Elixir Array Syntactic Sugar

Question

In Elixir, (linked) lists are in the format [head | tail] where head can be anything and tail is a list of the rest of the list, and [] - the empty list - is the only exception to this.

Lists can also be written like [1, 2, 3] which is equivalent to [1 | [2 | [3 | []]]]

Your task is to convert a list as described. The input will always be a valid list (in Elixir) containing only numbers matching the regex \[(\d+(, ?\d+)*)?\]. You may take the input with (one space after each comma) or without spaces. The output may be with (one space before and after each |) or without spaces.

For inputs with leading zeroes you may output either without the zeroes or with.

Input must be taken as a string (if writing a function), as does output.

Examples

[] -> []
[5] -> [5 | []]
[1, 7] -> [1 | [7 | []]]
[4, 4, 4] -> [4 | [4 | [4 | []]]]
[10, 333] -> [10 | [333 | []]]

related, not a duplicate as this in part involves adding mode ] to the end. Additionally, the Haskell answer here is quite different to the one there.

Answer 1

Haskell, 50 bytes

f.read
f(a:b)='[':show(a+0)++'|':f b++"]"
f _="[]"

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The +0 lets the Haskell type checker know that we are dealing with lists of numbers, so read will parse the input string for us.

Answer 2

Python 2, 50 bytes

r='%s'
for x in input():r%='[%s|%%s]'%x
print r%[]

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Answer 3

JavaScript (ES6), 50 bytes

s=>eval(s).map(v=>`[${p+=']',v}|`,p='[]').join``+p

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---

Recursive version, 51 bytes

f=(s,[v,...a]=eval(s))=>1/v?`[${v}|${f(s,a)}]`:'[]'

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Answer 4

Retina, 39 33 32 20 bytes

\b]
,]
+`,(.*)
|[$1]

Saved 13 bytes thanks to H.PWiz, ovs, ASCII-only, and Neil.
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Explanation

\b]
,]

If we don't have an empty list, add a trailing comma.

+`,(.*)
|[$1]

While there are commas, wrap things with |[ thing ].

Answer 5

Perl 5 -pl, 31 28 bytes

s/\d\K]/,]/;$\=']'x s/,/|[/g

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How?

-p                    # (command line) Implicit input/output via $_ and $\
s/\d\K]/,]/;          # insert a comma at the end if the list is not empty
$\=']'x s/,/|[/g      # At the end of the run, output as many ']' as there are
                      # commas in the input.  Replace the commas with "|["

Answer 6

Elixir, 111 85 bytes

f=fn[h|t],f->"[#{h}|#{f.(t,f)}]"
[],_->"[]"
h,f->f.(elem(Code.eval_string(h),0),f)end

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I have never used Elixir before. Defines a function that takes a string and a reference to itself and returns a string.

Answer 7

Ceylon, 113 bytes

String p(String s)=>s.split(" ,[]".contains).select((x)=>!x.empty).reversed.fold("[]")((t,h)=>"[``h`` | ``t``]");

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Here is it written out:

// define a function p mapping Strings to Strings.
String p(String s) =>
    // we split the string at all characters which are brackets, comma or space.
    s.split(" ,[]".contains)    // → {String+}, e.g.  { "", "1", "7", "" }
    // That iterable contains empty strings, so let's remove them.
    // Using `select` instead of `filter` makes the result a sequential instead of
    // an Iterable.
     .select((x)=>!x.empty)    // → [String*], e.g.   [1, 7]
    // now invert the order.
    // (This needs a Sequential (or at least a List) instead of an Iterable.)
     .reversed                 // → [String*], e.g.   [7, 1]
    // Now iterate over the list, starting with "[]", and apply a function
    // to each element with the intermediate result.
     .fold("[]")                       // → String(String(String, String))
    //    This function takes the intermediate result `t` (for tail) and an element
    //    `h` (for head), and puts them together into brackets, with a " | " in the
    //    middle. This uses String interpolation, I could have used `"+` and `+"`
    //    instead for the same length.
          ((t,h)=>"[``h`` | ``t``]");  // → String

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As noted by ovs in a (now deleted) comment: If one select the "without spaces" options for input and output indicated in the question, one can safe 3 more bytes (the obvious ones with spaces in them).

If we don't need to parse the input, but just could get a sequence as input, it gets much shorter (69 bytes).

String p(Object[]s)=>s.reversed.fold("[]")((t,h)=>"[``h`` | ``t``]");

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Answer 8

Python 3, 65 bytes

lambda k:"["+''.join(f"{u}|["for u in eval(k))+-~len(eval(k))*"]"

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If the input could be a list instead, then:

Python 3, 53 bytes

lambda k:"["+''.join(f"{u}|["for u in k)+-~len(k)*"]"

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Answer 9

SNOBOL4 (CSNOBOL4), 114 bytes

	I =INPUT
S	N =N + 1	
	I SPAN(1234567890) . L REM . I	:F(O)
	O =O '[' L ' | '	:(S)
O	OUTPUT =O '[' DUPL(']',N)
END

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	I =INPUT				;* read input
S	N =N + 1				;* counter for number of elements (including empty list)
	I SPAN(1234567890) . L REM . I	:F(O)	;* get value matching \d until none left
	O =O '[' L ' | '	:(S)		;* build output string
O	OUTPUT =O '[' DUPL(']',N)		;* print O concatenated with a '[' and N copies of ']'
END

Answer 10

Stax, 19 bytes

É▲²:WlÖ└%ï╪☺╒▓"We↨Φ

Run and debug it

My first Stax post, so probably not optimal.

Unpacked and commented:

U,                      Put -1 under input
  {                     Block
   i                      Push loop index, needed later
    '[a$'|++              Wrap the element in "[...|"
            m           Map
             '[+        Add another "["
                s2+     Get the latest loop index + 2
                   ']*+ Add that many "]"

Run and debug this one

Answer 11

Husk, 22 bytes

+:'[J"|[":msr¹₁*₁Lr
"]

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Answer 12

Befunge-98 (PyFunge), 22 21 bytes

'[,1;@j,]';#$&." |",,

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If there weren't weird restrictions on output, we could do this in 18:

'[,1;@j,]';#$&.'|,

Fun fact, this is technically a program that does nothing in Python.

Answer 13

Ruby -p, 39 bytes

Full program:

$_[$&]="|[#$2]"while/(,|\d\K(?=]))(.*)/

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Ruby, 48 45 bytes

Recursive function:

f=->s{(s[/,|\d\K(?=])/]&&="|[")&&f[s+=?]]||s}

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Answer 14

R, 84 71 69 bytes

function(x){while(x<(x=sub('(,|\\d\\K(?=]))(.+)','|[\\2]',x,,T)))1;x}

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• -15 bytes thanks to @KirillL.

Answer 15

Proton, 57 bytes

k=>"["+''.join(str(u)+"|["for u:(e=eval(k)))+-~len(e)*"]"

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Answer 16

Jelly, 19 bytes

ŒV©;€⁾|[”[;µ®L‘”]ẋṭ

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A non-recursive alternative to Erik's solution.

Answer 17

Jelly, 18 bytes

ŒVḢ,ŒṘÇ”|;ƲƊ¹¡⁾[]j

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Answer 18

Jelly, 18 bytes

ŒVµ⁾[]jj⁾|[ṫ3;”]ṁ$

A full program printing the result (as a monadic link it accepts a list of characters but returns a list of characters and integers).

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How?

ŒVµ⁾[]jj⁾|[ṫ3;”]ṁ$ - Main link: list of characters  e.g. "[10,333]"
ŒV                 - evaluate as Python code              [10,333]
  µ                - start a new monadic chain, call that X
   ⁾[]             - list of characters                   ['[',']']
      j            - join with X                          ['[',10,333,']']
        ⁾|[        - list of characters                   ['|','[']
       j           - join                                 ['[','|','[',10,'|','[',333,'|','[',']']
           ṫ3      - tail from index three                ['[',10,'|','[',333,'|','[',']']
                 $ - last two links as a monad (f(X)):
              ”]   -   character                          ']'
                ṁ  -   mould like X                       [']',']'] (here 2 because X is 2 long)
             ;     - concatenate                          ['[',10,'|','[',333,'|','[',']',']',']']
                   - implicit (and smashing) print        [10|[333|[]]]

Answer 19

Java 10, 107 bytes

s->{var r="[]";for(var i:s.replaceAll("[\\[\\]]","").split(","))r="["+i+"|"+r+"]";return s.length()<3?s:r;}

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Explanation:

s->{                       // Method with String as both parameter and return-type
  var r="[]";              //  Result-String, starting at "[]"
  for(var i:s.replaceAll("[\\[\\]]","") 
                           //  Removing trailing "[" and leading "]"
             .split(","))  //  Loop over the items
    r="["+i+"|"+r+"]";     //   Create the result-String `r`
  return s.length()<3?     //  If the input was "[]"
          s                //   Return the input as result
         :                 //  Else:
          r;}              //   Return `r` as result

Answer 20

Standard ML, 71 bytes

fun p[_]="]|[]]"|p(#","::r)="|["^p r^"]"|p(d::r)=str d^p r;p o explode;

Try it online! Uses the format without spaces. E.g. it "[10,333,4]" yields "[10|[333|[4]|[]]]]".

ungolfed

fun p [_]       = "]|[]]"          (* if there is only one char left we are at the end *)
  | p (#","::r) = "|[" ^ p r ^ "]" (* a ',' in the input is replaced by "|[" and an closing "]" is added to the end *)
  | p (d::r)    = str d ^ p r      (* all other chars (the digits and the initial '[') are converted to a string and concatenated to recursive result *)

val f = p o explode  (* convert string into list of chars and apply function p *)

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Answer 21

R, 140 136 bytes

Down 4 bytes as per Giuseppe's sound advice.

function(l,x=unlist(strsplit(substr(l,2,nchar(l)-1),", ")))paste(c("[",paste0(c(x,"]"),collapse=" | ["),rep("]",length(x))),collapse="")

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Answer 22

R, 108 bytes

function(l)Reduce(function(x,y)paste0("[",x,"|",y,"]"),eval(parse(t=sub("]",")",sub("\\[","c(",l)))),"[]",T)

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It took almost a year to find a better R solution than previous...should have known Reduce would be the answer! Outputs without spaces, input can be with or without spaces.

Answer 23

Python 2, 63 bytes

lambda s:'['+'|['.join(map(str,eval(s))+[']'])+']'*len(eval(s))

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Answer 24

sed + -E, 46 bytes

:
s/\[([0-9]+)(, ?([^]]*)|())\]/[\1 | [\3]]/
t

A fairly straightforward approach. The second line takes [\d+, ...] and changes it to [\d | [...]]. The third line jumps back to the first line, if the substitution was successful. The substitution repeats until it fails and then the program terminates. Run with sed -E -f filename.sed, passing input via stdin.

Answer 25

Red, 110 bytes

func[s][if s ="[]"[return s]replace append/dup replace/all b: copy s",""|[""]"(length? b)- length? s"]""|[]]"]

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Explanation of the ungolfed version:

f: func[s][                      
    if s = "[]" [return s]                    ; if the list is empty, return it    
    b: copy s                                 ; save a copy of the input in b 
    replace/all b "," "|["                    ; replace every "," with "|["  
    append/dup b "]" (length? b) - length? s  ; append as many "]" as there were ","
    replace b "]" "|[]]"                      ; replace the first "]" with "|[]]"     
]                                             ; the last value is returned implicitly

Red is so easily readable, that I doubt I needed adding the above comments :)

Answer 26

Perl 6, 38 bytes

{'['~S:g/\s|.<($/\|\[/~']'x$_+1}o&EVAL

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