将列表中的某些字符复制到序言中的另一个字符 [英] Copy certain characters from a list to another on prolog

问题描述

所以我有这段代码，可以将列表中的所有内容复制到另一个列表中. 为了复制，我应该如何修改它，比如说前两个字符.

So i have this code which copies everything from a list to another one. How should I modify it in order to copy, lets say the first two character.

$copy(L,R) :- 
    copy2(L,R).
copy2([X],[X]).
copy2([H|T1],[H|T2]) :-
    copy2(T1,T2).

我希望它是什么的示例:?-复制([a，b，c，d，e，f]，X， 2 ). -> X = [a，b]

example of what i want it to be: ?- copy([a,b,c,d,e,f],X,2). --> X = [a,b]

推荐答案

您可以统一复制列表:

?- [a,b,c,d,e] = List.
List = [a, b, c, d, e].

?- [a,b,c,d,e] = [V,W,X,Y,Z].
V = a,
W = b,
X = c,
Y = d,
Z = e.

?- [a,b,c,d,e] = [V,W|Rest].
V = a,
W = b,
Rest = [c, d, e].

可以定义一个像您描述的谓词一样的谓词，它复制了列表的前N个元素:

A predicate like the one you describe, copying the first N elements of a list, can be defined thus:

first_n(List, N, Xs) :-
    length(Xs, N),
    append(Xs _, List).

其工作方式如下:

?- first_n([a,b,c,d,e], 2, X).
X = [a, b].

有很多不同的方式来编写类似的谓词.我定义first_n/3的方式，如果N大于List的长度(在注释中@false指出)，它将失败.可以改写通用功能的类似物 take ，如果N的长度大于List的长度，则会完全返回List:

There are a bunch of different ways to write a similar predicate. The way I have defined first_n/3, it will fail if N is larger than the length of List (this was pointed to out by @false in the comments). One could instead write an analog of the common function take, which will return List in its entirety in the event that N is greater than List's length:

take_n(N, List, Taken) :-
    ( length(List, M),
      N > M
    ->
      Taken = List
    ;
      length(Taken, N),
      append(Taken, _, List)
    ).

---

这个答案在@false的有益批评的指导下得到了多次纠正.

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