实施 Clojure conditional/branching 转换器

Implementing Clojure conditional/branching transducer

我正尝试在 Clojure 中制作一个条件转换器,如下所示:

(defn if-xf
  "Takes a predicate and two transducers.
   Returns a new transducer that routes the input to one of the transducers
   depending on the result of the predicate."
  [pred a b]
  (fn [rf]
    (let [arf (a rf)
          brf (b rf)]
      (fn
        ([] (rf))
        ([result]
           (rf result))
        ([result input]
           (if (pred input)
             (arf result input)
             (brf result input)))))))

它非常有用,因为它可以让你做这样的事情:

;; multiply odd numbers by 100, square the evens.  
(= [0 100 4 300 16 500 36 700 64 900]
    (sequence
          (if-xf odd? (map #(* % 100)) (map (fn [x] (* x x))))
          (range 10)))

但是,此条件转换器不能很好地与在其 1-arity 分支中执行清理的转换器一起工作:

;; negs are multiplied by 100, non-negs are partitioned by 2
;; BUT! where did 6 go?
;; expected: [-600 -500 -400 -300 -200 -100 [0 1] [2 3] [4 5] [6]]
;;
(= [-600 -500 -400 -300 -200 -100 [0 1] [2 3] [4 5]]
 (sequence
  (if-xf neg? (map #(* % 100)) (partition-all 2))
  (range -6 7)))

是否可以调整 if-xf 的定义来处理带清理的换能器的情况?

我正在尝试这个,但有奇怪的行为:

(defn if-xf
  "Takes a predicate and two transducers.
   Returns a new transducer that routes the input to one of the transducers
   depending on the result of the predicate."
  [pred a b]
  (fn [rf]
    (let [arf (a rf)
          brf (b rf)]
      (fn
        ([] (rf))
        ([result]
           (arf result) ;; new!
           (brf result) ;; new!
           (rf result))
        ([result input]
           (if (pred input)
             (arf result input)
             (brf result input)))))))

具体来说,冲洗发生在最后:

;; the [0] at the end should appear just before the 100.
(= [[-6 -5] [-4 -3] [-2 -1] 100 200 300 400 500 600 [0]]
      (sequence
       (if-xf pos? (map #(* % 100)) (partition-all 2))
       (range -6 7)))

有没有一种方法可以使这个 branching/conditional 转换器无需将整个输入序列存储在该转换器的本地状态中(即在清理时在 1-arity 分支中进行所有处理)?

思路是每次换能器切换完成。 IMO 这是唯一无需缓冲的方法:

(defn if-xf
  "Takes a predicate and two transducers.
   Returns a new transducer that routes the input to one of the transducers
   depending on the result of the predicate."
  [pred a b]
  (fn [rf]
    (let [arf (volatile! (a rf))
          brf (volatile! (b rf))
          a? (volatile! nil)]
      (fn
        ([] (rf))
        ([result]
         (let [crf (if @a? @arf @brf)]
           (-> result crf rf)))
        ([result input]
         (let [p? (pred input)
               [xrf crf] (if p? [@arf @brf] [@brf @arf])
               switched? (some-> @a? (not= p?))]
           (if switched?
             (-> result crf (xrf input))
             (xrf result input))
           (vreset! a? p?)))))))
(sequence (if-xf pos? (map #(* % 100)) (partition-all 2)) [0 1 0 1 0 0 0 1])
; => ([0] 100 [0] 100 [0 0] [0] 100)

我认为你的问题定义不明确。当传感器有状态时,你到底想发生什么?例如,您希望这样做:

(sequence
  (if-xf even? (partition-all 3) (partition-all 2))
  (range 14))

此外,有时reducing函数在开始和结束时都有工作要做,不能随意重启。例如,这是一个计算均值的 reducer:

(defn mean
  ([] {:count 0, :sum 0})
  ([result] (double (/ (:sum result) (:count result))))
  ([result x]
   (update-in
     (update-in result [:count] inc)
     [:sum] (partial + x))))
(transduce identity mean [10 20 40 40]) ;27.5

现在我们取平均值,低于 20 的都算 20,但其他的都减 1:

(transduce
  (if-xf
    (fn [x] (< x 20))
    (map (constantly 20))
    (map dec))
  mean [10 20 40 40]) ;29.25

我的回答如下:我认为你原来的解决方案是最好的。它使用 map 效果很好,这就是您首先陈述条件传感器有用性的方式。