[英]Prolog predicate that compares values in facts
This is my first time using Prolog 这是我第一次使用Prolog
I have employees 我有员工
employee(eID,firstname,lastname,month,year)
example : 例如:
employee(1,liz,white,4,2000).
employee(2,ted,johnson,5,1998).
I want to make a predicate senior(X,Y)
that will return true if the first employee is older in the company. 我想做一个谓词
senior(X,Y)
,如果第一个员工在公司中年长,它将返回true。
I have this: 我有这个:
senior(X,Y) : -
employee(X,firstname,lastname,month,year),
employee(Y,firstname,lastname,month,year),
X.year < Y.year.
but this always return false. 但这总是返回false。 I can't understand the reason.
我不明白原因。
How can I make this predicate work? 如何使该谓词起作用?
In Prolog, variables start with either an underscore or an upper case letter. 在Prolog中,变量以下划线或大写字母开头。 Eg
firstname
is an atom , ie a constant, but FirstName
is a variable . 例如,
firstname
是一个原子 ,即一个常数,但是FirstName
是一个变量 。 But, in your specific question, you don't care about the employee names. 但是,在您的特定问题中,您无需关心员工姓名。 Thus, you can replace those arguments by the anonymous variable:
因此,您可以将这些参数替换为匿名变量:
senior(X,Y) : -
employee(X, _, _, Xmonth, Xyear),
employee(Y, _, _, Ymonth, Yyear),
...
Can you now complete the code by writing the necessary comparisons using the Xmonth
, Xyear
, Ymonth
, and Yyear
variables? 您现在可以通过使用
Xmonth
, Xyear
, Ymonth
和Yyear
变量编写必要的比较来完成代码吗?
Is it mandatory that you do this with one rule? 用一条规则来执行此操作是强制性的吗? You could use one rule for comparing employees who were hired in different years, and a second rule for comparing employees who were hired in the same year.
您可以使用一个规则比较不同年份的雇员,使用第二个规则比较相同年份的雇员。 To expand on this, let's say you have employees listed this way:
为了对此进行扩展,假设您以这种方式列出了员工:
employee(eid,year,month,day)
and, of course, a list of employees. 当然还有雇员名单。 You could use the following three rules:
您可以使用以下三个规则:
% For employees that were hired in different years.
senior(Eid1,Eid2) :-
employee(Eid1,X,_,_),
employee(Eid2,Y,_,_),
X<Y.
% For employees that were hired in the same year, different month.
senior(Eid1,Eid2) :-
employee(Eid1,Year,X,_);
employee(Eid2,Year,Y,_); % Notice how one common variable "Year" is used
X<Y.
% For employees that were hired in the same year, same month, different day,
% the rule is "expanded" from the previous one.
senior(Eid1,Eid2) :-
employee(Eid1,Year,Month,X);
employee(Eid2,Year,Month,Y);
X<Y.
Make sure you don't forget and replace "Year" and/or "Month" with underscores, because then somebody hired on 2010-01-01 (ISO 8601) would be shown as senior to someone hired on 2005-12-12 . 确保您不要忘记并用下划线代替“ Year”和/或“ Month”,因为这样,在2010-01-01(ISO 8601)雇用的人员将比在2005-12-12雇用的人员显示为高级。
Then again, perhaps you should catalogue all dates in ISO 8601:2004 . 再说一次,也许您应该在ISO 8601:2004中对所有日期进行分类。 No matter how big your employee list, you could write a small script to convert
无论您的员工人数多大,都可以编写一个小的脚本进行转换
employee(eID,firstname,lastname,month,year)
to 至
employee(eID,firstname,lastname,yyyymm)
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