我可以在for comprehension中使用map进行绑定吗？

Question

(defn get-pins [observed]
  (let [num (count observed)
        keypad {\1 [1 2 4] \2 [2 1 5 3] \3 [3 2 6]
                \4 [4 1 5 7] \5 [5 2 6 8 4] \6 [6 3 5 9]
                \7 [7 4 8] \8 [8 5 9 0 7] \9 [9 6 8] \0 [0 8]}
        observed-map (zipmap [:a :b :c :d] (map keypad observed))]
    (case num
      1 (for [a (:a observed-map)]
          (str a))
      2 (for [a (:a observed-map)
              b (:b observed-map)]
         (str a b))
      3 (for [a (:a observed-map)
              b (:b observed-map)
              c (:c observed-map)]
         (str a b c))
      4 (for [a (:a observed-map)
              b (:b observed-map)
              c (:c observed-map)
              d (:d observed-map)]
         (str a b c d))
       "default")))

Answer 1

以下操作将实现您所需的结果

(defn get-pins [observed]
  (let [num (count observed)
        keypad {\1 [1 2 4] \2 [2 1 5 3] \3 [3 2 6]
                \4 [4 1 5 7] \5 [5 2 6 8 4] \6 [6 3 5 9]
                \7 [7 4 8] \8 [8 5 9 0 7] \9 [9 6 8] \0 [0 8]}
        observed-map (zipmap [:a :b :c :d] (map keypad observed))]
    (for [a (:a observed-map) 
          b (or (:b observed-map) [nil]) 
          c (or (:c observed-map) [nil]) 
          d (or (:d observed-map) [nil])] 
      (str a b c d))))

[nil]的值将产生一个单个的“迭代”（带有的值），当调用str时，它会产生一个空字符串。

Comments

感谢帮助

Answer 2

在这里您提出了什么问题？能详细说明一下吗？

Comments

是否可以将情况语句更改为更灵活的解决方案，比如单个for语句，并且仍然能够处理各种长度的字符串。

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不可以，for不适用于这种情况。