clojure - 列出列表的所有排列 [英] clojure - list all permutations of a list
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问题描述
说我有这样的集合:
#{word1word2word3}
如何列出这些字可能排序的所有方式,例如
word1 word2 word3
word2 word3 word1
word3 word2 word1
$最简单的方法是使用 /github.com/clojure/math.combinatorics/\"> math.combinatorics :
user> (require'[clojure.math.combinatorics:as combo])
nil
user> (组合/排列#{word1word2word3})
((word1word2word3)(word1word3word2) word3)(word2word3word1)(word3word1word2)(word3word2word1))
编辑:我没有看过math.combinatorics的实现,但是这里是一个惰性版本,因为OP要求一些代码。
(defn permutations [s]
(lazy-seq
(if(seq(rest s))
(应用连接(for [xs]
(map#(cons x%)(permutations(remove#{x} s))))
[s]))
Say I have a set like this:
#{"word1" "word2" "word3"}
How could I list all ways that these words might be ordered, i.e.
word1 word2 word3 word2 word3 word1 word3 word2 word1
etc.
解决方案The easiest way is using math.combinatorics:
user> (require '[clojure.math.combinatorics :as combo]) nil user> (combo/permutations #{"word1" "word2" "word3"}) (("word1" "word2" "word3") ("word1" "word3" "word2") ("word2" "word1" "word3") ("word2" "word3" "word1") ("word3" "word1" "word2") ("word3" "word2" "word1"))
Edit: I haven't looked at the math.combinatorics implementation, but here's a lazy version because OP asked for some code to follow.
(defn permutations [s] (lazy-seq (if (seq (rest s)) (apply concat (for [x s] (map #(cons x %) (permutations (remove #{x} s))))) [s])))
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