Shorten (or not) text using run length encoding
Input:
heeeello woooorld
Output:
1h4e2l1o 1w4o1r1l1d
perl -pe's|((.)\2*)|@x=split//,$1;@x.$x[0]|eg'
All hail @ardnew for coming up with the idea of using the tr///c operator to count the number of characters in the matched string instead of splitting:
perl -pe's|((.)\2*)|$1=~y///c.$2|eg'
Degolfed:
while(defined($_ = <>)) {
$_ =~ s{((.)\2*)} # match 1 or more consecutive identical non-newlines
{
($1 =~ y///c ) # count the number of characters in $1
. # and concatenate it
$2 # with the first matched character
}eg; # execute substitution, match "global"
print $_; # print the modified line
}
Usage:
$ perl -pe's|((.)\2*)|$1=~y///c.$2|eg' infile
or via STDIN
$ perl -pe's|((.)\2*)|$1=~y///c.$2|eg'
heeeello
prints
1h4e2l1o
p option.
\$\endgroup\$
s|((.)\2*)|$1=~y///c.$2|eg, which sums to 27 total chars (using the same character counting rules as @Gareth)
\$\endgroup\$
-p) by eliminating the outer parens: Try it online!
\$\endgroup\$
ûèB☼å°╤
Run and debug it st staxlang.xyz!
m|RFNppz
m For each line of input:
|R Run-length encode: "heeeello" -> [[104,1],[101,4],[108,2],[111,1]]
F For each pair:
N Uncons-left: [104,1] -> push [104]; push 1
ppz Pop and print. Pop and print. Push "".
Implicit print (always an empty string) with a newline
∩l↨me
|RmEp] Unpacked
|R Run-length encode: "heeeello" -> [[104,1],[101,4],[108,2],[111,1]]
m Map block over input:
E Explode array: [104,1] -> push 104, push 1
p Pop and print with no newline
] Make a one-element list: 104 -> [104] (which is "h")
Implicit print with newline
Run and debug it at staxlang.xyz!
Perhaps not legal. This program prints each pair on a line of its own. A bit sketchy.
If that output format is illegal, I give you 6 bytes:
╡δôZ→╬
|RFEp]p Unpacked
F For each item in array, execute block:
p Pop and print with no newline
No implicit print in for-each block, so no extra newlines
Run and debug it at staxlang.xyz!
The language's creator recursive points out that uncons-right (N) can shorten this to six bytes unpacked, as it handles the E and the ] on its own. Programs this short, however, often get no shorter when packed, and this is am example. Still six bytes: |RFNpp Edit: Had to update my main answer; this form is what I used.
|RFNpp can give the specified output in 6 bytes unpacked, but unfortunately, doesn't pack.
\$\endgroup\$
io.Source.stdin.getLines.map(s=>println{val(x,y,z)=s.tail./:(("",s.head,1)){case((a,b,c),d)=>if(b==d)(a,b,c+1)else(a+c+b,d,1)};x+z+y})
UPDATE:
Here the pure Lambda de-golfed (97 bytes):
s => {
val (x,y,z) = s.tail./:(("", s.head, 1)) {
case ((a, b, c), d) =>
if (b == d)
(a, b, c + 1)
else
(a + c + b, d, 1)
}
x+z+y
}
scala. before io. Also I think some surrrounding code can be removed from the count. Only include the lambda to get 110 bytes: Try it online!
\$\endgroup\$
scala. is obsolete, which saves 6 bytes. Regarding the lambda: I agree that it is generally ok to provide just a lambda. However, this challenge mentions explicitly that the solution should read from stdin and print to stdout, and other solutions (e.g., Python) do also include this in their byte count, so I think it is fair to keep it.
\$\endgroup\$
map instead of foreach, and /: instead of foldLeft.
\$\endgroup\$
map for side-effects, but, hey, this is golfing, so it's a great tip :) Also I did not know about the /: alias, thanks!
\$\endgroup\$
D,g,@:,BGd€q$€bLzBFJ
D,f,@,10C$t€g
]getchar
y:''
Wx,`y,+x,`x,]getchar
$f>y
This could be half the size if we could take input from the command line
The first line (the function g) actually performs RLE. The rest of it just reads the input char by char, passes it to a function f, which splits on newlines, then runs g on each line.
g simply groups equal adjacent characters with BG, then gets the lengths (€bL) and deduplicates (€q). We then zip these together, flatten, Join and print.
,(](":@#,{.);.1~1,2~:/\])1!:1[1
Usage:
,(](":@#,{.);.1~1,2~:/\])1!:1[1
heeeello
1h4e2l1o
,(](":@#,{.);.1~1,2~:/\])1!:1[1
woooorld
1w4o1r1l1d
[:,(#,&":{.)/.~ for 15: Try it online!
\$\endgroup\$
|εÅγs.ιJ,
Or alternatively:
|εÅγøí˜J,
Explanation:
| # Read all lines of input as list
ε # For-each over the lines:
Åγ # Run-length encode, pushing the list of characters and lengths separately
s # Swap so the characters at at the top and lengths below it
.ι # Interleave the two lists
J # Join the list of characters and lengths together to a single string
, # And output it with trailing newline
|εÅγ # Same as above
ø # Zip/transpose; creating pairs of [character, length]
í # Reverse each pair to [length, character]
˜ # Deep flatten the pairs to a single list
J, # Join them together to a single string, and output it with trailing newline
m[0] with $1 for -2 and if you switch to ruby -p you can drop the ->s{s. ... } for another -7 :)
\$\endgroup\$
def f(s,c=1):i,*j=s;b=j[:1]==[i];print(end='%s%s'%(c,i)*(b^1));f(j,1+b*c)
f(input())
Checks if the first and second characters of the string are equal. If they are, increase the counter by 1. If they are not, print the counter and the first item and reset the counter to 1. In both cases, the function is called recursively with the first character removed.
Raises an error when end of string is reached.
f=lambda s:'%s%s'%(len(s)-len(t:=s.lstrip(p:=s[0])),p)+f(t)if s else''
Python 3 equivalent (77 bytes)
Strips all repeating characters off the start of the string. Then it returns a string containing (1.) the difference in lengths between the original string and the stripped string; (2.) the first character of the original string; (3.) the result of the recursive function applied to the stripped string. Recursion ends when an empty string is encountered.
Print[""<>StringCases[#,s:x_..:>ToString@StringLength@s<>x]]&/@StringSplit[$ScriptInputString,"
"]
A more flexible I/O format reduces this solution to 54 bytes:
""<>StringCases[#,s:x_..:>ToString@StringLength@s<>x]&
e`
e` is a built-in for run-length encoding. CJam's implicit output ignores array brackets, and so turns [[1 'h] [2 'e]] into "1h2e"
import Data.List
m@main=getLine>>=putStrLn.(>>=shows.length<*>take 1).group>>m
Åγsø˜
Åγsø˜ # full program
˜ # flatten each sublist of...
ø # list of all elements of...
Åγ # chars and corresponding run lengths of chars in...
# implicit input...
ø # with each element from...
s # second...
ø # sublist paired with corresponding element in...
s # first...
ø # sublist
# implicit output
-10 thanks to @Original Original Original VI
r=>r.split`
`.map(d=>d.replace(/(.)\1*/g,g=>g.length+g[0])).join`
`
My first time using \1 in a regular expression, which is a super overpowered way to approach this problem (looks like a few answers already use it). The rest of the code is basically just I/O.
_ split "\n"map{def d:Seq[_]=>Any={case e+:f=>val(g,h)=f span e.==
""+e+(g.size+1)+d(h)case x=>x}
d(_)}mkString "\n"
Is it short? No. Does it use best practices where it wouldn't make the code longer? No. Did I do it for ethical reasons? No. Is it creative? No. Am I proud of it? Yes.
ḅ⟨{lṫ}ch⟩ᵐc
(If output really has to be on stdout, add one byte for w at the end.)
c The output is the concatenation of
⟨ c ⟩ᵐ the concatenated pairs of
{lṫ} length converted to a string
h and first element
ḅ ᵐ for every run in the input.
while 1:
a=b='';k=0
for c in input():e=a!=c;b+=(str(k)+a)*e;k+=1-k*e;a=c
print(b[1:]+str(k)+a)
f=lambda r,c,s,k=1:s and(c==s[0]and f(r,c,s[1:],k+1)or f(r+str(k)+c,s[0],s[1:]))or r[1:]+str(k)+c
while 1:print(f('','',input()))
The iterative approach seems quite successful, while the recursive version has a lot of room for improvement.
while loop in your code is only there to demonstrate the input. Without it, your code is still valid. In that case, the loop does not need to be part of the code and you can reduce the first example to 85 bytes like so: Try it online!
\$\endgroup\$
(.)\1*
$.0$1
Explanation:
Get a part of 1 or more of the same character, capturing the character in capture group 1.
(.)\1*
Replace it with the length of the total match, concatted with the character from capture group 1:
$.0$1
foldl(((n,l),c)->(c==l||print(n,l);((c==l&&n)+1,c)),readline()*'\n',init=("",""))
'\n' to readline() instead of printing the last tuple manually
\$\endgroup\$
d, 2 bytesøe
That is,
øe # run length encode the implicit input
# the d flag deep sums the top of the stack before outputting
jksr8
Relatively self explanatory, just joins together Pyth's run length encode instruction
Edit: -1 byte because while working on something completely different I remembered flattening a 2D list is just summing the elements.
while read s;do e=;while [[ $s ]];do c=${s:0:1};n=${s##+($c)};e+=$[${#s}-${#n}]$c;s=$n;done;echo $e;done
Sample run:
bash-4.2$ while read s;do e=;while [[ $s ]];do c=${s:0:1};n=${s##+($c)};e+=$[${#s}-${#n}]$c;s=$n;done;echo $e;done <<END
heeeello
woooorld
END
1h4e2l1o
1w4o1r1l1d
while read s;do n=1;for i in {1..$#s};do if [[ $s[i] != $s[i+1] ]];then echo -n $n$s[i];n=0;fi;((n++));done;echo;done
Run it like this:
zsh script.zsh < infile
while read s; do # while stdin has more
n=1 # repeat counter
for i in {1..$#s}; do # for each character
if [[ $s[i] != $s[i+1] ]]; then # same as next one?
echo -n $n$s[i] # print if no
n=0
fi
((n++))
done
echo # newline between words
done
[[ construct is a command on it's own (like [) and has to be separated from other commands. As to using [ over [[, it requires the arguments to be quoted so four " need to be added.
\$\endgroup\$
,↑{(⍕⍴⍵),⊃⍵}¨B⊂⍨B≠¯1⌽B←⍞
{=[{J[-jL[Q}\m}WL
{=[{^^[~\/L[Sh}\m}WL
Older/Alternative and longer versions:
{=[{^^L[Sh\/-]Sh.+}m[\[}WL
{=[{^^L[Sh\/-][-.+}m[\[}WL
{=[{^^L[Sh\/-~.+}m[\[}WL
{=[{^^L[Sh\/-].+}\m}WL
{=[{^^L[Sh\/[~.+}\m}WL
{=[{^^L[Sh\/[~_+}\m}WL
{=[{^^L[Sh\/fc.+}\m}WL
{=[{^^L[Sh\/-~.+}\m}WL
{=[{^^L[Sh\/-]\/}\m}WL
This doesn't really count because I created rs way after this was posted...but it was fun anyway!
(.)(\1*)/(^^\1\2)\1
44 bytes: try it online. Based on this.
echo ${(s::)1}|xargs -n1|uniq -c|tr -d \\n\
70 bytes: try it online. Based on this.
for ((n=1;i++<#w;n++))[[ $w[i] != $w[i+1] ]]&&printf $n$w[i]&&n=0
echo
w4orldinstead of1w4o1r1l1d(you'd need to escape numerics, e.g. `f111 -> f3\1´). But then it would be a near-duplicate of this: codegolf.stackexchange.com/questions/6774 \$\endgroup\$22is a trivial fixpoint. \$\endgroup\$11is impossible.22must end there or be followed by another fixed point not beginning with2.333nnnis an impossible pattern, for you'll never find the same character at consecutive odd indices.4444and up fail for the same reason. \$\endgroup\$