从 Haskell 到 Clojure

Question

我将在 count down game 上复习这个 haskell 讲座，我不知道任何haskell，但我对这个问题很感兴趣，我正在尝试将他的代码移植到clojure。

这是我被卡住的部分必须是我在haskell中没有得到的东西，

split                :: [a] -> [([a],[a])]
split []              = [([],[])]
split (x:xs)          = ([],x:xs) : [(x:ls,rs) | (ls,rs)  [([a],[a])]
nesplit               = filter ne . split

ne                   :: ([a],[b]) -> Bool
ne (xs,ys)            = not (null xs || null ys)

exprs                :: [Int] -> [Expr]
exprs []              = []
exprs [n]             = [Val n]
exprs ns              = [e | (ls,rs) 

I have my own split given 1 2 3 4 it spits out,

(((1) (2 3 4)) ((1 2) (3 4)) ((1 2 3) (4)))

(defn split [v]
  (if (= (count v) 1)
    (list (first v))
    (map #(list (take % v) (drop % v)) (range 1 (count v)))))

(defn exprs [v]
  (if (= (count v) 1) 
    v
    (map #(concat (exprs (first %)) (exprs (second %))) v)))

(exprs (split [1 2 3 4]))

that gives me,

java.lang.IllegalArgumentException: Don't know how to create ISeq from: java.lang.Integer

谁能告诉我haskell代码中我缺少什么？

他的完整代码列表可用 here .

Answer 1

就我有限的 Haskell fu 允许我做的而言，这与 Haskell 实现密切相关......

(defn split
  [s]
  (map #(split-at % s) (range 1 (count s))))

(defn ne
  [s]
  (every? (complement empty?) s))

(defn nesplit
  [s]
  (filter ne (split s)))

(declare combine)

(defn exprs
  [s]
  (when-let [s (seq s)]
    (if (next s)
      (for [[ls rs] (nesplit s)
            l       (exprs ls)
            r       (exprs rs)
            e       (combine l r)]
        e)
      s)))

虽然没有测试过。

至于您的错误信息:我认为问题在于，您没有调用 split在 exprs 中递归.然后你得到 1是一个序列是预期的......

其他随机备注: count对于序列而言，时间上是线性的。由于我们只需要知道，我们是否有多个元素，我们可以检查 (next s) 的值。反对 nil .