我想制作一些套装，每次我的程序运行都是一样的：
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")]
但评估需要一段时间，而且看起来很浪费而且相当不优雅。
是否有某种方法可以直接生成shuffle 39038，而无需生成和消耗所有序列？
（我已经意识到我可以对它们进行硬编码，或者用宏来重新编译时间。这看起来有点垃圾。）
答案 0 :(得分：6)
clojure.core/shuffle使用java.util.Collection/shuffle，它带有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一个带有附加种子值参数的shuffle变体，并使用该种子值创建一个随机数生成器以传递给java.util.Collection/shuffle：
(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))))
答案 1 :(得分：3)
听起来像是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)))
我们试一试：
(def colours ["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"])
(nth-permutation colours 39038)
=> ["white" "magenta" "blue" "green" "cyan" "yellow" "red" "black"]
......如同问题一样，但没有产生任何其他排列。
嗯，但是我们能得到它们吗？
(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（按字典顺序排列）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"])
答案 2 :(得分：1)
我的建议：使用闭包并仅计算一次排列。然后重新使用这些排列从中选择一个元素。在您的函数colour-shuffle中，每次调用都会重新计算排列效果。
(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的第二次运行速度更快。这是因为延迟序列的结果在计算后被缓存。在lazy-seq中非常深入地请求元素将导致所有前面的元素也被计算。这就是第一次运行需要更长时间的原因。当从新的lazy-seq请求第39039个元素时，将导致至少计算39040个元素（以32个为单位）。
顺便说一句，如果您的随机数字无论如何都要硬编码，您也可以对上面检索的排列进行硬编码。