模仿Python产量的惯用Clojure方法 [英] Idiomatic Clojure way of mimicking Python's yield

问题描述

我正在遍历列表，逐步建立状态，偶尔遇到某个哨兵时，我会返回结果。如果我在Python中执行此操作，则会懒惰地 yield 结果，并在执行过程中跟踪函数本地范围内的状态：

I'm iterating through a list, building up state as I go, and occasionally when I encounter a certain sentinel, I return a result. If I was doing this in Python, I would lazily yield the results, tracking state in the function's local scope as I go:

# this is simplified for illustration
def yielder(input_list):
    state = 0
    for item in input_list:
        if item = 'SENTINEL':
            yield state * 2
            state = 0
        else:
            state += item

yielder([1, 5, 2, 5, 'SENTINEL', 4, 6, 7]) # [26, 34]

我的第一个实现使用 reduce ，但是不如 yield 那样好，因为：

My first implementation uses reduce, but that's not as good as yield because:

• 我在两次迭代之间传递的值既具有循环状态，又具有要产生的项，这似乎很笨拙

iterate 可用于减轻后者，但我不愿意实际上想为每个输入项返回一些内容，因此将需要更多操作。

iterate could be used to mitigate the latter, but i don't actually want to return something for every input item, so it would require more munging.

在Clojure中执行此操作的惯用方式是什么？

推荐答案

您可以构建这可以使用您提到的lazy-seq来解决，也可以使用 partition 和 reduce 将问题分为多个阶段，然后进行线程化他们在一起。我将使用thread-last宏单独显示每个步骤：

You can build this yourself using lazy-seq as you mention or you could use partition and reduce to split the problem into phases then thread them together. I'll use the thread-last macro to show each step on it's own:

user> (->> [1, 5, 2, 5, :SENTINEL, 4, 6, 7] ;; start with data
           (partition-by #(= :SENTINEL %))  ;; ((1 5 2 5) (:SENTINEL) (4 6 7))
           (take-nth 2)                     ;; ((1 5 2 5) (4 6 7))
           (map #(* 2 (reduce + %))))       ;; the map here keeps it lazy
(26 34)

，这里是usin lazy-直接seq：

and here it is usin lazy-seq directly:

user>  (defn x [items]
         (when (seq items)
           (lazy-seq (cons (* 2 (reduce + (take-while #(not= :SENTINEL %) items)))
                           (x (rest (drop-while #(not= :SENTINEL %) items)))))))
#'user/x
user> (x [1, 5, 2, 5, :SENTINEL, 4, 6, 7])
(26 34)

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