基于多个事实的聚合解决方案 [英] Aggregate solution over multiple facts

问题描述

尝试创建一个谓词 (timePeriod/2) 来计算特定事实的两个日期之间的时间段.我自己设法做到了这一点，但是当同一列表中存在其他答案"时会遇到问题(即更容易用示例解释).

Trying to create a predicate (timePeriod/2) that calculates the time period between two dates for a specific fact. I've managed to do this by myself, but face issues when 'other answers' exist in the same list (i.e. easier to explain with examples).

我有以下知识库事实；

 popStar('Jackson',1987,1991).
 popStar('Jackson',1992,1996).
 popStar('Michaels',1996,2000).
 popStar('Newcastle',2000,2007).
 popStar('Bowie',2008,2010).

以下函数计算特定事实的日期之间的时间(如下所示).

And the following function, calculates the time between dates for a specific fact (as per below).

谓词(timePeriod/2)-

 timePeriod(PS,X) :-
    bagof((Name,Start,End),popStar(Name,Start,End),PSs),X is End-Start+1)

以鲍伊为例；它返回 X=3 (这是正确的).

Using Bowie as an example; it returns X=3 (which is correct).

但是，当列表中存在重复且有多个答案可用时，谓词只会声明假".以事实杰克逊"为例，我希望能够计算两个事实的两个时间段；同时.

However, when there is repetition in the list, with more than one answer available, the predicate just states 'false'. Using the facts 'Jackson' as an example, I want to be able to calculate both of the time periods for both facts; at the same time.

因此，如果谓词适用于两个杰克逊事实，则谓词 timePeriod 将声明 X=10.

So, if the predicate would work for both of the Jackson facts, the predicate timePeriod would state X=10.

如果有人能建议改变什么以使其正常工作，我将不胜感激.

Would really appreciate if anyone could suggest what to change in order for this to work correctly.

谢谢.

推荐答案

你可能不太明白 foreach/3 是做什么的.我也不认为我完全理解 foreach/3 .我确定这与说 不 相同:

You probably don't quite understand what foreach/3 does. I don't think I fully understand foreach/3 either. I know for sure that it is not the same as say:

for each x in xs:
    do foo(x)

另一件事:Prolog 中的元组"不是您所期望的，它来自 Python 或 Haskell 等语言.这:(a,b,c) 实际上是这样的:','(a,','(b,c)).更好的是使用扁平术语，通用形式是 triple(a,b,c).对于一对，成语是First-Second.

Another thing: "tuples" in Prolog are not what you might expect, coming from a language like Python or Haskell. This: (a,b,c) is actually this: ','(a,','(b,c)). Much better is to use a flat term, the generic form would be triple(a,b,c). For a pair, the idiom is First-Second.

因此，您可以将对 bagof/3 的调用简化为:

So, you can simplify your call to bagof/3 to this:

?- bagof(From-To, pop_star(Name, Start, End), Ts).
Name = 'Bowie',
Ts = [2008-2010] ;
Name = 'Jackson',
Ts = [1987-1991, 1992-1996] ;
Name = 'Michaels',
Ts = [1996-2000] ;
Name = 'Newcastle',
Ts = [2000-2007].

一旦你有了上面的列表，你需要总结差异，这可能是这样的:

Once you have a list as above, you need to sum the differences, which would be maybe something like:

periods_total(Ps, T) :-
    maplist(period_length, Ps, Ls),
    sum_list(Ls, T).

period_length(From-To, Length) :-
    Length is To - From + 1.

然后你可以这样查询:

?- bagof(From-To, pop_star('Jackson', From, To), Ps), periods_total(Ps, T).
Ps = [1987-1991, 1992-1996],
T = 10.

?- bagof(From-To, pop_star(Name, From, To), Ps), periods_total(Ps, T).
Name = 'Bowie',
Ps = [2008-2010],
T = 3 ;
Name = 'Jackson',
Ps = [1987-1991, 1992-1996],
T = 10 ;
Name = 'Michaels',
Ps = [1996-2000],
T = 5 ;
Name = 'Newcastle',
Ps = [2000-2007],
T = 8.

这篇关于基于多个事实的聚合解决方案的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！