clojure 将集合分成越来越大的块

Question

嗨，我是 clojure 新手，

我正在尝试创建一个函数，该函数将集合拆分为大小不断增加的块

(apply #(#(take %) (range 1 n)) col)

其中 n 是块的数量

预期输出示例：n = 4 且 col =（范围 1 4）

(1) (2 3) (4)

n = 7 和 col =（范围 1 7）

(1) (2 3) (4 5 6) (7)

Answer 1

你可以使用这样的东西：

(defn partition-inc
  "Partition xs at increasing steps of n" 
  [n xs] 
  (lazy-seq 
    (when (seq xs) 
      (cons (take n xs) 
            (partition-inc (inc n) (drop n xs))))))

; (println (take 5 (partition-inc 1 (range))))
; ? ((0) (1 2) (3 4 5) (6 7 8 9) (10 11 12 13 14))

或者如果你想有更多的影响，你也可以提供一个大小的序列（行为与上面相同，如果传递(iterate inc 1)给sizes：

(defn partition-sizes
  "Partition xs into chunks given by sizes"
  [sizes xs]
  (lazy-seq
    (when (and (seq sizes) (seq xs))
      (let [n (first sizes)]
        (cons (take n xs) (partition-sizes (rest sizes) (drop n xs)))))))

; (println (take 5 (partition-sizes (range 1 10 2) (range))))
; ? ((0) (1 2 3) (4 5 6 7 8) (9 10 11 12 13 14 15) (16 17 18 19 20 21 22 23 24))