如何在 Clojure 中为Burrows-Wheeler 变换惯用地旋转字符串？
我想出了这个，它使用(cycle "string")，但感觉有点必要：
(let [s (str "^" "banana" "|")
l (count s)
c (cycle s)
m (map #(take l (drop % c)) (range l))]
(apply map str m))
=> ("^banana|" "banana|^" "anana|^b" "nana|^ba" "ana|^ban" "na|^bana" "a|^banan" "|^banana")
我不确定这是否符合代码高尔夫的条件。有没有更清洁的方法来做到这一点？
我会做：
(defn bwrot [s]
(let [s (str "^" s "|")]
(for [i (range (count s))]
(str (subs s i) (subs s 0 i)))))
或者：
(defn bwrot [s]
(let [n (+ 2 (count s))
s (str "^" s "|^" s "|")]
(for [i (range n)]
(subs s i (+ i n)))))
第二个应该分配更少（一个字符串而不是每次迭代三个）。
曾经有一个rotations功能clojure.contrib.seq可能值得寻找灵感。来源转载如下：
(defn rotations
"Returns a lazy seq of all rotations of a seq"
[x]
(if (seq x)
(map
(fn [n _]
(lazy-cat (drop n x) (take n x)))
(iterate inc 0) x)
(list nil)))
然后你可以做类似的事情：
(apply map str (rotations "^banana|"))
; => ("^banana|" "banana|^" "anana|^b" "nana|^ba" "ana|^ban" "na|^bana" "a|^banan" "|^banana")
逐步调用partition作品：
(defn bwt[s]
(let [s' (str "^" s "|")
c (cycle s')
l (count s')]
(map last (sort (apply map str (take l (partition l 1 c)))))))
(apply str (bwt "banana"))
=> "|bnn^aaa"
如果我不关心效率或字符数，我会写如下内容：
(defn rotate-string
[s]
(apply str (concat (drop 1 s) (take 1 s))))
(defn string-rotations
[s]
(->> s
(iterate rotate-string)
(take (count s))))
(rotate-string "^banana|") ; "banana|^"
(string-rotations "^banana|") ; ("^banana|" "banana|^" "anana|^b" "nana|^ba" "ana|^ban" "na|^bana" "a|^banan" "|^banana")
特别是，将单个旋转分解为它自己的函数。
完成旋转的另一种方法是使用“双字符串”（即将字符串连接到自身）并使用子字符串。
(defn rotations [strng]
(let [indices (range (count strng))
doublestr (str strng strng)]
(map #(subs doublestr % (+ % (count strng))) indices)))
(rotations "^banana|")
;;(^banana| banana|^ anana|^b nana|^ba ana|^ban na|^bana a|^banan |^banana)
“foo”的旋转：
n“foo”的长度= 3n“foofoo”的所有以indices0、1、2 开头并且长度相同的子字符串n