algorithm - Clojure Group 连续出现 - 改进功能

Question

我正在尝试对直接相邻出现的项目进行分组，只要它们都在给定的“白名单”中即可。分组必须至少包含两个或更多项目。

例如，第一个参数是集合，第二个参数是白名单。

(group-sequential [1 2 3 4 5] [2 3])
>> ((2 3))

(group-sequential ["The" "quick" "brown" "healthy" "fox" "jumped" "over" "the" "fence"] 
                  ["quick" "brown" "over" "fox" "jumped"])
>> (("quick" "brown") ("fox" "jumped" "over"))

(group-sequential [1 2 3 4 5 6 7] [2 3 6])
>> ((2 3))

这是我想出的:

(defn group-sequential
  [haystack needles]
  (loop [l haystack acc '()]
    (let [[curr more] (split-with #(some #{%} needles) l)]
      (if (< (count curr) 2)
        (if (empty? more) acc (recur (rest more) acc))
        (recur (rest more) (cons curr acc))))))

它有效，但非常丑陋。我想知道在 Clojure 中是否有更简单的惯用方法？ (在我发现 split-with 之前你应该已经看过 fn :)

我打赌有一个很好的单行代码，带有 partition-by 之类的东西，但是已经晚了，我似乎无法让它工作。

Answer 1

(defn group-sequential [coll white] 
  (->> coll
       (map (set white))
       (partition-by nil?)
       (filter (comp first next))))

... Diego Basch's method 的更简洁版本.