Python 3, 50 bytes

We can simply omit the lambda name since we don't refer to it anywhere in the lambda definition.

lambda l:[l[:i+1].count(j)for i,j in enumerate(l)]

Example usage in the terminal:

>>>(lambda l:[l[:i+1].count(j)for i,j in enumerate(l)])([1,1,2,2,2,1,1,1,3,3])
[1, 2, 1, 2, 3, 3, 4, 5, 1, 2]
>>>(lambda l:[l[:i+1].count(j)for i,j in enumerate(l)])([3,7,5,4,9,2,3,2,6,6])
[1, 1, 1, 1, 1, 1, 2, 2, 1, 2]

I tested this on version 3.8.5 but this should work on earlier as well as newer versions of Python 3.

posted 11 months ago

CC BY-SA 4.0

CrSb0001‭

244 reputation 6 17 26 25

APL (Dyalog Unicode), 11 bytes (SBCS)

1 1∘⍉+\∘.=⍨

Try it online!

This was a fun APL exercise. Working on figuring out how to get it down to 7 bytes.

posted almost 5 years ago

CC BY-SA 4.0

hyperneutrino‭

331 reputation 0 11 33 9

Jelly, 4 bytes

¹Ƥċ"

Try it online!

   "    For each element of the input,
  ċ     how many times does it occur in
¹Ƥ "    the corresponding prefix of the input?

posted almost 5 years ago

CC BY-SA 4.0

5y ago

Unrelated String‭

221 reputation 0 6 22 2

BQN, 5 bytes

1+⊒

Try it here!

3 characters, but, as there's no SBCS codepage for BQN, it must be counted as UTF-8.

Two of the three characters are just adding one to the built-in that almost solves the challenge too.

posted almost 5 years ago

CC BY-SA 4.0

dzaima‭

779 reputation 6 10 92 14

APL (Dyalog Unicode), 11 7 bytes (SBCS)

Razetime and rak1507 came up with 7 byte equivalents of my original dfn (this one's rak1507's). See their solutions below.

+/¨⊢=,\

Try it online!

+/¨⊢=,\
     ,\ ⍝ Prefixes of the list
    =   ⍝ Compare every prefix
   ⊢   ⍝ to the corresponding element in the original list
+/     ⍝ Sum each to get a count of how many elements in each prefix match

posted almost 5 years ago

CC BY-SA 4.0

5y ago

user‭

1591 reputation 9 59 179 89

Japt, 14 11 8 bytes

£¯YÄ è¥X

Try it

-3 bytes thanks to @Shaggy

posted almost 5 years ago

CC BY-SA 4.0

4y ago

Quintec‭

2406 reputation 17 46 253 49

Husk, 4 bytes

Sz#ḣ

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Explanation

Sz#ḣ
Sz   zip the input
   ḣ with its prefixes
  #  using the count function

posted almost 5 years ago

CC BY-SA 4.0

Razetime‭

2960 reputation 42 65 331 83

Scala 3, 50 44 bytes

? => ?.indices zip?map(?take _+1 count _.==)

Try it in Scastie!

This is an annoying, stupid approach. The more elegant one using inits was much longer (x=>x.inits.toSeq.reverse.tail zip x map(_ count _.==)).

? =>   //The input
?.indices //The indices
  zip ?   //Zip them with the elements of the input
  .map(   //For every index i and the corresponding element x,
    ? take _+1   //Take the first i+1 elements of the input
    count        //Count how many of them
    _.==)        //Are equal to the element x

posted almost 5 years ago

CC BY-SA 4.0

4y ago

user‭

1591 reputation 9 59 179 89

JavaScript (Node.js), 34 29 26 bytes

-5 bytes thanks to Shaggy's suggestion on a similar challenge

-3 bytes again thanks to Shaggy!

a=>a.map(d=v=>d[v]=-~d[v])

Try it online!

posted over 4 years ago

CC BY-SA 4.0

3y ago

Moshi‭

1762 reputation 17 58 208 80

Jelly, 7 bytes

=þ`ÄŒDḢ

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posted almost 5 years ago

CC BY-SA 4.0

hyperneutrino‭

331 reputation 0 11 33 9

Python 3.8 (pre-release), 69 bytes

def f(x,y={},z=[]):
 for i in x:y[i]=y.get(i,0)+1;z+=[y[i]]
 return z

Try it online!

Bonus: theoretical answer if python allowed named assignment with subscript (54 bytes)

lambda x,y={}:[y[i]:=y[i]+1if i in y else 1for i in x]

posted over 3 years ago

CC BY-SA 4.0

3y ago

dino‭

151 reputation 0 10 15 31

Ruby, 36 bytes

->a{i=-1;a.map{a[0..i+=1].count _1}}

Try it online!

The code in the TIO link is 2 bytes longer because _n block parameter names are not supported on TIO's Ruby instance yet. It will work the same, however. The code is:

->a{i=-1;a.map{a[0..i+=1].count a[i]}}

posted over 4 years ago

CC BY-NC 4.0

snail_‭

691 reputation 4 28 74 18

Vyxal, 4 bytes

KƛtO

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Explained

KƛtO
Kƛ     # For each prefix of the input
  tO   #   How many times does the tail occur in the prefix?

posted over 3 years ago

CC BY-SA 4.0

lyxal‭

616 reputation 2 25 65 21

Python 3, 74 bytes

def f(a):
 d={x:0 for x in a};r=[]
 for x in a:d[x]+=1;r+=[d[x]]
 return r

Try it online!

posted over 3 years ago

CC BY-SA 4.0

celtschk‭

1514 reputation 28 51 191 74

J, 9 bytes

1#.]=&><\

This is the 7 byte APL solve but makes use of #. in place of +/"1. I came up with 1#.]=[\ first, but bubbler pointed out it breaks when non zeros are present.

Attempt it online!

posted over 3 years ago

CC BY-SA 4.0

south‭

371 reputation 0 44 37 22

Python 3, 70 bytes

def f(a):
 d={};r=[]
 for x in a:d[x]=d.get(x,0)+1;r+=[d[x]]
 return r

Try it online!

posted over 3 years ago

CC BY-SA 4.0

__blackjack__‭

81 reputation 0 5 8 4

Factor, 122 bytes

USING: kernel sequences sequences.windowed ;
IN: c
: c ( s -- s ) dup length [ dup last [ = ] curry count ] rolling-map ;

posted over 3 years ago

CC BY-SA 4.0

gifti‭

41 reputation 0 5 4 5

Ruby, 31 bytes

->a{a.map{$*[_1]=1.+$*[_1]||0}}

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$* is a global variable, so calling this lambda multiple times (in a single process) would give wrong result. A 32 bytes version that does not rely on a global state:

->a,*c{a.map{c[_1]=1.+c[_1]||0}}

If array consists of positive integers from a known range (lets say 0...1e3, then 30 bytes version is:

->a{b=[0]*1e3;a.map{b[_1]+=1}}

posted about 4 years ago

CC BY-SA 4.0

3y ago

radarek‭

239 reputation 0 31 23 42

Haskell + hgl, 10 bytes

mpn$ce~<gj

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Explanation

• mpn map across all non-empty prefixes of the input ...
• ce count the number of elements in each prefix equal to ...
• gj the last element of that prefix.

Reflection

I'm happy with this. It doesn't take any extra steps, and the glue is cheap.

posted over 2 years ago

CC BY-SA 4.0

WheatWizard‭

525 reputation 18 6 50 39

J, 24 char

([: +/ [: (* +/\ )"1 ~. ="0 1 ])

Sample Runs

   ([:+/[:(*+/\)"1~.="0 1]) 1 1 2 2 2 1 1 1 3 3
1 2 1 2 3 3 4 5 1 2
   
   ([:+/[:(*+/\)"1~.="0 1]) 3 7 5 4 9 2 3 2 6 6
1 1 1 1 1 1 2 2 1 2

posted over 3 years ago

CC0

torres‭

131 reputation 0 17 13 28