Chunk + Enumerate a list of digits

Question

I have a list of decimal digits:

4, 4, 4, 7, 7, 9, 9, 9, 9, 2, 2, 2, 4, 4

The list of decimal digits are known as items. We can form "chunks" from these items by grouping together identical and adjacent numbers. I want to assign each chunk a unique number, starting from 1, and increasing by 1 in the order the chunks appear in the original list. So, the output for the given example would look like this:

1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5

Input format

A list of digits. (0-9) You may use your language built-ins to read this list however you want. Encoding: ASCII

Output format

A series of decimal numbers, separated by a delimiter. Your program must always use the same delimiter. The delimiter must be longer than 0 bits. Encoding: ASCII

Standard loopholes apply.

Answer 1

Python 3.8 (pre-release), 41 bytes

lambda l,n=0:[n:=n+(l!=(l:=x))for x in l]

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Praise the magic walrus := of assignment expressions.

---

Python 2, 42 bytes

n=0
for x in input():n+=x!=id;id=x;print n

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Answer 2

Python 2, 44 bytes

l=input()
n=0
for i in l:n+=i!=l;l=i;print n

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Answer 3

APL (dzaima/APL), 7 bytes^SBCS

Anonymous tacit prefix function. Prints space-separated.

+\1,2≠/

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2≠/ pair-wise inequality

1, prepend 1

+\ cumulative sum

Answer 4

Jelly, 6 5 bytes

ŒɠµJx

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Saved one byte thanks to UnrelatedString!

Inputs and outputs as array's (with opening/closing brackets)

How it works

ŒɠµJx - Main link, takes one argument:                       [7, 7, 5, 5, 5, 1]
Œɠ    - Get the lengths of consecutive elements:             [2, 3, 1]
  µ   - Call these lengths A
   J  - range(length(A))                                     [1, 2, 3]
    x - Repeat each element by the corresponding value in A: [1, 1, 2, 2, 2, 3]

Answer 5

Jelly, 4 bytes

ŒgƤẈ

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How?

ŒgƤẈ - Link: list of integers  e.g. [7,7,2,7,7]
  Ƥ  - for prefixes:     [7]   [7,7]   [7,7,2]      [7,7,2,7]        [7,7,2,7,7]
Œg   -   group runs      [[7]] [[7,7]] [[7,7],[2]]  [[7,7],[2],[7]]  [[7,7],[2],[7,7]]
   Ẉ - length of each    [1,   1,      2,           3,               3]

Answer 6

Octave / MATLAB, 25 bytes

@(x)cumsum([1 ~~diff(x)])

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Answer 7

Wolfram Language (Mathematica), 29 bytes

Join@@(i=1;0#+i++&/@Split@#)&

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Answer 8

05AB1E, 5 bytes

¥Ā.¥>

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Answer 9

Haskell, 40 bytes

f(a:t)=1:map(+sum[1|a/=t!!0])(f t)
f e=e

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Answer 10

Perl 6, 21 bytes

{+<<[\+] $,|$_ Zne$_}

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Anonymous code block that takes a list and returns a list. This works by comparing whether each pair of adjacent elements are not equal, than taking the cumulative sum of the list.

Answer 11

05AB1E, 4 bytes

γdƶ˜

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γ       # group adjacent equal digits together
 d      # replace all digits with 1
  ƶ     # multiply each group by its 1-based index
   ˜    # flatten

Answer 12

R, 33 bytes

function(x)cumsum(c(1,!!diff(x)))

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Uses the same cumulative sum method as Luis Mendo and others.

Answer 13

MATL, 8 bytes

Y'wn:wY"

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Explanation:

    Y'      % run-length encoding
    w       % swap elements in stack
    n       % number of elements in array / size along each dimension
    :       % range; vector of equally spaced values
    w       % swap elements in stack
    Y"      % replicate elements of array
            % (implicit) convert to string and display

Answer 14

JavaScript (ES6), 30 bytes

Takes input as an array of integers.

a=>a.map(p=n=>i+=p!=(p=n),i=0)

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Commented

a =>                // a[] = input array
  a.map(p =         // initialize p to a non-numeric value
  n =>              // for each value n in a[]:
    i +=            //   increment i if:
      p != (p = n), //     p is not equal to n; and update p to n
    i = 0           //   start with i = 0 (chunk counter)
  )                 // end of map()

Answer 15

Oracle SQL, 121 bytes

select m from t match_recognize(order by i measures match_number()m all rows per match pattern(p+)define p as x=first(x))

Test in SQL*PLus.

SQL> with t(i,x) as (select rownum,value(v) from table(sys.odcinumberlist(4, 4, 4, 7, 7, 9, 9, 9, 9, 2, 2, 2, 4, 4))v)
  2  select m from t match_recognize(order by i measures match_number()m all rows per match pattern(p+)define p as x=first(x))
  3  /

         M
----------
         1
         1
         1
         2
         2
         3
         3
         3
         3
         4
         4
         4
         5
         5

14 rows selected.

Answer 16

Stax, 10 bytes

▓ª2ªmD?Ä╧╖

Run and debug it

The output uses space as a delimiter. The input follows the precise specifications using commas as separators, and now enclosing braces.

Answer 17

C (gcc), 62 61 bytes

This is one of the few entries I've done where a complete program is shorter than a function submission!

On the first pass, I don't care about the previous value, so I get to rely on the fact that argv is a pointer to somewhere and is extremely unlikely to be between [0..9]!

s;main(i,j){for(;~scanf("%d,",&i);j=i)printf("%d ",s+=j!=i);}

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Answer 18

Haskell, 46 43 bytes

scanl(+)1.map fromEnum.(zipWith(/=)=<<tail)

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Anonymous pointfree function that takes a list and returns a list

Answer 19

J, 12 bytes

1+/\@,2~:/\]

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Similar to Adám's APL answer

      2~:/\]  - pair-wise inequality
1    ,        - prepend 1
    @         - and   
+/\           - find the cumulative sum

Answer 20

Perl 5, 27 bytes

s/\d/$i+=$&!=$p;$p=$&;$i/ge

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The command line option -p makes perl read the input line from STDIN into the "default variable" $_. It then search-replaces all digits in $_ with the counter $i. And $i is increased for each digit which is different than the previous digit, which it also is at the first digit so the counter starts at 1. The previous digit is stored in $p.

Answer 21

Pyth, 13 11 bytes

s.e*]hkhbr8

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         r8  # Run-length encode (implicit argument is the input) (-> [[frequency, char], ...]
 .e          # Enumerated map (current element is b, index is k) over rQ8
   *]hk      # [ k+1 ] *
       hb    #           b[0]
s            # Reduce list on + ([a]+[b] = [a,b])

-2 bytes thanks to Mr. Xcoder

Answer 22

Scala, 114 bytes

s.split(", ").zipWithIndex.scan(s.head,0){(a,b)=>if(a._1==b._1)a else b._1->(a._2+1)}.tail.unzip._2.mkString(", ")

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Answer 23

Scala, 75 bytes

s=>s.scanLeft(("",0))((x,y)=>(y,x._2+(if(x._1!=y)1 else 0))).tail.map(_._2)

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If input and output must be comma separated String (and not List) then 102 bytes.

s=>s.split(", ").scanLeft(("",0))((x,y)=>(y,x._2+(if(x._1!=y)1 else 0))).tail.map(_._2).mkString(", ")

Answer 24

Jelly, 5 bytes

nƝÄŻ‘

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I initially aimed for a 4-byter (the same program but without the Ż) but then quickly realized that a 1 had to be prepended every time due to an oversight... Even though there is another 5-byter in Jelly, I'll actually keep this because it uses a different method.

For each pair of neighbouring items of the input list \$L\$, test if \$L_i\ne L_{i+1}, \forall 1\le i<|L|\$ and save these results in a list. Then take the cumulative sum of this list and increment them by 1 to match the chunk indexing system. TL;DR. Whenever we encounter different neighbouring items, we increment the chunk index by 1.

Answer 25

C (gcc), 62 bytes

f(_,l)int*_;{printf("%d ",l=--l?f(_,l)+(_[l]!=_[l-1]):1);_=l;}

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A function that takes the list and its length as arguments.

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C (gcc), 60 bytes

f(_,l)int*_;{_=printf("%*d",--l?f(_,l)+(_[l]!=_[l-1]):2,0);}

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Outputs in unary, delimited by 0s

Answer 26

x86-16, 12 bytes

Unassembled listing:

        NUM_LOOP: 
AC          LODSB                   ; load [SI] into AL, advance SI 
3A C2       CMP  AL, DL             ; is this char same as last? 
74 01       JZ   SAME               ; if so, same index 
43          INC  BX                 ; otherwise increment index 
        SAME: 
92          XCHG AX, DX             ; save current char for next compare 
8A C3       MOV  AL, BL             ; put index value in AL to store
AA          STOSB                   ; store AL into [DI], advance DI
E2 F4       LOOP NUM_LOOP           ; keep looping 

Input is byte array at SI, length in CX. Output is byte array at DI. Of course, this is if default I/O rules apply now.

Example: SI = 120H, DI = 130H

Answer 27

Ruby, 38 bytes

->a{i=0;a.map{|x,y|p i+=(a==a=x)?0:1}}

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On the first iteration, a is equal to the input list, so we always increment the chunk count. On subsequent iterations it's equal to the previous element.

Answer 28

PHP, 52 bytes

while(''<$d=$argv[++$x])echo$i+=$argv[$x-1]!=$d,' ';

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Input via command line, output to STDOUT.

Thx to @Night2 for the pesky '0' == 0 comparison bugfix!

Answer 29

Julia 1.0, 56 bytes

l->foldl(l,init=(0,0))do(p,i),c
println(i+=p!=c)
c,i
end

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Answer 30

R, 37 bytes

function(x,r=rle(x)$l)rep(seq(a=r),r)

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Not as golfy as this answer, but an alternative method using rep.