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!

Answer 1

Python 2, 23 bytes

lambda l:map(l.count,l)

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Answer 2

MATL, 3 bytes

&=s

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Explanation

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

Answer 3

JavaScript (ES6), 36 bytes

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

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Answer 4

APL(Dyalog Unicode), 6 bytes ^SBCS

+/∘.=⍨

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

Answer 5

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 }

Answer 6

Ruby, 27 bytes

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

Testing:

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

Answer 7

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.

Answer 8

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.

Answer 9

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

Answer 10

Python 3, 26 bytes

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

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-5 bytes (indirectly) thanks to xnor!

Answer 11

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

Answer 12

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]

Answer 13

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.

Answer 14

Racket, 42 bytes

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

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Answer 15

J, 6 bytes

1#.=/~

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K (oK), 9 bytes

{+/x=\:x}

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Answer 16

Haskell, 27 bytes

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

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Answer 17

05AB1E, 1 byte

¢

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

Answer 18

Perl 5 -pa, 22 bytes

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

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Answer 19

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.

Answer 20

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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Answer 21

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.

Answer 22

Julia 1.0, 24 21 bytes

[email protected](==(a),[a])

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

Answer 23

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)

Answer 24

Japt, 4 bytes

£è¥X

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maps input by returning number of occurrence in input

Answer 25

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

Answer 26

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

Nothing special

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

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Answer 27

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

Answer 28

C (clang), ^{84 \$\cdots\$ 81} 75 bytes

Saved 6 bytes thanks to AZTECCO!!!

b;i;j;f(*a,l){for(i=-1;++i<l;printf("%d ",b))for(b=j=l;j--;)b-=a[i]!=a[j];}

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Inputs a pointer to an array and its length (because array pointers in C don't carry any length info) and prints the occurrence counts.

Answer 29

Zsh, 23 bytes

xargs -I. grep -c . f<f

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Answer 30

Pyt, 2 bytes

Đɔ

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Đ      implicit input; duplicate
 ɔ     for each item in input, count occurrences in input; implicit print