如何在SWI-Prolog中从事实生成姓名与国家的配对列表？

Hey there! Let's work through this Prolog problem together. You've got a set of person/3 facts storing each person's name, age, and country, and you want a predicate filter/1 that spits out a list of (Name, Country) pairs. Here's a straightforward solution:

Full Code Implementation #

% Predefined person facts
person(jessica,19,usa).
person(james,18,uk).
person(eric,34,italy).
person(jake,24,france).

% Predicate to generate name-country pairs
filter(ResultList) :-
    findall( (Name, Country), person(Name, _, Country), ResultList ).

Breakdown of the Solution #

The magic here comes from Prolog's built-in findall/3 predicate—it's perfect for collecting all matching results into a list. Let's break down each part:

• First argument of findall/3: (Name, Country) is the structure we want to collect for each matching person. This creates a tuple pairing the name and country.
• Second argument: person(Name, _, Country) is our query. The _ is an anonymous variable—we use it to ignore the age value since we don't need it in our final list.
• Third argument: ResultList is the variable that will hold the final list of pairs once the query completes.

Example Usage & Output #

When you run this in your Prolog interpreter:

?- filter(L).

You'll get exactly the result you're looking for:

L = [(jessica, usa), (james, uk), (eric, italy), (jake, france)]

Bonus: Extending the Predicate #

If you ever need to filter for specific conditions (like only adults), you can easily modify the query inside findall/3. For example:

% Filter only people over 20
filter_adults(ResultList) :-
    findall( (Name, Country), (person(Name, Age, Country), Age > 20), ResultList ).

Calling filter_adults(L) would return L = [(eric, italy), (jake, france)].

内容的提问来源于stack exchange，提问作者Angelrina