我试图将直接出现在彼此旁边的项目分组,只要它们分别位于给定的"白名单"中。分组必须至少包含两个或更多项目。
例如,第一个arg是集合,第二个是白名单。
(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))
这就是我提出的:
(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,但非常难看。我想知道在Clojure中用它做一个更简单的惯用方法吗? (在我发现拆分之前你应该看过fn :)。
我敢打赌,这是一个不错的单行内容,有分区或者其他东西,但是它已经很晚了,我似乎无法让它发挥作用。
答案 0 :(得分:3)
(defn group-sequential [coll white]
(->> coll
(map (set white))
(partition-by nil?)
(filter (comp first next))))
...... Diego Basch's method更整洁的版本。
答案 1 :(得分:0)
这是我的第一次尝试:
(defn group-sequential [xs wl]
(let [s (set wl)
f (map #(if (s %) %) xs)
xs' (partition-by nil? f)]
(remove #(or (nil? (first %)) (= 1 (count %))) xs')))
答案 2 :(得分:0)
好的,我意识到partition-by非常接近我正在寻找的东西,所以我创建了这个函数,它似乎更符合核心内容。
(defn partition-if
"Returns a lazy seq of partitions of items that match the filter"
[pred coll]
(lazy-seq
(when-let [s (seq coll)]
(let [[in more0] (split-with pred s)
[out more] (split-with (complement pred) more0)]
(if (empty? in)
(partition-if pred more)
(cons in (partition-if pred more)))))))
(partition-if #(some #{%} [2 3 6]) [1 2 3 4 5 6 7])
>> ((2 3))
答案 3 :(得分:0)
(defn group-sequential
[coll matches]
(let [matches-set (set matches)]
(->> (partition-by (partial contains? matches-set) coll)
(filter #(clojure.set/subset? % matches-set))
(remove #(< (count %) 2)))))