从序言中的列表中过滤掉大量数字

发布于 2024-10-06 00:44:22 字数 243 浏览 2 评论 0原文

我想编写一个函数，通过删除小于或等于特定数字的所有内容来过滤数字列表。该函数将采用两个参数：数字列表和要过滤的数字。该函数应返回一个列表，其中包含大于过滤器编号的所有数字。

有时像这样：

filter_num_list(L1,N,L2) :- ...

test_filter_num_list :- filter_num_list([1,2,3,4,5,6,7,8,9],5,[5,6,7,8,9]).

原文

I want to write a function which filters a list of numbers by removing everything less than or equal to a specific number. The function will take two parameters: a list of numbers and the number to filter. The function should returns a list which has all the numbers larger than the filter number.

Sometime like this:

filter_num_list(L1,N,L2) :- ...

test_filter_num_list :- filter_num_list([1,2,3,4,5,6,7,8,9],5,[5,6,7,8,9]).

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薄荷→糖丶微凉 2024-10-13 00:44:22

另请参见库谓词，例如 include/3 和 except/3：

?- include(=<(5), [1,2,3,4,5,6,7,8,9], Is).
Is = [5, 6, 7, 8, 9].

See also library predicates like include/3 and exclude/3:

?- include(=<(5), [1,2,3,4,5,6,7,8,9], Is).
Is = [5, 6, 7, 8, 9].

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梦幻的味道 2024-10-13 00:44:22

使用 meta-predicate tfilter/3 和具体化的clpfd 约束(# <)/3，您可以跟上逻辑纯度并立即表达您想要的内容！

:- use_module(library(clpfd)).

这是我使用 SWI-Prolog 版本 7.1.37 运行的查询：

?- tfilter(#<(5),[1,2,3,4,5,6,7,8,9],Xs).
Xs = [6,7,8,9].                             % succeeds deterministically
false.

因为代码是单调，我们还可以提出更一般性的问题并得到逻辑上合理的答案。

?- tfilter(#<(7),[A,B,C],Xs).
Xs = [],      A in inf..7, B in inf..7, C in inf..7 ;
Xs = [C],     A in inf..7, B in inf..7, C in 8..sup ;
Xs = [B],     A in inf..7, B in 8..sup, C in inf..7 ;
Xs = [B,C],   A in inf..7, B in 8..sup, C in 8..sup ;
Xs = [A],     A in 8..sup, B in inf..7, C in inf..7 ;
Xs = [A,C],   A in 8..sup, B in inf..7, C in 8..sup ;
Xs = [A,B],   A in 8..sup, B in 8..sup, C in inf..7 ;
Xs = [A,B,C], A in 8..sup, B in 8..sup, C in 8..sup ;
false.

With meta-predicate tfilter/3 and the reified clpfd constraint (#<)/3, you can keep up logical-purity and express what you want in no time!

:- use_module(library(clpfd)).

Here's a query that I ran with SWI-Prolog version 7.1.37:

?- tfilter(#<(5),[1,2,3,4,5,6,7,8,9],Xs).
Xs = [6,7,8,9].                             % succeeds deterministically
false.

As the code is monotone, we can also ask more general queries and get logically sound answers.

?- tfilter(#<(7),[A,B,C],Xs).
Xs = [],      A in inf..7, B in inf..7, C in inf..7 ;
Xs = [C],     A in inf..7, B in inf..7, C in 8..sup ;
Xs = [B],     A in inf..7, B in 8..sup, C in inf..7 ;
Xs = [B,C],   A in inf..7, B in 8..sup, C in 8..sup ;
Xs = [A],     A in 8..sup, B in inf..7, C in inf..7 ;
Xs = [A,C],   A in 8..sup, B in inf..7, C in 8..sup ;
Xs = [A,B],   A in 8..sup, B in 8..sup, C in inf..7 ;
Xs = [A,B,C], A in 8..sup, B in 8..sup, C in 8..sup ;
false.

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叹梦 2024-10-13 00:44:22

尝试类似的东西：

filter_num_list([],N,[]) :- true.  
filter_num_list([H|T],N,[H|S]) :- H > N,filter_num_list(T,N,S).  
filter_num_list([H|T],N,S) :- N >= H, filter_num_list(T,N,S).

try something like:

filter_num_list([],N,[]) :- true.  
filter_num_list([H|T],N,[H|S]) :- H > N,filter_num_list(T,N,S).  
filter_num_list([H|T],N,S) :- N >= H, filter_num_list(T,N,S).

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