Replace 0s In a String With Their Consecutive Counts

Question

Challenge

Given a string of any length which contains only digits from 0 to 9, replace each consecutive run of the digit 0 with its length.

Test Cases

1. 1234500362000440 → 1234523623441
2. 123450036200044 → 123452362344
3. 000000000000 → 12
4. 0123456789 → 1123456789
5. 1234567890 → 1234567891
6. 123456789 → 123456789
7. 010203004050 → 11121324151

Note

The shortest answer in bytes wins as per code-golf rules

• Noodle9: What is a run of \10ドル\$ 0s replaced with? 11?
• Rhaixer: Not OP, but I think it will be replaced with 10 as no modifications to the output were also needed, as most answers suggest

Answer 1

Retina 0.8.2, 6 bytes

0+
$.&

Try it online! Link includes test cases. Explanation: Unary to decimal conversion.

• Luis Mendo: That & is neat! What does it do?
• Neil: @LuisMendo It refers to the entire match, like 0ドル does in Perl.
• Neil: @LuisMendo Actually my original source was wrong; PCRE, PHP and Java use 0ドル but Perl actually uses $&, as does JavaScript. (Retina actually supports both.)

Answer 2

JavaScript, 31 bytes

s=>s.replace(/0+/g,x=>x.length)

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Answer 3

Pip, 7 bytes

-1 byte by jezza_99 / DLosc

aR+X0#_

Wow pip is really good at regex

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aR+X0#_
aR+X0 Replace all occurrences of (regexified zero with a +)
 #_ with it's length

• jezza_99: @DLosc might have another method, but I got this 7-byter

Answer 4

Japt, 7 bytes

r+iT Èl

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r+iT Èl :Implicit input of string
r :Replace
 +i : "+" prepended with
 T : 0, giving the RegEx /0+/g
 È : Pass each match through a function
 l : Length

Answer 5

QuadR, 5 bytes

0+
⍵L

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0+ Replace any runs of zeros

⍵L with the match Length

Answer 6

Vyxal s, 7 bytes

ĠṠƛ0cßL

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Explained

ĠṠƛ0cßL
Ġ # Group on consecutive items
 Ṡ # join each into a single string
 ƛ # To each group:
 0c # if it contains 0:
 ßL # push the length of the string
 # the s flag joins into a single string

An alternate 8 byter that uses regex match replacement

‛0+$(Løṙ

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Explained

‛0+$(Løṙ
‛0+ # The string "0+"
 $ # pushed under the input
 (L # a function object that returns the length of its argument
 øṙ # replace instances of runs of 0 with their length

Answer 7

><>, 55 bytes

i:'0'= ?v>:0(?;o
='0':i<1<\n$v?
 \+1~ /

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Answer 8

Python 3, (削除) 91 (削除ここまで) 90 bytes

lambda x:''.join(l>"0"and l or f'{len([*g])}'for l,g in groupby(x))
from itertools import*

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-1 byte thanks to The Thonnu

I use the itertools module's groupby function to group consecutively.

• The Thonnu: You can replace the != with > to save a byte: Try it online!
• U13-Forward: @TheThonnu Thanks!
• Shaggy: As this is a recursive lambda, you'll need to include the f= in your byte count.
• U13-Forward: @Shaggy This isn't a recursive lambda, I am not calling the function within it.

Answer 9

Perl 5 -p, 16 bytes

s/0+/length$&/eg

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Answer 10

05AB1E, 7 bytes

0ÃηRāR:

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• -6 thanks to Kevin Cruijssen

Explanation

0ÃηRāR: # Implicit input
0Ã # List intersection with [0]
 ηR # Reversed prefixes
 āR # Reversed length range
 : # Infinite replacement

Previous 13 byte answer

.γ}εD0.åig}}J # Implicit input
.γ} # Group by consecutive items
 ε } # For each group:
 D # Duplicate the group
 0.åi } # If 0 is in the group:
 g # Push its length
 J # Join everything into a single string

• Kevin Cruijssen: .γ} can be γ; 0.å can be _ and }} can be ] for 8 bytes: try it online. Although 0ÃηRāR: is 1 byte shorter: try it online.
• The Thonnu: @KevinCruijssen thanks

Answer 11

R, (削除) 64 (削除ここまで) (削除) 58 (削除ここまで) 57 bytes

Edit: -1 byte thanks to @Dominic van Essen .

\(x){regmatches(x,t)=Map(attr,t<-gregexpr("0+",x),"m")
x}

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R has some weird string manipulation functions...

• Dominic van Essen: Can't you swap the <- for simply = in the 2-liner to save a byte?
• pajonk: @DominicvanEssen, well, to my surprise, I can. Thanks! I thought that ...<- functions need the actual <-, but after a while of thought [<- works well with =, so why not other functions like that?

Answer 12

Raku, 18 bytes

{S:g[0+]=$/.chars}

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• Jo King: still annoyed this can't be .=chars

Answer 13

Bash - (削除) 181 (削除ここまで) 158 chars

Edit 1: I used this trick for the for loop, remove whitespace, condence an if statement

n=0;for ((i=0;i<${#1};i++));{ c="${1:$i:1}";if [ $c = '0' ]; then ((n++)); else [ $n -ne 0 ] && { echo -n $n;n=0;};echo -n $c;fi;};[ $n -ne 0 ] && echo -n $n

Explanation

# a counter of how far into a sequence of zeros we are
# this is zero if we're not in a 'zero-sequence'
n=0
# iterate over all the characters
# 1ドル is the first function argument (0ドル would be the script name)
for (( i=0; i<${#1}; i++ )); do
 # get the current char
 c="${1:$i:1}"
 # if it is a zero, then increment the counter
 if [ $c = '0' ]; then
 ((n++));
 # if it is NOT a zero
 else
 # if a zero sequence is over (given that the counter is
 # not equal to zero and the current char isn't zero)
 if [ $n -ne 0 ]; then
 # print the number of zeros in the sequence,
 # and reset the counter to zero
 echo -n $n;
 n=0;
 fi;
 echo -n $c;
 fi
done
# check if there was a zero sequence terminating the string,
# as we wouldn't otherwise check as there wouldn't be a non-zero
# char initiating the check
if [ $n -ne 0 ]; then echo -n $n; fi

I referred to this SO answer for how to iterate over a string in Bash and this AU answer for a short way to increment a variable

Answer 14

Vim, 28 bytes

:s/0\+/\=len(submatch(0))/g Enter

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We cannot use 0* because it matches the empty string between digits.

• DLosc: Here's a TIO link for you. The % isn't necessary because the input "contains only digits from 0 to 9" and therefore contains no newlines. However, if your language is Vim and not V, I think you have to count the Enter keypress after the replace command (otherwise, the replacement doesn't happen), so that makes it 28 bytes. If you want to call it V instead of Vim, you can skip the newline and claim 27 bytes.
• D. Ben Knoble: Thanks @DLosc, edited.

Answer 15

Retina 0.8.2, 8 bytes

(0+)
$.1

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How it works

Match runs of zeros (0+), capture each match in a group (( )), replace it with the length of the most recent capture ($.1).

• Neil: Oh, I see you had already posted your answer while I was writing mine. Sorry to outgolf you like that.
• Luis Mendo: @Neil No problem at all, and great answer!

Answer 16

><>, 28 bytes

0i:"0"=?v{:?n}ol?!
~1+00. >

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Answer 17

Ruby, 16 bytes

gsub /0+/,&:size

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Answer 18

Python, (削除) 62 (削除ここまで) 57 bytes

lambda n:re.sub("0+",lambda s:str(len(s[0])),n)
import re

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-5 bytes thanks to @xnor

Regex solution, port of @mathcat's Pip answer

• xnor: I think you can do s[0] in place of s.group()

Answer 19

Brachylog, 11 bytes

ḅ{ị0&lṫ|}mc

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Explanation

I was hoping for some automatic number -> string conversion, but no dice.

ḅ{ị0&lṫ|}mc
ḅ Break the input string into blocks of identical characters
 { }m Map this predicate to each block:
 ị Convert to integer
 0 Assert that the result is zero
 &l If so, get the length of the block
 ṫ and convert to string
 | If that failed (because the number wasn't zero), return the block unchanged
 c Concatenate the results together into a single string

Answer 20

Excel VBA, (削除) 100 (削除ここまで) 77 bytes

^{Saved 23 bytes thanks to a translation by Taylor Raine}

For i=2^15-1To 1Step-1:[B1]=i:[A1]=[Substitute(A1,Rept(0,B1),B1)]:Next:[B1]="

Input is in the cell A1 of the active sheet. Output is in the same cell. The code is run from the immediate window in VBA. The only clever bit is that Excel only allows 32,767 characters in a cell and counting down from there is less bytes than counting down from the length of the input.

• Taylor Raine: You could make this into an immediate window function something like For i=2^15-1To 1Step-1:[B1]=i:[A1]=[Substitute(A1,Rept(0,B1),B1)]:Next:[B1]=" for a pretty significant drop in bytes. (Also reminder that using ^ with no leading space makes this a 32-bit only solution as ^ is the Longlong type literal in 64-bit installs of excel)

Answer 21

K (ngn/k), 25 bytes

{,/$(.'x)|#'x:(&~=':x)_x}

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• x:(&~=':x)_x split the input where it changes, and reassign to x
• (.'x)|#'x take the maximum of the chunk of input (converted to its corresponding integer) and its count
• ,/$ convert to strings and flatten (and implicitly return)

Answer 22

C (gcc), 70 bytes

n;f(char*a){*a-48?n=!printf(n?"%d":"%c",n?:*a++),*a&&f(a):f(a+!!++n);}

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C (gcc), 62 bytes by c--

n;f(int*a){*a-48?n=!printf(n?"%d":a++,n),*a&&f(a):f(a+!!++n);}

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Answer 23

Julia 1.0, 27 bytes

!x=replace(x,r"0+"=>length)

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Answer 24

Jelly, 8 bytes

Œgċ"0ȯƊ€

A full program that accepts a string of digit characters and prints the result.

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How?

Œgċ"0ȯƊ€ - Main Link: list of characters, S
Œg - group runs
 € - for each group:
 Ɗ - last three links as a monad - f(group):
 "0 - literal '0' character
 ċ - count occurrences
 ȯ - logical OR (group)
 - implicit, smashing print

Answer 25

Husk, 8 bytes

ṁ?IosLig

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Answer 26

Factor, 42 bytes

[ R/ 0+/ [ length >dec ] re-replace-with ]

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Answer 27

Zsh --extendedglob, 24 bytes

<<<${1//(#m)0##/$#MATCH}

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Answer 28

Java, 88 bytes

s->java.util.regex.Pattern.compile("0+").matcher(s).replaceAll(r->r.group().length()+"")

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Answer 29

><>, (削除) 23 (削除ここまで) 18 bytes

\&l:?n&o]
\i:'0'-?

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Answer 30

Java 19, 156 (削除) 161 (削除ここまで) (削除) 162 (削除ここまで) bytes

interface A{static void main(String[] a){var i=0;var s="";for(var c:a[0].toCharArray())if(c==48)i++;else{s+=i>0?i+""+c:c;i=0;}System.out.print(i>0?s+i:s);}}

Without Java's golfing tax, 113 bytes

var i=0;var s="";for(var c:a[0].toCharArray())if(c==48)i++;else{s+=i>0?i+""+c:c;i=0;}System.out.print(i>0?s+i:s);

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Edit: replaced '0' with 48 as suggested in the comments. Thanks!

Edit: String concatentation is shorter than doing var o=System.out;

• rydwolf: Welcome to Code Golf, and nice answer!