:13Z

5

C (clang) 175 117 95 bytes

c(*a,*b){return*a-*b;}r(*l,m){qsort(l,m,4,c);return((!m||l[1]-*l)&l[-1]==*l)+(m?r(l+1,m-1):0);}

Try it online!

This is the first time I've submitted one of these, so let me know if there are any issues with formatting or anything.

Updates from the comments:

-58 to 117 bytes from Jo King
-80 to 95 bytes from ASCII-only

original submission

edited Feb 25, 2019 at 1:27

answered Feb 25, 2019 at 0:09

Collin Phillips

512 bronze badges

5

\$\begingroup\$ Welcome, nice start. I'm not a C person but here's a link to a tips for golfing C page \$\endgroup\$

MickyT
– MickyT

2019-02-25 00:27:52 +00:00
Commented Feb 25, 2019 at 0:27
2

\$\begingroup\$ 117 bytes => d,i;c(*a,*b){return*a-*b;}r(l[],m){qsort(l,m,4,c);for(i=d=0;++i<m;)d+=((l[i+1]-l[i]||i>m-2)&&l[i-1]==l[i]);return d;}. As @ASCII-only noted, the includes don't affect the compilation of your program \$\endgroup\$

Jo King
– Jo King

2019-02-25 00:38:13 +00:00
Commented Feb 25, 2019 at 0:38
2

\$\begingroup\$ @JoKing 100: d;c(*a,*b){return*a-*b;}r(*l,m){qsort(l,m,4,c);for(d=0;~m--;)d+=(!m||l[1]-*l)&l[-1]==*l++;return d;} \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-25 00:41:42 +00:00
Commented Feb 25, 2019 at 0:41
1

\$\begingroup\$ @CollinPhillips yes. as you can see in the link i posted, it still compiles fine without the includes \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-25 00:42:11 +00:00
Commented Feb 25, 2019 at 0:42
2

\$\begingroup\$ 95: c(*a,*b){return*a-*b;}r(*l,m){qsort(l,m,4,c);return((!m||l[1]-*l)&l[-1]==*l)+(m?r(l+1,m-1):0);} \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-25 01:12:27 +00:00
Commented Feb 25, 2019 at 1:12

| Show 3 more comments

Emigna · Accepted Answer · 2019-02-24 19:48:41Z

4

05AB1E, 4 bytes

Ù¢≠O

Try it online! or as a Test Suite

Explanation

 O # sum ≠ # the false values ¢ # in the count Ù # of each unique digit in input

edited Feb 24, 2019 at 19:48

answered Feb 24, 2019 at 19:43

Emigna

53.2k5 gold badges44 silver badges164 bronze badges

\$\begingroup\$ So all values that are not 1 are false? \$\endgroup\$

Adám
– Adám

2019-02-24 20:20:00 +00:00
Commented Feb 24, 2019 at 20:20
\$\begingroup\$ @Adám: Yes, that is correct. \$\endgroup\$

Emigna
– Emigna

2019-02-24 21:01:07 +00:00
Commented Feb 24, 2019 at 21:01

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Kirill L. · Accepted Answer · 2019-02-24 20:07:58Z

4

Python 3, 38 bytes

lambda a:sum(a.count(x)>1for x in{*a})

Try it online!

answered Feb 24, 2019 at 20:07

Kirill L.

14.3k2 gold badges17 silver badges54 bronze badges

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Jonathan Allan · Accepted Answer · 2019-02-24 21:49:38Z

Jelly, 4 bytes

ĠITL

Try it online!

...Or ĠIƇL

How?

ĠITL - Link: list of integers e.g. [234, 2, 12, 234, 5, 10, 1000, 2, 99, 234] Ġ - group indices by value [[2,8],5,6,3,9,[1,4,10],7] I - incremental differences [[6],[],[],[],[],[3,6],[]] T - truthy indices [1,6] L - length 2

IƇ would filter to keep only truthy results of I ([[6],[3,6]]) which also has the desired length.

Galen Ivanov · Accepted Answer · 2019-02-25 07:18:46Z

4

J, 11 9 bytes

-2 bytes thanks to Jonah!

1#.1<1#.=

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Original solution:

1#.(1<#)/.~

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Explanation:

 /.~ group the list by itself ( ) for each group 1<# is the length greater than 1 1#. sum by base-1 conversion

edited Feb 25, 2019 at 7:18

answered Feb 24, 2019 at 19:30

Galen Ivanov

21.5k3 gold badges26 silver badges62 bronze badges

\$\begingroup\$ Hey Galen. 1#.1<1#.= for 9 bytes + good ol' self-classify fun. \$\endgroup\$

Jonah
– Jonah

2019-02-24 23:15:44 +00:00
Commented Feb 24, 2019 at 23:15
1

\$\begingroup\$ @Jonah Thanks! Honestly, I wasn't aware of this. \$\endgroup\$

Galen Ivanov
– Galen Ivanov

2019-02-25 07:20:22 +00:00
Commented Feb 25, 2019 at 7:20
1

\$\begingroup\$ @Jonah Nice! \$\endgroup\$

Adám
– Adám

2019-02-25 07:21:25 +00:00
Commented Feb 25, 2019 at 7:21
\$\begingroup\$ @Adám and here i was pleased that i'd gotten J to tie with APL. Foiled again :) \$\endgroup\$

Jonah
– Jonah

2019-02-25 18:14:10 +00:00
Commented Feb 25, 2019 at 18:14

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 74 73 bytes

L->L.stream().filter(i->L.indexOf(i)<L.lastIndexOf(i)).distinct().count()

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Explanation:

L-> // Method with ArrayList parameter and integer return-type L.stream() // Create a stream of the input-list .filter(i-> // Filter it by: L.indexOf(i) // Where the first index of a value <L.lastIndexOf(i)) // is smaller than the last index of a value .distinct() // Deduplicate this filtered list .count() // And return the count of the remaining values

chunes · Accepted Answer · 2023-10-13 23:29:44Z

Uiua, 7 bytes

/+>1⍘⊚⊛

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/+>1⍘⊚⊛ ⊛ # classify (i.e. map integers -> naturals) ⍘⊚ # inverse where (occurrences) >1 # where are they greater than one? /+ # sum

Kirill L. · Accepted Answer · 2019-02-24 19:48:17Z

3

Ruby, 34 bytes

->a{a.uniq.count{|x|a.count(x)>1}}

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answered Feb 24, 2019 at 19:48

Kirill L.

14.3k2 gold badges17 silver badges54 bronze badges

1

\$\begingroup\$ Ruby 2.7, 24 bytes: ->a{a.tally.count{_2>1}} \$\endgroup\$

Value Ink
– Value Ink

2023-10-14 01:27:46 +00:00
Commented Oct 14, 2023 at 1:27

Add a comment |

Jo King · Accepted Answer · 2019-02-24 22:07:42Z

Perl 6, 15 bytes

+*.repeated.Set

Try it online!

Pretty self explanatory. An anonymous code block that gets the count (+) of the Set of elements among the repeated elements of the input (*).

I've realised I've posted almost the exact same solution for a related question.

pizzapants184 · Accepted Answer · 2019-02-25 06:05:57Z

3

Python 3, 63 bytes

lambda l:len(C(l)-C({*l})) from collections import Counter as C

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answered Feb 25, 2019 at 6:05

pizzapants184

3,55816 silver badges18 bronze badges

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Adám · Accepted Answer · 2019-02-25 14:58:20Z

APL (Dyalog Extended), 8 7 bytes^SBCS

Anonymous tacit prefix function using Jonah's method.

+/1<∪⍧⊢

Try it online!

+/ the total number occurrences
^{literally the sum of Truths}

1< where one is less than

∪ the unique elements'

⍧ count in

⊢ the unmodified argument

xnor · Accepted Answer · 2019-02-26 01:38:25Z

Haskell, 41 bytes

f(h:t)=sum[1|filter(==h)t==[h]]+f t f _=0

Try it online!

Count suffixes where the first element h appears exactly once in the part t that comes after.

Haskell, 40 bytes

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

Try it online!

Stealing the method from other answers.

\$\begingroup\$ Dammit, we had the exact same answer \$\endgroup\$

proud haskeller
– proud haskeller

2019-02-28 16:29:24 +00:00
Commented Feb 28, 2019 at 16:29 — proud haskeller
– proud haskeller, Commented Feb 28, 2019 at 16:29

Community · Accepted Answer · 2020-06-17 09:04:33Z

Haskell, 41 bytes

f[]=0 f(a:s)=sum[1|filter(==a)s==[a]]+f s

This solution basically counts how many elements of the list have the same element appear exactly once later in the list.

Community · Accepted Answer · 2020-06-17 09:04:33Z

PHP, 39 bytes

a nice occasion to use variable variables:

foreach($argv as$v)$r+=++$$v==2;echo$r;

takes input from command line arguments. Run with -nr or try it online.

$argv[0] is - and that appears only once in the arguments, so it does not affect the result.

Wheat Wizard · Accepted Answer · 2019-02-24 19:28:55Z

Haskell, 47 bytes

f[]=0 f(a:b)|x<-filter(/=a)b,x/=b=1+f x|1>0=f b

Try it online!

This is the naïve approach. There is likely something that could be done to improve this.

f[]=0

We return 0 for the empty list

f(a:b)

In the case of a non-empty list starting with a and then b.

|x<-filter(/=a)b,x/=b=1+f x

If filtering a out of b is different from b (that is a is in b) then we return 1 more than f applied to b with the as filtered out.

|1>0=f b

If filtering as doesn't change b then we just run f across the rest.

Here is another similar approach that has the same length:

f[]=0 f(a:b)|elem a b=1+f(filter(/=a)b)|1>0=f b

Try it online!

Arnauld · Accepted Answer · 2019-02-24 19:38:41Z

2

JavaScript (ES6), 40 bytes

a=>a.map(o=x=>n+=(o[x]=-~o[x])==2,n=0)|n

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answered Feb 24, 2019 at 19:38

Arnauld

206k21 gold badges187 silver badges670 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Wolfram Language 34 bytes

 Length@DeleteCases[Gather@#,{x_}]&

Gather groups identical integers into lists. DeleteCases[...{x_}] eliminates lists containing a single number. Length returns the number of remaining lists (each containing two or more identical integers.

\$\begingroup\$ Count[{_,__}]@*Gather \$\endgroup\$

alephalpha
– alephalpha

2019-02-26 03:10:08 +00:00
Commented Feb 26, 2019 at 3:10 — alephalpha
– alephalpha, Commented Feb 26, 2019 at 3:10

Quintec · Accepted Answer · 2019-02-25 13:06:16Z

2

Japt, 12 11 9 8 6 bytes

ü èÈÊÉ

With lots of help from @ASCII-Only, and suggestions from @Shaggy and @Luis felipe De jesus Munoz.

Try it online!

edited Feb 25, 2019 at 13:06

answered Feb 24, 2019 at 23:00

Quintec

2,8891 gold badge16 silver badges28 bronze badges

\$\begingroup\$ 11 \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-24 23:09:46 +00:00
Commented Feb 24, 2019 at 23:09
\$\begingroup\$ 9 9 9? \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-24 23:19:20 +00:00
Commented Feb 24, 2019 at 23:19
\$\begingroup\$ 8 8 \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-24 23:30:42 +00:00
Commented Feb 24, 2019 at 23:30
2

\$\begingroup\$ 6 \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-24 23:45:42 +00:00
Commented Feb 24, 2019 at 23:45
2

\$\begingroup\$ 6 6 \$\endgroup\$

ASCII-only
– ASCII-only

2019-02-24 23:51:17 +00:00
Commented Feb 24, 2019 at 23:51

| Show 4 more comments

user48543 · Accepted Answer · 2019-02-25 19:17:27Z

Pyth, 6 bytes

l{.-Q{

Try it here

Explanation

l{.-Q{ {Q Deduplicate the (implicit) input. .-Q Remove the first instance of each from the input. l{ Count unique.

Unrelated String · Accepted Answer · 2019-02-26 08:41:36Z

Brachylog, 7 bytes

ọzt;1xl

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Explanation:

ọ For every unique element E of the input, [E, how many times E occurs] zt The last elements of the previous value. ;1x With every 1 removed, l how many there are.

Giuseppe · Accepted Answer · 2019-03-30 16:56:58Z

Gaia, 6 bytes

e:uDul

Try it online!

e	| eval as a list :	| duplicate u	| uniquify D	| multiset difference; keep only repeated elements u	| uniquify l	| find length

chunes · Accepted Answer · 2023-02-11 00:58:39Z

Arturo, 49 31 bytes

$=>[tally&|enumerate[k,v]->v>1]

Try it

-18 bytes due to Arturo 0.9.83 providing tally and enumerate.

$ => [ ; a function tally & ; create a dictionary of elements and their occurrences from the input | ; then... enumerate[k,v]-> ; count the number of key-value pairs where... v>1 ; the value (occurrences) is greater than one ] ; end function

Kip the Malamute · Accepted Answer · 2023-02-11 17:30:12Z

Pyt, 7 bytes

ĐỤ⇹ɔ⁻žŁ

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Đ implicit input; Đuplicate on the stack Ụ get Ụnique elements ⇹ɔ ɔount occurrences of each in original list ⁻žŁ decrement; remove žeros; get Łength; implicit print

The Thonnu · Accepted Answer · 2023-07-14 08:57:13Z

Thunno 2 `L`, 5 bytes

Uæc1Q

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Explanation

Uæc1Q # Implicit input U # Uniquify the input æ # Filter it by: c # Count in the input 1Q # Is not equal to 1 # Push the length # Implicit output

PhiNotPi · Accepted Answer · 2019-02-24 22:00:10Z

Element, 40 bytes

_(#'{"2:0+4:'~1+";~2=[''1+""]$2+'[(#]'}`

Try it online!

This requires input to be in a precise format like [234, 2, 1000, 2, 99, 234] (enclosed with [] with a comma and space between integers).

Explanation:

_ input (# delete the [ at start of input '{" '} WHILE the string is non-empty '{"2: '} duplicate it '{" 0+ '} add 0 to coerce to integer (gets next number in array) '{" 4: '} make 3 additional copies '{" ' '} temporarily move 1 copy to control stack '{" ~ '} fetch the current map value for given integer '{" 1+ '} increment map value '{" " '} retrieve temporary copy of integer (the key for the map) '{" ; '} store updated map value '{" ~ '} fetch map value again (1 if 1st instance, 2 if 2nd, etc.) '{" 2= '} test for map value = 2, this is the first duplication '{" [ ] '} IF '{" ['' ] '} move stuff from main stack to control stack '{" [ 1+ ] '} increment the counter of duplicate (bottom of stack) '{" [ ""] '} move stuff back to main stack '{" $ '} take length of current integer '{" 2+ '} add 2 (for the comma and space) '{" '[ ]'} FOR loop with that number '{" '[(#]'} trim those many characters from front of input string ` output result

Stack Exchange Network

Count repetitions of an array

54 Answers 54

R, 20 bytes

Bash + coreutils, 18

Haskell, 42 bytes

APL (Dyalog Unicode), 9 8 bytes^SBCS

C# (Visual C# Interactive Compiler), 40 bytes

C (clang) 175 117 95 bytes

05AB1E, 4 bytes

Python 3, 38 bytes

Jelly, 4 bytes

How?

J, 11 9 bytes

Original solution:

Java 8, 74 73 bytes

Uiua, 7 bytes

Ruby, 34 bytes

Perl 6, 15 bytes

Python 3, 63 bytes

APL (Dyalog Extended), 8 7 bytes^SBCS

Haskell, 41 bytes

Haskell, 40 bytes

Haskell, 41 bytes

PHP, 39 bytes

Haskell, 47 bytes

JavaScript (ES6), 40 bytes

Wolfram Language 34 bytes

Japt, 12 11 9 8 6 bytes

Pyth, 6 bytes

Explanation

Brachylog, 7 bytes

Gaia, 6 bytes

Arturo, 49 31 bytes

Pyt, 7 bytes

Thunno 2 `L`, 5 bytes

Explanation

Element, 40 bytes

Linked

Hot Network Questions

Count repetitions of an array

54 Answers 54

R, 20 bytes

Bash + coreutils, 18

Haskell, 42 bytes

APL (Dyalog Unicode), 9 8 bytesSBCS

C# (Visual C# Interactive Compiler), 40 bytes

C (clang) 175 117 95 bytes

05AB1E, 4 bytes

Python 3, 38 bytes

Jelly, 4 bytes

How?

J, 11 9 bytes

Original solution:

Java 8, 74 73 bytes

Uiua, 7 bytes

Ruby, 34 bytes

Perl 6, 15 bytes

Python 3, 63 bytes

APL (Dyalog Extended), 8 7 bytesSBCS

Haskell, 41 bytes

Haskell, 40 bytes

Haskell, 41 bytes

PHP, 39 bytes

Haskell, 47 bytes

JavaScript (ES6), 40 bytes

Wolfram Language 34 bytes

Japt, 12 11 9 8 6 bytes

Pyth, 6 bytes

Explanation

Brachylog, 7 bytes

Gaia, 6 bytes

Arturo, 49 31 bytes

Pyt, 7 bytes

Thunno 2 L, 5 bytes

Explanation

Element, 40 bytes

Linked

Related

Hot Network Questions

APL (Dyalog Unicode), 9 8 bytes^SBCS

APL (Dyalog Extended), 8 7 bytes^SBCS

Thunno 2 `L`, 5 bytes