clojure - 计算集合中元素的连续出现次数

Question

获得类似东西的惯用方法是什么？

(fn [coll] (function body) [1 0 -1 -1 -1 0 0 0 0 1 1 1 1 1 0 1 -1])
-> [1 1 3 2 1 4 3 2 1 5 4 3 2 1 1 1 1]

只能这样了

(fn [coll] 
  (loop [i 0 r []]
    (if (< i (count coll))
      (let [elem (nth coll i)
            cc (loop [j 1]
                 (if (< (+ i j) (dec (count coll)))
                   (let [nelem (nth coll (+ i j))]
                     (if (= elem nelem)
                       (recur (inc j))
                       j))
                   j))]
        (recur (inc i) (conj r cc)))
      r)))

Answer 1

针对我们认为的问题修改@noisesmith 的简洁解决方案:

(defn countdown-runs [s]
  (->> s
       (partition-by identity)
       (map count)
       (mapcat #(range % 0 -1))))

例如，

(countdown-runs [1 0 -1 -1 -1 0 0 0 0 1 1 1 1 1 0 1 -1])
; (1 1 3 2 1 4 3 2 1 5 4 3 2 1 1 1 1)