Question

我对def如何以不同的方式绑定变量感到困惑。有人可以向我解释为什么会有效：

(def leven  
  (memoize (fn [x y]
  (cond (empty? x) (count y)
      (empty? y) (count x)
       :else (min (+ (leven (rest x) y) 1)
                  (+ (leven x (rest y)) 1)
                  (+ (leven (rest x) (rest y)) (if (= (first x) (first y)) 0 1))
              )
  )))
)

但是当我尝试将函数声明为无法编译时：

(def leven  
    (let [l (memoize (fn [x y]
    (cond (empty? x) (count y)
          (empty? y) (count x)
           :else (min (+ (l (rest x) y) 1)
                      (+ (l x (rest y)) 1)
                      (+ (l (rest x) (rest y)) (if (= (first x) (first y)) 0 1))
                  )
    )
    ))]
    (l x y)
    )
)

编辑：这是有效的，使用Ankur展示的技术。

(defn leven [x y] 
(let [l (memoize (fn [f x y]
(cond (empty? x) (count y)
      (empty? y) (count x)
       :else (min (+ (f f (rest x) y) 1)
                  (+ (f f x (rest y)) 1)
                  (+ (f f (rest x) (rest y)) (if (= (first x) (first y)) 0 1))
              )
)
))
magic (partial l l)]
(magic x y)
)
)

Answer 1

下面是这样一个例子来做你要求的。我只是为了简单而使用factorial，并在factorial中添加println以确保memoization工作正常

(let [fact (memoize (fn [f x] 
                       (println (str "Called for " x))
                       (if (<= x 1) 1 (* x  (f f (- x 1))))))
      magic (partial fact fact)] 
     (magic 10)
     (magic 11))

首先计算阶乘10然后再计算11，在这种情况下，它应该不会再次将因子称为10到1，因为它已被记忆化。

Called for 10
Called for 9
Called for 8
Called for 7
Called for 6
Called for 5
Called for 4
Called for 3
Called for 2
Called for 1
Called for 11
39916800

Answer 2

let形式按顺序绑定名称，因此在第二个函数定义中，当您尝试引用它时，名称l不存在。你可以使用letfn（带有一些小的mod）或者给定义的函数命名，而是改为引用它，如下所示：

(def leven  
  (let [l (memoize (fn SOME-NAME [x y]
    (cond 
      (empty? x) (count y)
      (empty? y) (count x)
      :else (min (+ (SOME-NAME (rest x) y) 1)
                 (+ (SOME-NAME x (rest y)) 1)
                 (+ (SOME-NAME (rest x) (rest y)) (if (= (first x) (first y)) 0 1))))))]
l))

您可能会注意到我将let的返回值更改为l本身，因为这是您希望leven绑定的内容。 (l x y)存在问题，因为它仅引用了函数本地的绑定，let无法访问。

里面的递归让函数

2 个答案: