将列表转换为序言中的另一个列表 [英] Translating a list to another list in prolog

问题描述

我试图在 Prolog 中编写一个简单的代码，将一个列表转换为另一个列表.例如，如果我们调用 listtrans([a,b,c],L)，L 将变成 [1,2,3].(a,b,c 替换为 1,2,3).但是我在最后一行遇到了语法错误.问题是什么?这是我的代码:

I Tried to write a simple code in Prolog which translate a list to another list. for instance, if we call listtrans([a,b,c],L), L will become [1,2,3]. (a,b,c is replaced with 1,2,3). But i faced with a syntax error in last line. what is the problem? here is my code:

trans(a,1).
trans(b,2).
trans(c,3).

listtrans([],L).
listtrans([H|T],L1):-
   trans(H,B),
   append(B,L,L2),
   listtrans(T,L2).

推荐答案

该错误很可能是因为在您的代码中:

The error is very likely because in your code:

listtrans([H|T],L1):-
   trans(H,B),
   append(B,L,L2),
   listtrans(T,L2).

变量 L1 在头部声明，但没有在任何地方引用:你拼错了什么?

the variable L1 is declared in the head, but not referenced anywhere: you mispelled something?

无论如何，你的代码是行不通的.

Anyway, your code is not going to work.

此外，将 append/3 用于此类易于通过递归定义的任务被认为糟糕(也是因为您从中获得的性能不佳).

Moreover, using append/3 for this kind of tasks which are easily defined by recursion is considered terrible (also because of the bad performance you get out of it).

将函数应用于列表很简单.你已经知道在序言中你不写 Y = f(X) 而是声明 X 和 之间的函数关系Y 为:f(X, Y)..(这基本上就是你对 trans(X,Y) 所做的).

Applying a function to a list is straightforward. You already know that in prolog you don't write Y = f(X) but rather declare the functional relation between X and Y as: f(X, Y).. (That's basically what you did with trans(X,Y)).

现在是(简单的)递归公式:

Now the (easy) recursive formulation:

• 转换后的空列表就是空列表
• [X|Xs] 的变换是 [Y|Ys] 如果 trans(X,Y) 并且我们递归地变换 Xs 到 Ys

• the transformed empty list is the empty list
• the transformation of [X|Xs] is [Y|Ys] if trans(X,Y) and we recursively transform Xs into Ys

或在序言中表示:

listtrans([],[]).
listtrans([X|Xs],[Y|Ys]) :- trans(X,Y), listtrans(Xs,Ys).

我建议您阅读立即学习 Prolog 的前 4 章以更好地了解这些概念.

I recommend you reading the first 4 chapters of Learn Prolog Now to better understand these concepts.

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