Replace all items with their counts

Question

Task:

Given an array of numbers as input (you can choose what subset, such as integers or natural numbers), replace all items with the number of times they appear within the array. As an example, [1, 2, 2, 1, 4, 8, 1] would become [3, 2, 2, 3, 1, 1, 3].

You can take input/produce output as arrays, lists, tuples, or any other reasonable representation of some numbers.

Test cases:

[1]                     [1]
[1, 2]                  [1, 1]
[1, 1]                  [2, 2]
[1, 4, 4]               [1, 2, 2]
[4, 4, 2]               [2, 2, 1]
[4, 4, 4, 4]            [4, 4, 4, 4]
[10, 20, 10, 20]        [2, 2, 2, 2]
[1, 2, 2, 4, 4, 4, 4]   [1, 2, 2, 4, 4, 4, 4]
[1, 2, 2, 1, 4, 8, 1]   [3, 2, 2, 3, 1, 1, 3]

Other:

This is code-golf, shortest answer (in bytes) per language wins!

I feel like this is a dupe of something but that's probably just because it's a convenient sub-problem of a bunch of other things — Unrelated String, Commented Apr 21, 2021 at 23:37
A siller question on that same note, could the input be restricted to prime numbers? I don't see it being relevant to anything as it is, but if this exact question were posted in the dark days of Ð€, Jelly would still have had a 2-byter in Pọ :P — Unrelated String, Commented Jun 3, 2021 at 7:20

xnor · Accepted Answer · 2021-04-21 23:48:56Z

16

Python 2, 23 bytes

lambda l:map(l.count,l)

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answered Apr 21, 2021 at 23:48

xnor

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\$\begingroup\$ ...I forgot map was a thing, thanks for helping me save 5 bytes on my Python 3 answer! \$\endgroup\$
– caird coinheringaahin g ♦
Commented Apr 21, 2021 at 23:50

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Luis Mendo · Accepted Answer · 2021-04-22 00:00:07Z

9

MATL, 3 bytes

&=s

Try it online! Or verify all test cases.

Explanation

     % Implicit input: numeric row vector
&=   % Matrix of all pairwise equality comparisons
s    % Sum of each column
     % Implicit display

answered Apr 22, 2021 at 0:00

Luis Mendo

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Arnauld · Accepted Answer · 2021-04-21 23:45:22Z

8

JavaScript (ES6), 36 bytes

a=>a.map(x=>a.map(y=>t+=x==y,t=0)|t)

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answered Apr 21, 2021 at 23:45

Arnauld

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\$\begingroup\$ The one answer that doesn't use a counting builtin... \$\endgroup\$
– emanresu A
Commented Apr 25, 2021 at 9:38

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hyperneutrino · Accepted Answer · 2021-04-22 00:02:02Z

7

APL(Dyalog Unicode), 6 bytes ^SBCS

+/∘.=⍨

Try it on APLgolf!

+/∘.=⍨  dfn submission
  ∘.=   product table using equality
     ⍨  applied to the input on the left and the right
+/      reduce by addition / sum

answered Apr 22, 2021 at 0:02

hyperneutrino♦

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3

\$\begingroup\$ Another take +/¨⊢=⊂ \$\endgroup\$
– Andrew Ogden
Commented Apr 22, 2021 at 4:16

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chunes · Accepted Answer · 2021-04-22 00:40:00Z

Factor, 28 bytes

[ dup histogram substitute ]

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It's a bit shorter than the version for squares:

[ dup [ '[ _ = ] count ] with map ]

Explanation:

It's a quotation (anonymous function) that takes a sequence from the data stack as input and leaves a sequence on the data stack as output. Assuming { 1 4 4 } is on the data stack when this quotation is called...

dup Duplicate an object.

Stack: { 1 4 4 } { 1 4 4 }
histogram Create a histogram from a sequence.

Stack: { 1 4 4 } H{ { 1 1 } { 4 2 } }
substitute Take a sequence and an associative array and substitute elements in the sequence that have keys in the assoc with their values.

Stack: { 1 2 2 }

Sony Santos · Accepted Answer · 2021-04-22 02:28:21Z

5

Ruby, 27 bytes

f=->*a{a.map{|e|a.count e}}

Testing:

p f[4,3,4]  #=> [2, 1, 2]

edited Apr 22, 2021 at 2:28

answered Apr 22, 2021 at 0:51

Sony Santos

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1

\$\begingroup\$ Welcome to Code Golf, and nice first answer! \$\endgroup\$
– caird coinheringaahin g ♦
Commented Apr 22, 2021 at 0:56
\$\begingroup\$ Similar to the other answer, you need to wrap this in a function or lambda. \$\endgroup\$
– Razetime
Commented Apr 22, 2021 at 1:38

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Sean · Accepted Answer · 2021-04-22 02:57:00Z

5

Raku, 10 bytes

{.Bag{$_}}

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.Bag generates a Bag (a set with multiplicity) from the input argument $_. Then {$_} slices into that Bag with the original list, producing a list of the multiplicities of the elements of that list, in order.

answered Apr 22, 2021 at 2:57

Sean

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Galen Ivanov · Accepted Answer · 2021-04-22 07:43:24Z

5

Racket, 42 bytes

(λ(b)(map(λ(y)(count(λ(x)(= x y))b))b))

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answered Apr 22, 2021 at 7:43

Galen Ivanov

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Unrelated String · Accepted Answer · 2021-04-22 21:48:10Z

Brachylog, 8 bytes

∋;?{∈ᵈ}ᶜ

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Generates the output through the output variable.

∋;          Pair some element of the input with
  ?         the input.
   {  }ᶜ    In how many ways is
    ∈ᵈ      the element an element of the input?

Using ọ comes out one byte longer:

Brachylog, 9 bytes

⟨ọ⟨∋h⟩∋⟩t

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Also generates the output through the output variable.

      ∋      Choose an element of the input.
  ⟨ h⟩       It is the first element of
   ∋         an element of
⟨ọ     ⟩     the list of pairs [element of input, how many times it occurs in input]
        t    the last element of which is the output.

caird coinheringaahin g · Accepted Answer · 2021-04-21 23:39:29Z

4

Jelly, 2 bytes

ċⱮ

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How it works

ċⱮ - Main link. Takes a list L on the left
 Ɱ - For each element in L
ċ  -   Count the times it appears in L

answered Apr 21, 2021 at 23:39

caird coinheringaahin g♦

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caird coinheringaahin g · Accepted Answer · 2021-04-21 23:46:25Z

4

Python 3, 26 bytes

lambda l:[*map(l.count,l)]

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-5 bytes (indirectly) thanks to xnor!

answered Apr 21, 2021 at 23:46

caird coinheringaahin g♦

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hakr14 · Accepted Answer · 2021-04-21 23:46:49Z

4

Pyth, 3 bytes

/LQ

Test suite

Explanation:

/LQ  | Full program
/LQQ | with implicit variables
-----+-------------------------------------
 L Q | replace each element d in input with
/ Q  | the count of d in input

answered Apr 21, 2021 at 23:46

hakr14

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lyxal · Accepted Answer · 2021-04-21 23:51:08Z

4

Vyxal, 2 bytes

vO

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Wow y'all using unicode in your golfing languages while I'm chilling in the ASCII zone.

Explained

vO  # vectorise count over the input
    # essentially, [input.count(n) for n in input]

edited Apr 21, 2021 at 23:51

answered Apr 21, 2021 at 23:42

lyxal

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tsh · Accepted Answer · 2021-04-22 03:52:56Z

4

JavaScript (ES2021), 35 bytes (not quite reasonable I/O)

a=>a.map(n=>a[++(a[n]||=[0])[0],n])

Accept an array of ~~1 element array of~~ negative numbers. Return an array of 1 element array...

f([-1, -2, -2, -1, -4, -8, -1]) // [[3], [2], [2], [3], [1], [1], [3]]
f([[-1], [-2], [-2], [-1], [-4], [-8], [-1]]) // [[3], [2], [2], [3], [1], [1], [3]]

~~Just consider input / output as column vectors...~~

JavaScript (ES2021), 42 bytes

Add an extra .flat() to make it reasonable would cost +7 bytes (42 bytes in total)

a=>a.map(n=>a[++(a[n]||=[0])[0],n]).flat()

That's too long.

edited Apr 22, 2021 at 3:52

answered Apr 22, 2021 at 2:21

tsh

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\$\begingroup\$ Is ||= a ES2021 feature? I don't remember seeing that before. \$\endgroup\$
– rydwolf
Commented Apr 22, 2021 at 2:29
\$\begingroup\$ @RedwolfPrograms Yes. ||=, &&= and ??= are ES2021 features. Logical ||= and &&= are also supported by ActionScript (ES4) long long ago. \$\endgroup\$
– tsh
Commented Apr 22, 2021 at 2:34
1

\$\begingroup\$ Can't you use flatMap? \$\endgroup\$
– rydwolf
Commented Apr 22, 2021 at 3:43
1

\$\begingroup\$ @RedwolfPrograms flatMap will flat as soon as the array returns, not flat after map. <del>maybe we need a proposal about Array#mapFlat API to JavaScript.</del> \$\endgroup\$
– tsh
Commented Apr 22, 2021 at 3:46
\$\begingroup\$ a=>a.map(x=>a.filter(y=>x==y).length) (37 bytes) would be a better substitute for the second solution. \$\endgroup\$
– kamoroso94
Commented Jan 18, 2023 at 4:45

| Show 1 more comment

Galen Ivanov · Accepted Answer · 2021-04-22 07:54:54Z

4

J, 6 bytes

1#.=/~

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K (oK), 9 bytes

{+/x=\:x}

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answered Apr 22, 2021 at 7:54

Galen Ivanov

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1

\$\begingroup\$ An alternative 9-byte answer is possible in ngn/k: {(#'=x)x} \$\endgroup\$
– coltim
Commented Apr 22, 2021 at 16:26
\$\begingroup\$ @coltim Thank you! So you make a dictionary with numbers/frequencies for keys/values and request the values according to the input keys? \$\endgroup\$
– Galen Ivanov
Commented Apr 22, 2021 at 19:17
1

\$\begingroup\$ Yep, exactly! It's possible to implement it in oK as well, but you would need to use {(#:'=x)x}. \$\endgroup\$
– coltim
Commented Apr 22, 2021 at 20:05

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Delfad0r · Accepted Answer · 2021-04-21 23:44:03Z

3

Haskell, 27 bytes

f a=[sum[1|y<-a,y==x]|x<-a]

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answered Apr 21, 2021 at 23:44

Delfad0r

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Makonede · Accepted Answer · 2021-04-22 00:34:00Z

3

05AB1E, 1 byte

¢

Try it online! Beats all other answers.

¢  # full program
¢  # number of times...
   # (implicit) each element in...
   # implicit input...
¢  # appears in...
   # implicit input
   # implicit output

answered Apr 22, 2021 at 0:34

Makonede

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Xcali · Accepted Answer · 2021-04-22 16:38:14Z

3

Perl 5 `-pa`, 22 bytes

s/\S+/grep$_==$&,@F/ge

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answered Apr 22, 2021 at 16:38

Xcali

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Axuary · Accepted Answer · 2021-04-22 22:54:27Z

3

Excel 17, bytes

=COUNTIF(A1#,A1#)

Assuming A1 = { .... } then this works. It's a longer, less flexible formula if the data in entered in individual cells.

answered Apr 22, 2021 at 22:54

Axuary

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Unmitigated · Accepted Answer · 2021-04-23 21:41:15Z

3

Java, 58 bytes

l->l.stream().map(x->java.util.Collections.frequency(l,x))

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Java, 56 bytes

l->l.stream().map(x->l.stream().filter(y->y==x).count())

This only works for integers in the Integer Cache (-128 to 127, by default).

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edited Apr 23, 2021 at 21:41

answered Apr 23, 2021 at 21:24

Unmitigated

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pxeger · Accepted Answer · 2021-04-24 14:48:09Z

3

Zsh, 23 bytes

xargs -I. grep -c . f<f

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answered Apr 24, 2021 at 14:48

pxeger

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noodle person · Accepted Answer · 2023-01-27 21:07:47Z

3

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

ec

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ec : implicit input array
e  : map the input to the following:
 c :   count the occurrences of the number in the input
   : implicit output

Thought I'd try an easy and short problem for my first Thunno answer.

answered Jan 27, 2023 at 21:07

noodle person

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1

\$\begingroup\$ Nice! I didn't see your answer before answering, but I came up with another 2 character solution: Dc. \$\endgroup\$
– The Thonnu
Commented Jan 28, 2023 at 9:39
\$\begingroup\$ @TheThonnu Can you explain how your version works? My understanding is: it duplicates the array. Then c automatically maps over both arrays at once. Is that correct? \$\endgroup\$
– noodle person
Commented Jan 28, 2023 at 14:01
\$\begingroup\$ Yeah, c will vectorise over a list. Your one isn't supposed to work, since e should pop the list, but somehow it does... \$\endgroup\$
– The Thonnu
Commented Jan 28, 2023 at 17:35

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bigyihsuan · Accepted Answer · 2023-01-28 06:31:57Z

3

Go, 103 bytes

func f(l[]int)[]int{m:=make(map[int]int)
for _,e:=range l{m[e]++}
for i,e:=range l{l[i]=m[e]}
return l}

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answered Jan 28, 2023 at 6:31

bigyihsuan

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DLosc · Accepted Answer · 2023-09-14 18:38:02Z

3

Pip `-p`, 5 bytes

_NgMg

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Explanation

    g  ; List of command-line args
   M   ; Map to each:
_N     ;  Number of occurrences of that argument in...
  g    ;  ... the list of command-line args

answered Sep 14, 2023 at 18:38

DLosc

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Ashlin Harris · Accepted Answer · 2024-07-12 20:58:38Z

3

Julia 1.0, 19 bytes

[email protected](==(a),[a])

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Saved 2 bytes by substituting sum for count.

-3 bytes thanks to MarcMush: replace Ref(a) with [a]

edited Jul 12 at 20:58

answered Sep 14, 2023 at 17:34

Ashlin Harris

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Giuseppe · Accepted Answer · 2021-04-22 00:52:15Z

2

R, 26 bytes

ave(a,a<-scan(),FUN=table)

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ave takes a vector x, an arbitrary number of grouping variables ..., and a function FUN, and replaces each element of x with the result of applying FUN to the group containing that element.

I've also found a number of 26 byte variants with FUN=sum; they differ only in the way they generate a vector of ones with length length(a).

ave(a^0,a<-scan(),FUN=sum)
ave(a,a<-scan(),FUN=sum)/a
ave(a|1,a<-scan(),FUN=sum)

answered Apr 22, 2021 at 0:52

Giuseppe

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2

\$\begingroup\$ Great, and a nice showcase of a previously-unused-by-me R function \$\endgroup\$
– Dominic van Essen
Commented Apr 22, 2021 at 7:50

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AZTECCO · Accepted Answer · 2021-04-22 09:26:04Z

2

Japt, 4 bytes

£è¥X

Try it

maps input by returning number of occurrence in input

answered Apr 22, 2021 at 9:26

AZTECCO

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Kaddath · Accepted Answer · 2021-04-22 12:37:00Z

2

PHP, 56 bytes

fn($a)=>array_map(fn($e)=>array_count_values($a)[$e],$a)

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As usual, those "array_" PHP prefixes are ruining the golfing :P

answered Apr 22, 2021 at 12:37

Kaddath

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LiefdeWen · Accepted Answer · 2021-04-22 14:09:22Z

2

C# (Visual C# Interactive Compiler), 32 bytes

Nothing special

a=>a.Select(x=>a.Count(y=>x==y))

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answered Apr 22, 2021 at 14:09

LiefdeWen

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LiefdeWen · Accepted Answer · 2021-04-22 14:45:30Z

2

Neim, 2 bytes

Right language, right time.

Ψ𝕠

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Explanation:

Ψ  # Apply next token to all in list
 𝕠 # Count element in list

edited Apr 22, 2021 at 14:45

answered Apr 22, 2021 at 14:14

LiefdeWen

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Replace all items with their counts

55 Answers 55

Python 2, 23 bytes

MATL, 3 bytes

Explanation

JavaScript (ES6), 36 bytes

APL(Dyalog Unicode), 6 bytes SBCS

Factor, 28 bytes

Explanation:

Ruby, 27 bytes

Raku, 10 bytes

Racket, 42 bytes

Brachylog, 8 bytes

Brachylog, 9 bytes

Jelly, 2 bytes

How it works

Python 3, 26 bytes

Pyth, 3 bytes

Vyxal, 2 bytes

Explained

JavaScript (ES2021), 35 bytes (not quite reasonable I/O)

JavaScript (ES2021), 42 bytes

J, 6 bytes

K (oK), 9 bytes

Haskell, 27 bytes

05AB1E, 1 byte

Perl 5 -pa, 22 bytes

Excel 17, bytes

Java, 58 bytes

Java, 56 bytes

Zsh, 23 bytes

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

Go, 103 bytes

Pip -p, 5 bytes

Explanation

Julia 1.0, 19 bytes

R, 26 bytes

Japt, 4 bytes

PHP, 56 bytes

C# (Visual C# Interactive Compiler), 32 bytes

Neim, 2 bytes

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