Simple nth1 predicate in Prolog
With SWI Prolog, there's a predicate that finds the nth item in a list called nth1. I want to implement my own version of the predicate but SWI's is so complicated if you look at the listing(nth1) code. Is开发者_JAVA技巧 there a simpler way of doing it?
Thank you :).
Consider using finite domain constraints for general (reversible) integer arithmetic:
:- use_module(library(clpfd)).
nth1(1, [E|_], E).
nth1(N, [_|Xs], E) :-
N #> 1,
N #= N1 + 1,
nth1(N1, Xs, E).
The SWI code is a bit complex because the predicate can be used to generate from a variable index:
?- nth1(Idx,[a,b,c],X).
Idx = 1,
X = a ;
Idx = 2,
X = b ;
Idx = 3,
X = c ;
false.
If you don't want that behavior, nth1/3
can be implemented easily in terms of nth0
:
nth1(Idx,List,X) :-
Idx0 is Idx-1,
nth0(Idx0,List,X).
Edit: it's also possible to do without nth0
in just a few lines of code:
nth1(1,[X|_],X) :- !.
nth1(Idx,[_|List],X) :-
Idx > 1,
Idx1 is Idx-1,
nth1(Idx1,List,X).
I didn't mean to be contradictory or get someone else to do my work actually; I just wanted some advice, sorry for not being clearer.
I've implemented it myself now but could you guys possibly suggest improvements or better ways of doing it? What I often find myself doing in Prolog is writing a predicate with say a counter or set of counters and getting a predicate with fewer arguments to call the clauses with extra arguments. This often ends up producing quite a bit of code. Anyway, here's my implementation I just did:
item_at( N, L, Item ) :-
item_at( N, 0, L, Item ).
item_at( N, Count, [H|_], Item ) :-
CountNew is Count + 1,
CountNew = N,
Item = H.
item_at( N, Count, [_|T], Item ) :-
CountNew is Count + 1,
item_at( N, CountNew, T, Item ).
Any comments? Thanks :). Usage:
?- item_at(3,[a,b,c,d,e],Item).
Item = c ;
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