Multisplit

Multisplit is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

It is often necessary to split a string into pieces based on several different (potentially multi-character) separator strings, while still retaining the information about which separators were present in the input. This is particularly useful when doing small parsing tasks.

Write code to demonstrate this. The function (or procedure or method, as appropriate) should take an input string and an ordered collection of separator strings, and split the string into pieces representing the various substrings. Note that the order of the separators is significant; where there would otherwise be an ambiguity as to which separator to use at a particular point (e.g., because one separator is a prefix of another) the first separator in the collection should be used. The result of the function should be an ordered sequence of substrings.

Extra Credit: include match information that indicates which separator was matched at each separation point and where in the input string that separator was matched.

Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.

D

<lang d>import std.stdio, std.array, std.algorithm, std.string; void main() {

   auto s = `a!===b=!=c==!=d`;
   auto divs = [ `==`, `!=`, `=`, `!` ];
   auto lst = MultiSplit(s, divs);
   foreach(int i, string p; lst) {
       write(p);
       if (i < lst.length-1) write(` <`~ divs[i] ~`> `);
   }

}

string[] MultiSplit(string s, string[] divisors) {

   if (s.empty) return [];
   string[] res;
   char[] rest = s.dup;
   foreach(string div; divisors){
       auto p = findSplit(rest, div);
       res.length = res.length + 1;
       res[$-1] = p[0].idup;
       rest = p[2];
       if (p[1].empty || rest.empty) // divisor is not found OR it was last in string
           break;
   }
   if (!rest.empty) {
       res.length = res.length + 1;
       res[$-1] = rest.idup;
   }
   return res;

}</lang> Output (separators are in angle brackets):

a! <==> =b= <!=> c <=> = <!> =d

F#

If we ignore the "Extra Credit" requirements and skip 'ordered separators' condition (i.e. solving absolute different task), this is exactly what one of the overloads of .NET's String.Split method does. Using F# Interactive: <lang fsharp>> "a!===b=!=c".Split([|"=="; "!="; "="|], System.StringSplitOptions.None);; val it : string [] = [|"a"; ""; "b"; ""; "c"|]

> "a!===b=!=c".Split([|"="; "!="; "=="|], System.StringSplitOptions.None);; val it : string [] = [|"a"; ""; ""; "b"; ""; "c"|]</lang>

System.StringSplitOptions.None specifies that empty strings should be included in the result.

J

<lang j>multisplit=:4 :0

 'sep begin'=.|:t=. y /:~&.:(|."1)@;@(i.@#@[ ,.L:0"0 I.@E.L:0) x
 end=. begin + sep { #@>y
 last=.next=.0
 r=.2 0$0
 while.next<#begin do.
   r=.r,.(last}.x{.~next{begin);next{t
   last=.next{end
   next=.1 i.~(begin>next{begin)*.begin>:last
 end.
 r=.r,.;~last}.x

)</lang>

Explanation:

First find all potentially relevant separator instances, and sort them in increasing order, by starting location and separator index. sep is separator index, and begin is starting location. end is ending location.

Then, loop through the possibilities, skipping over those separators which would overlap with previously used separators.

The result consists of two rows: The first row is the extracted substrings, the second row is the "extra credit" part -- for each extracted substring, the numbers in the second row are the separator index (0 for the first index, 1 for the seconde, ...), and the location in the original string where the separator appeared. Note that the very last substring does not have a separator following it, so the extra credit part is blank for that substring.

Example use:

<lang j> S=:'a!===b=!=c'

  S multisplit '==';'!=';'='

┌───┬───┬───┬───┬─┐ │a │ │b │ │c│ ├───┼───┼───┼───┼─┤ │1 1│0 3│2 6│1 7│ │ └───┴───┴───┴───┴─┘

  S multisplit '=';'!=';'=='

┌───┬───┬───┬───┬───┬─┐ │a │ │ │b │ │c│ ├───┼───┼───┼───┼───┼─┤ │1 1│0 3│0 4│0 6│1 7│ │ └───┴───┴───┴───┴───┴─┘

  'X123Y' multisplit '1';'12';'123';'23';'3'

┌───┬───┬─┐ │X │ │Y│ ├───┼───┼─┤ │0 1│3 2│ │ └───┴───┴─┘</lang>

PicoLisp

<lang PicoLisp>(de multisplit (Str Sep)

  (setq Sep (mapcar chop Sep))
  (make
     (for (S (chop Str) S)
        (let L
           (make
              (loop
                 (T (find head Sep (circ S))
                    (link
                       (list
                          (- (length Str) (length S))
                          (pack (cut (length @) 'S)) ) ) )
                 (link (pop 'S))
                 (NIL S (link NIL)) ) )
           (link (pack (cdr (rot L))))
           (and (car L) (link @)) ) ) ) )

(println (multisplit "a!===b=!=c" '("==" "!=" "="))) (println (multisplit "a!===b=!=c" '("=" "!=" "==")))</lang> Output:

("a" (1 "!=") NIL (3 "==") "b" (6 "=") NIL (7 "!=") "c")
("a" (1 "!=") NIL (3 "=") NIL (4 "=") "b" (6 "=") NIL (7 "!=") "c")

Python

Using Regular expressions

<lang python>>>> import re >>> def ms2(txt="a!===b=!=c", sep=["==", "!=", "="]): if not txt or not sep: return [] ans = m = [] for m in re.finditer('(.*?)(?:' + '|'.join('('+re.escape(s)+')' for s in sep) + ')', txt): ans += [m.group(1), (m.lastindex-2, m.start(m.lastindex))] if m and txt[m.end(m.lastindex):]: ans += [txt[m.end(m.lastindex):]] return ans

>>> ms2() ['a', (1, 1), , (0, 3), 'b', (2, 6), , (1, 7), 'c'] >>> ms2(txt="a!===b=!=c", sep=["=", "!=", "=="]) ['a', (1, 1), , (0, 3), , (0, 4), 'b', (0, 6), , (1, 7), 'c']</lang>

Not using RE's

<lang python>>>> def ms(txt="a!===b=!=c", sep=["==", "!=", "="]): if not txt or not sep: return [] size = [len(s) for s in sep] ans, pos0 = [], 0 def getfinds(): return [(-txt.find(s, pos0), -sepnum, size[sepnum]) for sepnum, s in enumerate(sep) if s in txt[pos0:]]

finds = getfinds() while finds: pos, snum, sz = max(finds) pos, snum = -pos, -snum ans += [ txt[pos0:pos], [snum, pos] ] pos0 = pos+sz finds = getfinds() if txt[pos0:]: ans += [ txt[pos0:] ] return ans

>>> ms() ['a', [1, 1], , [0, 3], 'b', [2, 6], , [1, 7], 'c'] >>> ms(txt="a!===b=!=c", sep=["=", "!=", "=="]) ['a', [1, 1], , [0, 3], , [0, 4], 'b', [0, 6], , [1, 7], 'c']</lang>

Alternative version <lang python>def min_pos(List): return List.index(min(List))

def find_all(S, Sub, Start = 0, End = -1, IsOverlapped = 0): Res = [] if End == -1: End = len(S) if IsOverlapped: DeltaPos = 1 else: DeltaPos = len(Sub) Pos = Start while True: Pos = S.find(Sub, Pos, End) if Pos == -1: break Res.append(Pos) Pos += DeltaPos return Res

def multisplit(S, SepList): SepPosListList = [] SLen = len(S) SepNumList = [] ListCount = 0 for i, Sep in enumerate(SepList): SepPosList = find_all(S, Sep, 0, SLen, IsOverlapped = 1) if SepPosList != []: SepNumList.append(i) SepPosListList.append(SepPosList) ListCount += 1 if ListCount == 0: return [S] MinPosList = [] for i in range(ListCount): MinPosList.append(SepPosListList[i][0]) SepEnd = 0 MinPosPos = min_pos(MinPosList) Res = [] while True: Res.append( S[SepEnd : MinPosList[MinPosPos]] ) Res.append([SepNumList[MinPosPos], MinPosList[MinPosPos]]) SepEnd = MinPosList[MinPosPos] + len(SepList[SepNumList[MinPosPos]]) while True: MinPosPos = min_pos(MinPosList) if MinPosList[MinPosPos] < SepEnd: del SepPosListList[MinPosPos][0] if len(SepPosListList[MinPosPos]) == 0: del SepPosListList[MinPosPos] del MinPosList[MinPosPos] del SepNumList[MinPosPos] ListCount -= 1 if ListCount == 0: break else: MinPosList[MinPosPos] = SepPosListList[MinPosPos][0] else: break if ListCount == 0: break Res.append(S[SepEnd:]) return Res

S = "a!===b=!=c" multisplit(S, ["==", "!=", "="]) # output: ['a', [1, 1], , [0, 3], 'b', [2, 6], , [1, 7], 'c'] multisplit(S, ["=", "!=", "=="]) # output: ['a', [1, 1], , [0, 3], , [0, 4], 'b', [0, 6], , [1, 7], 'c']</lang>

Ruby

The simple method, using a regular expression to split the text.

<lang ruby>text = 'a!===b=!=c' separators = ['==', '!=', '=']

def multisplit_simple(text, separators)

 sep_regex = Regexp.new(separators.collect {|sep| Regexp.escape(sep)}.join('|'))
 text.split(sep_regex)

end

p multisplit_simple(text, separators)

1. => ["a", "", "b", "", "c"]</lang>

The version that also returns the information about the separations.

<lang ruby>def multisplit(text, separators)

 sep_regex = Regexp.new(separators.collect {|sep| Regexp.escape(sep)}.join('|'))
 separator_info = []
 pieces = []
 i = prev = 0
 while i = text.index(sep_regex, i)
   separator = Regexp.last_match(0)
   pieces << text[prev .. i-1]
   separator_info << [separator, i]
   i = i + separator.length
   prev = i
 end
 pieces << text[prev .. -1]
 [pieces, separator_info]

end

p multisplit(text, separators)

1. => [["a", "", "b", "", "c"], [["!=", 1], ["==", 3], ["=", 6], ["!=", 7]]]</lang>

Also demonstrating a method to rejoin the string given the separator information.

<lang ruby>def multisplit_rejoin(multisplit_info)

 pieces, separator_info = multisplit_info
 pairs = pieces.zip(separator_info)
 str = pairs.take(pairs.length - 1).inject("") do
         |str, (piece, (separator, index))| 
         str << piece << separator
       end
 str << pieces.last

end

p multisplit_rejoin(multisplit(text, separators)) == text

1. => true</lang>

Tcl

This simple version does not retain information about what the separators were: <lang tcl>proc simplemultisplit {text sep} {

   set map {}; foreach s $sep {lappend map $s "\uffff"}
   return [split [string map $map $text] "\uffff"]

} puts [simplemultisplit "a!===b=!=c" {"==" "!=" "="}]</lang>

Output:

a {} b {} c

However, to keep the match information a more sophisticated technique is best. Note that the most natural model of result here is to return the split substrings as a separate list to the match information (because the two collections of information are of different lengths). <lang tcl>proc multisplit {text sep} {

   foreach s $sep {lappend sr [regsub -all {\W} $s {\\&}]}
   set sepRE [join $sr "|"]
   set pieces {}
   set match {}
   set start 0
   while {[regexp -indices -start $start -- $sepRE $text found]} {

lassign $found x y lappend pieces [string range $text $start [expr {$x-1}]] lappend match [lsearch -exact $sep [string range $text {*}$found]] $x set start [expr {$y + 1}]

   }
   return [list [lappend pieces [string range $text $start end]] $match]

}</lang> Demonstration code: <lang tcl>set input "a!===b=!=c" set matchers {"==" "!=" "="} lassign [multisplit $input $matchers] substrings matchinfo puts $substrings puts $matchinfo</lang> Output:

a {} b {} c
1 1 0 3 2 6 1 7