[英]How to count and sum recursively in Prolog
I'm trying to figure out how to use recursion on count and sum rules.我试图弄清楚如何在计数和求和规则上使用递归。
I usually do it with lists, using findall and length or findall and sum_list, but I'm not sure if that's my best option on all cases.我通常使用列表,使用 findall 和 length 或 findall 和 sum_list,但我不确定这是否是所有情况下的最佳选择。
This is my approach with lists:这是我的列表方法:
%person(name, surname, age)
person('A', 'H', 22).
person('B', 'G', 24).
person('C', 'F', 20).
person('D', 'E', 44).
person('E', 'D', 45).
person('F', 'C', 51).
person('G', 'B', 40).
person('H', 'A', 51).
count_person(Total_count) :- % rule to count how many person are.
findall(N, person(N, _, _), List),
length(List, Total_count).
sum_ages(Total_sum) :- % rule to sum all the ages.
findall(Age, person(_, _, Age), List),
sum_list(List, Total_sum).
or here: https://swish.swi-prolog.org/p/cswl.pl或在这里: https : //swish.swi-prolog.org/p/cswl.pl
How should I do this using recursion?我应该如何使用递归来做到这一点?
I do not have an elegant solution.我没有一个优雅的解决方案。 But with
retract
and assert
you can control the recursion:但是使用
retract
和assert
你可以控制递归:
:- dynamic([person/3,person1/3]).
count_person(N) :-
count_person(0,N).
count_person(Acc,N) :-
retract(person(A,B,C)),
!,
assert(person1(A,B,C)),
N1 is Acc+1,
count_person(N1,N).
count_person(N,N) :-
clean_db.
clean_db :-
retract(person1(A,B,C)),
assert(person(A,B,C)),
fail.
clean_db.
You should take a look at library( aggregate ).你应该看看 library( aggregate )。
For instance:例如:
count_person(Total_count) :-
aggregate(count, A^B^C^person(A,B,C), Total_count).
or the simpler form (try to understand the difference, it's aa good way to learn the basic about variables quantification )或者更简单的形式(尝试理解差异,这是学习变量量化基础的好方法)
count_person(Total_count) :-
aggregate_all(count, person(_,_,_), Total_count).
The library has grown out of the necessity to simplify the implementation of typical aggregation functions available in SQL (since Prolog is relational at heart):该库的发展是为了简化 SQL 中可用的典型聚合函数的实现(因为 Prolog 是关系型的):
sum_ages(Total_sum) :-
aggregate(sum(Age), A^B^person(A,B,Age), Total_sum).
You can also get combined aggregates in a step.您还可以在一个步骤中获得组合聚合。 Average is readily implemented:
平均值很容易实现:
ave_ages(Ave) :-
aggregate(t(count,sum(Age)), A^B^person(A,B,Age), t(Count,Sum)), Ave is Sum/Count.
If you implement using count_person/1 and sum_ages/1 the interpreter would scan twice the goal...如果您使用 count_person/1 和 sum_ages/1 实现,解释器将扫描目标的两倍......
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