Clojure 如何在core.logic中分解映射?
我相信在core.logic中解构映射时遇到问题。我有以下代码:Clojure 如何在core.logic中分解映射?,clojure,clojure-core.logic,Clojure,Clojure Core.logic,我相信在core.logic中解构映射时遇到问题。我有以下代码: ... used clojure.core.logic ... required clojure.core.logic.arithmetic as logic.arithmetic. (def hand ({:rank 9, :suit :hearts} {:rank 13, :suit :clubs} {:rank 6, :suit :spades}
... used clojure.core.logic
... required clojure.core.logic.arithmetic as logic.arithmetic.
(def hand ({:rank 9, :suit :hearts}
{:rank 13, :suit :clubs}
{:rank 6, :suit :spades}
{:rank 8, :suit :hearts}
{:rank 12, :suit :clubs}))
(run* [q]
(fresh [v w x y z] ;;cards
(== q [v w x y z])
(membero v hand)
(membero w hand)
(membero x hand)
(membero y hand)
(membero z hand)
(fresh [a b c d e] ;;ranks
(== {:rank a} v)
(== {:rank b} w)
(== {:rank c} x)
(== {:rank d} y)
(== {:rank e} z)
(logic.arithmetic/>= a b)
(logic.arithmetic/>= b c)
(logic.arithmetic/>= c d)
(logic.arithmetic/>= d e))
(distincto q)))
它返回空列表(),表示未找到匹配项。我认为这是代码(={:rank a}v)部分的一个问题。我试图简单地返回q,其中q是一个按:降序排列的映射向量。显然,您必须在映射上进行精确匹配。这意味着您可以创建垃圾变量来捕获您不感兴趣的任何内容的值。看起来很奇怪,但还好
(run* [q]
(fresh [v w x y z] ;;cards
(== q [v w x y z])
(membero v hand)
(membero w hand)
(membero x hand)
(membero y hand)
(membero z hand)
(fresh [a b c d e, f g h i j] ;;ranks, garbage
(== {:rank a :suit f } v)
(== {:rank b :suit g } w)
(== {:rank c :suit h } x)
(== {:rank d :suit i } y)
(== {:rank e :suit j } z)
(logic.arithmetic/>= a b)
(logic.arithmetic/>= b c)
(logic.arithmetic/>= c d)
(logic.arithmetic/>= d e))
(distincto q)))
最后,这里是一个更简洁、更快、更混乱的版本
(run* [q]
(fresh [v w x y z] ;;cards
(permuteo hand q)
(== q [v w x y z])
(fresh [a b c d e, f g h i j] ;;ranks, garbage
(== {:rank a :suit f } v)
(== {:rank b :suit g } w)
(logic.arithmetic/>= a b)
(== {:rank c :suit h } x)
(logic.arithmetic/>= b c)
(== {:rank d :suit i } y)
(logic.arithmetic/>= c d)
(== {:rank e :suit j } z)
(logic.arithmetic/>= d e))))
如果不需要引用逻辑变量,则实际上不需要将其命名为:
(== {:rank a :suit (lvar)} v)
我发现自己越来越多地使用
(lvar)
。它通常使代码比一次性使用的变量更清晰,但我真希望有更好的方法来表达这一点。现在可以使用最新的core.logic 0.8.3版编写更简洁的解决方案:
(ns cards
(:refer-clojure :exclude [==])
(:use [clojure.core.logic])
(:require [clojure.core.logic.fd :as fd]))
(def hand
[{:rank 9, :suit :hearts}
{:rank 13, :suit :clubs}
{:rank 6, :suit :spades}
{:rank 8, :suit :hearts}
{:rank 12, :suit :clubs}])
(defn ranko [card rank]
(featurec card {:rank rank}))
(run* [v w x y z :as q]
(permuteo hand q)
(fresh [a b c d e]
(ranko v a) (ranko w b) (ranko x c)
(fd/>= a b) (fd/>= b c)
(ranko y d) (ranko z e)
(fd/>= c d) (fd/>= d e)))
现在可以使用
featurec
只匹配地图的一部分。我还建议使用新的有限域名称空间,而不是算术名称空间。