从 Prolog 中的列表构建复合谓词

问题描述

如果我在 Prolog 中有一个谓词列表，例如 [flies,swims]，我如何构建一个谓词，该谓词是列表中所有谓词的合取，即 fliesAndSwims(X) :- 苍蝇(X)，游泳(X).?

If I have a list of predicates in Prolog like [flies, swims], how can I build a predicate that is the conjunction of all the predicates in the list, ie fliesAndSwims(X) :- flies(X), swims(X).?

或者，有没有更好的方法可以在运行时构建这样的谓词，而无需将组件谓词放在列表中并在需要时从它们构建复合谓词?

Alternatively, is there a better way of building up a predicate at runtime like this without putting the component predicates in a list and building the compound one from them when needed?

所以事实证明这是Prolog中的谓词列表的副本.我以前找到了那个答案，但我认为它只返回给定原子是否与列表中的每个谓词匹配.我没有意识到你可以传递一个变量而不是一个原子，并让它返回每个匹配的案例.

So it turns out this is a duplicate of List of predicates in Prolog. I had previously found that answer, but I thought that it only returned whether a given atom matched every predicate in the list. I didn't realise that you can pass a variable instead of an atom and have it return every case that matches as well.

推荐答案

library(lambda) 功能强大，但也有代价.如果您认为越简单越好"(特别是调试，特别是...)，请考虑

library(lambda) is powerful, but it has a cost. If you think 'simpler is better' (wrt debugging, specially...) consider

call_unary_list([], _).
call_unary_list([P|Ps], X) :-
    call(P, X),
    call_unary_list(Ps, X).

让我们比较一下性能:

compare_call_list :-
    findall(flies, between(1,100000,_), L),
    time(call_unary_list(L, _)),
    time(maplist(call_unary(_), L)),
    time(maplist(X+\Pred^call(Pred,X), L)).

call_unary(X, P) :- call(P, X).

?- compare_call_list.
% 200,000 inferences, 0.123 CPU in 0.123 seconds (100% CPU, 1629657 Lips)
% 300,000 inferences, 0.145 CPU in 0.149 seconds (98% CPU, 2064184 Lips)
% 1,000,001 inferences, 1.286 CPU in 1.297 seconds (99% CPU, 777362 Lips)
true .

call_unary/2 突出显示了 maplist 元谓词所需的参数交换

the call_unary/2 highlights the arguments swap that's required by the maplist meta predicate

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