filter list into separate lists

问题

I need to filter the list [#,d,e,#,f,g] such that I get the output as [[d,e],[f,g]] ,
I am stuck while creating a new list every time I encounter '#' is there a way to do this? I tried the code below,

filterL([],List) :-[].
filterL([Head|Tail],X) :-
   (  Head \='#'->
      append(X,Head,List),
      filterL(Tail,List)
   ;  filterL(Tail,X)
   ).

回答1:

Your problem is not very well defined. Are empty sequences allowed or not? Shall [#] be related to [[],[]] (there is an empty sequence before and after) or []? You say it should be []. So:

list_splitbyhash(Xs, Xss) :-
   phrase(splitby(Xss,#), Xs).

splitby([],_E) -->
    [].
splitby(Xss,E) -->
    [E],
    splitby(Xss,E).
splitby([Xs|Xss],E) -->
    {Xs = [_|_]},
    all_seq(dif(E),Xs),
    splitby(Xss,E).

all_seq(_, []) --> [].
all_seq(C_1, [C|Cs]) -->
   [C],
   {call(C_1,C)},
   all_seq(C_1, Cs).

回答2:

Here is another version, which uses an even more general approach:

list_splitbyhash(Xs, Xss) :-
   phrase(by_split(=(#), Xss), Xs).

=(X,X,true).
=(X,Y,false) :- dif(X,Y).

by_split(_C_2, []) --> [].
by_split(C_2, Xss) -->
   [E],
   {call(C_2,E,T)},
   (  { T = true },
      by_split(C_2, Xss)
   |  { T = false, Xss = [[E|Xs]|Xss1] },
      all_seq(callfalse(C_2),Xs),
      el_or_nothing(C_2),
      by_split(C_2, Xss1)
   ).

callfalse(C_2,E) :-
   call(C_2,E,false).

el_or_nothing(_) -->
   call(nil).
el_or_nothing(C_2), [E] -->
   [E],
   {call(C_2,E,true)}.

nil([], []).

With lambdas, this can be expressed more compactly. Instead of

   all_seq(callfalse(C_2),Xs)

and the definition for callfalse/3, one can now write

   all_seq(C_2+\F^call(C_2,F,false))

回答3:

With meta-predicate splitlistIf/3 and the reified equality predicate (=)/3, the task at hand becomes a one-liner---that is both efficient and logically pure!

?- splitlistIf(=(#),[#,d,e,#,f,g],Xs).
Xs = [[d,e],[f,g]].                      % succeeds deterministically

As the code is monotone, logical soundness is ensured even for quite general queries:

?- Xs = [A,B,C], splitlistIf(=(X),Xs,Yss).
Xs = [A,B,C],     X=A ,     X=B ,     X=C , Yss = [       ] ;
Xs = [A,B,C],     X=A ,     X=B , dif(X,C), Yss = [    [C]] ;
Xs = [A,B,C],     X=A , dif(X,B),     X=C , Yss = [  [B]  ] ;
Xs = [A,B,C],     X=A , dif(X,B), dif(X,C), Yss = [  [B,C]] ;
Xs = [A,B,C], dif(X,A),     X=B ,     X=C , Yss = [[A]    ] ;
Xs = [A,B,C], dif(X,A),     X=B , dif(X,C), Yss = [[A],[C]] ;
Xs = [A,B,C], dif(X,A), dif(X,B),     X=C , Yss = [[A,B]  ] ;
Xs = [A,B,C], dif(X,A), dif(X,B), dif(X,C), Yss = [[A,B,C]].

来源：https://stackoverflow.com/questions/26439058/filter-list-into-separate-lists

标签

list

prolog

prolog-dif