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.

Code to split string with several separators.

Input: string, list of separators
Output: [Sub0, [Sep0Num, Sep0Pos], Sub1, [Sep1Num, Sep1Pos], ..., SubN]
Note: Sub - substring, SepNum - separator number in input list, SepPos - separator position in input string.
Input order of separators is important: they are considered in that order.

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 which conflict with the currently selected sequence.

Example use:

<lang j> 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>

Python

<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 1: 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 in range(len(SepList)): Sep = SepList[i] 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 1: Res.append( S[SepEnd : MinPosList[MinPosPos]] ) Res.append([SepNumList[MinPosPos], MinPosList[MinPosPos]]) SepEnd = MinPosList[MinPosPos] + len(SepList[SepNumList[MinPosPos]]) while 1: 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>