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社区首页 >问答首页 >我可以在clojure中进行确定性的混洗吗？

问我可以在clojure中进行确定性的混洗吗？
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Stack Overflow用户

提问于 2013-02-12 15:54:48

回答 3查看 1.4K关注 0票数 9

我想对集合进行一些打乱，每次我的程序运行时都是一样的：

这是一种方法：

(def colours ["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"])

(defn colour-shuffle [n] 
  (let [cs (nth (clojure.math.combinatorics/permutations colours) n)]
    [(first cs) (drop 1 cs)]))

; use (rand-int 40320) to make up numbers, then hard code:
(def colour-shuffle-39038 (colour-shuffle 39038))
(def colour-shuffle-28193 (colour-shuffle 28193))
(def colour-shuffle-5667  (colour-shuffle 5667))
(def colour-shuffle-8194  (colour-shuffle 8194))
(def colour-shuffle-13895 (colour-shuffle 13895))
(def colour-shuffle-2345  (colour-shuffle 2345))

colour-shuffle-39038 ; ["white" ("magenta" "blue" "green" "cyan" "yellow" "red" "black")]

但它需要一段时间来评估，而且看起来很浪费，而且相当不优雅。

有没有什么方法可以直接生成混洗39038，而不用生成和消耗所有的序列？

(我已经意识到我可以对它们进行硬编码，或者使用宏将工作带回编译时。这似乎也有点垃圾。)

clojure

permutation

shuffle

deterministic

random

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回答 3

Stack Overflow用户

回答已采纳

发布于 2013-02-12 19:58:06

听起来你想要number permutations

(def factorial (reductions * 1 (drop 1 (range))))

(defn factoradic [n] {:pre [(>= n 0)]}
   (loop [a (list 0) n n p 2]
      (if (zero? n) a (recur (conj a (mod n p)) (quot n p) (inc p)))))

(defn nth-permutation [s n] {:pre [(< n (nth factorial (count s)))]}
  (let [d (factoradic n)
        choices (concat (repeat (- (count s) (count d)) 0) d)]
    ((reduce 
        (fn [m i] 
          (let [[left [item & right]] (split-at i (m :rem))]
            (assoc m :rem (concat left right) 
                     :acc (conj (m :acc) item))))
      {:rem s :acc []} choices) :acc)))

让我们试一下：

(def colours ["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"])

(nth-permutation colours 39038)
=> ["white" "magenta" "blue" "green" "cyan" "yellow" "red" "black"]

...as，但不生成任何其他排列。

很好，但是我们能把它们都买下来吗？

(def x (map (partial nth-permutation colours) (range (nth factorial (count colours)))))

(count x)
=> 40320
(count (distinct x))
=> 40320
(nth factorial (count colours))
=> 40320

请注意，排列是按(按索引排列的字典)顺序生成的：

user=> (pprint (take 24 x))
(["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"]
 ["red" "blue" "green" "yellow" "cyan" "magenta" "white" "black"]
 ["red" "blue" "green" "yellow" "cyan" "black" "magenta" "white"]
 ["red" "blue" "green" "yellow" "cyan" "black" "white" "magenta"]
 ["red" "blue" "green" "yellow" "cyan" "white" "magenta" "black"]
 ["red" "blue" "green" "yellow" "cyan" "white" "black" "magenta"]
 ["red" "blue" "green" "yellow" "magenta" "cyan" "black" "white"]
 ["red" "blue" "green" "yellow" "magenta" "cyan" "white" "black"]
 ["red" "blue" "green" "yellow" "magenta" "black" "cyan" "white"]
 ["red" "blue" "green" "yellow" "magenta" "black" "white" "cyan"]
 ["red" "blue" "green" "yellow" "magenta" "white" "cyan" "black"]
 ["red" "blue" "green" "yellow" "magenta" "white" "black" "cyan"]
 ["red" "blue" "green" "yellow" "black" "cyan" "magenta" "white"]
 ["red" "blue" "green" "yellow" "black" "cyan" "white" "magenta"]
 ["red" "blue" "green" "yellow" "black" "magenta" "cyan" "white"]
 ["red" "blue" "green" "yellow" "black" "magenta" "white" "cyan"]
 ["red" "blue" "green" "yellow" "black" "white" "cyan" "magenta"]
 ["red" "blue" "green" "yellow" "black" "white" "magenta" "cyan"]
 ["red" "blue" "green" "yellow" "white" "cyan" "magenta" "black"]
 ["red" "blue" "green" "yellow" "white" "cyan" "black" "magenta"]
 ["red" "blue" "green" "yellow" "white" "magenta" "cyan" "black"]
 ["red" "blue" "green" "yellow" "white" "magenta" "black" "cyan"]
 ["red" "blue" "green" "yellow" "white" "black" "cyan" "magenta"]
 ["red" "blue" "green" "yellow" "white" "black" "magenta" "cyan"])

票数 3

Stack Overflow用户

发布于 2015-09-08 08:26:53

clojure.core/shuffle使用java.util.Collection/shuffle，它使用一个可选的随机数生成器。clojure.core/shuffle不使用此参数，但您可以使用它来创建带有附加种子值参数的shuffle变体，并使用该种子值创建传递给java.util.Collection/shuffle的随机数生成器

(defn deterministic-shuffle
  [^java.util.Collection coll seed]
  (let [al (java.util.ArrayList. coll)
        rng (java.util.Random. seed)]
    (java.util.Collections/shuffle al rng)
    (clojure.lang.RT/vector (.toArray al))))

票数 9

Stack Overflow用户

发布于 2013-02-12 16:24:36

我的建议是:使用闭包，并且只计算一次排列。然后重用这些排列来从中选择一个元素。在你的函数colour-shuffle中，每次调用都会重新计算排列，效率不是很高。

(use 'clojure.math.combinatorics)

(def colours ["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"])

(def select-permutation
  (let [perms (permutations colours)]
    (fn [n]
      (nth perms n))))

(defn colour-shuffle [n] 
  (let [cs (nth (permutations colours) n)]
    [(first cs) (drop 1 cs)]))

(time (do  (def colour-shuffle-39038 (colour-shuffle 39038))
           (def colour-shuffle-28193 (colour-shuffle 28193))
           (def colour-shuffle-5667  (colour-shuffle 5667))
           (def colour-shuffle-8194  (colour-shuffle 8194))
           (def colour-shuffle-13895 (colour-shuffle 13895))
           (def colour-shuffle-2345  (colour-shuffle 2345))))

(time (do (def select-permutation-39038 (select-permutation 39038))
          (def select-permutation-28193 (select-permutation 28193))
          (def select-permutation-5667  (select-permutation 5667))
          (def select-permutation-8194  (select-permutation 8194))
          (def select-permutation-13895 (select-permutation 13895))
          (def select-permutation-2345  (select-permutation 2345))))

(time (do  (def colour-shuffle-39038 (colour-shuffle 39038))
           (def colour-shuffle-28193 (colour-shuffle 28193))
           (def colour-shuffle-5667  (colour-shuffle 5667))
           (def colour-shuffle-8194  (colour-shuffle 8194))
           (def colour-shuffle-13895 (colour-shuffle 13895))
           (def colour-shuffle-2345  (colour-shuffle 2345))))

(time (do (def select-permutation-39038 (select-permutation 39038))
          (def select-permutation-28193 (select-permutation 28193))
          (def select-permutation-5667  (select-permutation 5667))
          (def select-permutation-8194  (select-permutation 8194))
          (def select-permutation-13895 (select-permutation 13895))
          (def select-permutation-2345  (select-permutation 2345))))

输出：

"Elapsed time: 129.023 msecs"
"Elapsed time: 65.472 msecs"
"Elapsed time: 182.226 msecs"
"Elapsed time: 5.715 msecs"

请注意，使用select-permutation的第二次运行时间甚至更快。这是因为惰性序列的结果在计算后被缓存。在lazy-seq中非常深入地请求一个元素将导致所有前面的元素也被计算出来。这就是为什么第一次运行需要更长的时间。当从一个新的lazy-seq请求第39039个元素时，将导致至少计算39040个元素(以32个元素为单位)。

顺便说一句，如果你的随机数无论如何都要硬编码，你也可以硬编码上面检索到的排列。

票数 1

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/14836414

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