用于拆分数组的 Prolog 谓词 [英] Prolog predicate to split an array

问题描述

当 Res 是一个列表时，谓词 splitEvery(N,List,Res) 成功，其中每个列表List中的N个连续元素被分组在一个保持顺序的列表中

The predicate splitEvery(N,List,Res) succeeds when Res is a list in-which every N consecutive elements in the list List are grouped in a list maintaining the order

示例:

?- splitEvery(2,[a,b,c,d,e,f,g,h],R).
    R = [[a, b], [c, d], [e, f], [g, h]] ;
    false.
?- splitEvery(4,[a,b,c,d,e,f,g,h],R).
    R = [[a, b, c, d], [e, f, g, h]] ;
    false.
?- splitEvery(8,[a,b,c,d,e,f,g,h],R).
    R = [[a, b, c, d, e, f, g, h]] ;
    false.

这是我目前正在解决的一个问题，这是我每次都返回的尝试:

That is a problem I am currently working on and that was my attempt that return everytime:

splitEvery(A,L,L):-
    length(L,A).
    splitEvery(A,[],[]).

splitEvery(A,[H|T],R):-
    helper(A,B,[H|T],R).

helper(A,B,[H|T],R):-
    B\=0,
    B=<A,
    B1 is B-1,
    helper(A,B1,T,[[H|Z]|R]).

helper(A,0,[H|T],R):-
    helper(A,A,T,[[H|Z]|R]).

推荐答案

假设列表必须被分成大小均匀的子列表，一个可能的解决方案是:

Supposing the list must be splitted into evenly sized sublists, a possible solution is:

% split_every(+N, +List, -ListOfLists)

  split_every(_, [], []).
  split_every(N, [X|Xs], [Y|Ys]) :-
      length(Y, N),
      append(Y, L, [X|Xs]), 
      split_every(N, L, Ys).  

示例:

?- split_every(2, [1,2,3,4,5,6,7,8], L).
L = [[1, 2], [3, 4], [5, 6], [7, 8]] ;
false.

?- split_every(4, [1,2,3,4,5,6,7,8], L).
L = [[1, 2, 3, 4], [5, 6, 7, 8]] ;
false.

?- split_every(3, [1,2,3,4,5,6,7,8,9], L).
L = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] ;
false.

?- split_every(4, [1,2,3,4,5,6,7,8,9], L).
false.

为了避免虚假的选择点，我们可以改变这个谓词的前两个参数的顺序:

To avoid spurious choice points, we can change the order of the first two arguments of this predicate:

% split_every(+List, +N, -ListOfLists)

  split_every([], _, []).
  split_every([X|Xs], N, [Y|Ys]) :-
      length(Y, N),
      append(Y, L, [X|Xs]),
      split_every(L, N, Ys).

示例:

?- split_every([1,2,3,4,5,6,7,8], 2, L).
L = [[1, 2], [3, 4], [5, 6], [7, 8]].

?- split_every([1,2,3,4,5,6,7,8], 4, L).
L = [[1, 2, 3, 4], [5, 6, 7, 8]].

?- split_every([1,2,3,4,5,6,7,8,9], 3, L).
L = [[1, 2, 3], [4, 5, 6], [7, 8, 9]].

?- split_every([1,2,3,4,5,6,7,8,9], 4, L).
false.

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