从列表中的一个索引元素删除到另一个索引元素

Question

我需要从列表中的一个元素索引删除另一个索引元素。所以它应该看起来像：

?-delm(2,4,[5,6,-3,6,11,56,81],L),write(L),nl. 
L = [5,11,56,81] 

所以我做了这个。

delm(A,B,C,D):-A>B,delm(B,A,C,D). 
del1(1,[_|T],T):-!. 
del1(N,[X|T],[X|L]):-N1 is N - 1,del1(N1,T,L). 
delm(N,2,L,R):-del1(N,L,R),!. 
delm(N,M,L,R):-M1 is M - 1,del1(N,L,Buf),delm(N,M1,Buf,R). 
length([],0). 
length([_|T],N):- length(T,N1),N is N1+1. 
?-delm(2,4,[5,6,-3,6,11,56,81],L),write(L),nl. 

但我还需要补充的是，如果其中一个数字小于1或大于列表长度，写入消息（“错误”）。所以它应该看起来像

?-delm(-2,4,[5,6,-3,6,11,56,81],L),write(L),nl. 
"Error" 
?-delm(2,-4,[5,6,-3,6,11,56,81],L),write(L),nl. 
"Error" 
?-delm(2,40,[5,6,-3,6,11,56,81],L),write(L),nl. 
"Error" 

我不知道该怎么做。请帮忙！

Answer 1

你只需要添加一对夫妇的规则，即检查该A <= B规则后：

delm(A, _, _, 'Error') :- A < 1, !. 
delm(_, B, C, 'Error') :- length(C, L), B > L, !. 

所以整个代码是（只用你的代码）：

del1(1,[_|T],T):-!. 
del1(N,[X|T],[X|L]):-N1 is N - 1,del1(N1,T,L). 

delm(A,B,C,D):-A>B,delm(B,A,C,D). 
delm(A,_,_,'Error'):-A<1,!. 
delm(_,B,C,'Error'):-length(C, L), B>L, !. 
delm(N,2,L,R):-del1(N,L,R),!. 
delm(N,M,L,R):-M1 is M - 1,del1(N,L,Buf),delm(N,M1,Buf,R). 

一般我会打电话给这样一个例程slice。顺便提一句，length/2通常内置于大多数prolog环境。

Answer 2

你想要可以实现什么样的“核心”部分如下：

:- use_module(library(clpfd)). 

list_from_to_skipped(Xs,From,To,Ys) :- 
    From #= From0 + 1, 
    From #=< To, 
    N_Skip #= To - From0, 
    append(Prefix,Xs0,Xs), 
    append(Skip,Suffix,Xs0), 
    length(Prefix,From0), 
    length(Skip,N_Skip), 
    append(Prefix,Suffix,Ys). 

这里是你的查询示例：

?- list_from_to_skipped([5,6,-3,6,11,56,81],2,4,Ls). 
Ls = [5,11,56,81]. 

现在，让我们有一个更一般的查询：

?- list_from_to_skipped([a,b,c,d],From,To,Xs). 
From = To, To = 1, Xs = [ b,c,d] ; 
From = 1, To = 2, Xs = [ c,d] ; 
From = 1, To = 3, Xs = [  d] ; 
From = 1, To = 4, Xs = [  ] ; 
From = To, To = 2, Xs = [a, c,d] ; 
From = 2, To = 3, Xs = [a, d] ; 
From = 2, To = 4, Xs = [a  ] ; 
From = To, To = 3, Xs = [a,b, d] ; 
From = 3, To = 4, Xs = [a,b ] ; 
From = To, To = 4, Xs = [a,b,c ] ; 
false.