Prolog:取第一个“N"；列表的元素 [英] Prolog: Take the first "N" elements of a list

问题描述

我需要编写一个 Prolog 谓词 take(L, N, L1) 如果列表 L1 包含第一个 N 元素以相同的顺序列出 L.例如:

I need to write a Prolog predicate take(L, N, L1) which succeeds if list L1 contains the first N elements of list L, in the same order. For example:

?- take([5,1,2,7], 3, L1).
L1 = [5,1,2]
?- take([5,1,2,7], 10, L1).
L1 = [5,1,2,7] 

到目前为止，Prolog 对我来说意义不大，我很难分解它.这是我目前所拥有的:

Prolog thus far is making little sense to me, and I'm having a hard time breaking it down. Here is what I have so far:

take([H|T], 0, []).
take([H|T], N, L1) :-
   take(T, X, L2),
   X is N-1.

你能解释一下我在这里做错了什么吗?

Can you please explain what I did wrong here?

推荐答案

这是一个在 Haskell¹ 等函数式语言中实现 take 的对应关系的定义.首先，参数顺序应该不同，这有利于部分应用.有一个切入点，但只有在错误检查内置 (=<)/2 产生 instantiation_error 参数应该包含变量之后.

Here is a definition that implements the relational counterpart to take in functional languages like Haskell¹. First, the argument order should be different which facilitates partial application. There is a cut, but only after the error checking built-in (=<)/2 which produces an instantiation_error should the argument contain a variable.

take(N, _, Xs) :- N =< 0, !, N =:= 0, Xs = [].
take(_, [], []).
take(N, [X|Xs], [X|Ys]) :- M is N-1, take(M, Xs, Ys).

?- take(2, Xs, Ys).
   Xs = [], Ys = []
;  Xs = [_A], Ys = [_A]
;  Xs = [_A,_B|_C], Ys = [_A,_B].

注意上面的查询是如何读取的:

Note how above query reads:

如何从Xs中取出2个元素得到Ys?

How can one take 2 elements from Xs to get Ys?

并且有 3 个不同的答案.如果 Xs 为空，那么 Ys 也是如此.如果 Xs 是一个包含一个元素的列表，那么 Ys 也是如此.如果 Xs 至少有 2 个元素，那么这两个是 Ys.

And there are 3 different answers. If Xs is empty, then so is Ys. If Xs is a list with one element, then so is Ys. If Xs has at least 2 elements, then those two are Ys.

^{1) 唯一的区别是 take(-1, Xs,Ys) 失败(对于所有 Xs, Ys).可能最好的方法是发出类似于 arg(-1,s(1),2) 的 domain_error}

^{1) The only difference being that take(-1, Xs,Ys) fails (for all Xs, Ys). Probably the best would be to issue a domain_error similar to arg(-1,s(1),2)}

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