[英]clojure update value during doseq
在doseq期间更新/计算变量的最佳方法是什么?
显然,我可以执行以下操作:
(doseq [x xs]
(println (string/join " " x)))
(println "total:" (reduce + 0 (map count xs)))
但是我想避免必须在最后映射整个列表来计算值...在迭代时只更新一个计数更有意义。
我发现这很有效 ,但似乎有点麻烦。
(defn display-xs [xs]
;; all I want to do is update a count while I print,
;; and have that value available afterwards!
(let [n (ref 0)]
(do
(doseq [x xs]
(dosync
(ref-set n (+ @n (count x)))
(println (string/join " " x))))
(println "total:" @n))))
我知道doseq
允许:let
但是我需要在doseq完成之后的值。
要么
(println "total:" (reduce (fn [m x] (do (println x) (+ m (count x)))) 0 xs))
您不必担心循环的最小开销。 分离不同的任务更为重要。 我非常喜欢您的第一个版本。
您的第二个版本可能看起来更好,但我仍然希望第一个版本。
(defn display-xs [xs]
(let [n (atom 0)]
(doseq [x xs]
(swap! n + (count x))
(println (string/join " " x)))
(println "total: " @n)))
我稍微重写了您的代码,所以请展示我该怎么做:
(def sample-vals (repeatedly 10 #(range (+ 5 (rand-int 10)))))
(defn display-xs [xs]
(let [n (atom 0)]
(doseq [x xs]
(swap! n + (count x))
(println x))
(println "total:" @n)))
(newline)
(display-xs sample-vals)
(defn print-n-sum [cum coll]
(let [cnt (count coll) ]
(printf "%4d: " cnt)
(println coll)
(+ cum cnt)))
(defn calc-sum [xs]
(let [n (reduce print-n-sum 0 xs) ]
(println "total:" n)))
(newline)
(calc-sum sample-vals)
结果
> rs; lein run
(0 1 2 3 4 5 6 7 8 9)
(0 1 2 3 4 5 6 7 8 9 10 11)
(0 1 2 3 4 5 6 7 8 9 10)
(0 1 2 3 4 5 6 7 8 9 10 11)
(0 1 2 3 4)
(0 1 2 3 4 5 6 7 8 9 10 11)
(0 1 2 3 4 5)
(0 1 2 3 4 5 6 7 8 9 10)
(0 1 2 3 4 5 6)
(0 1 2 3 4 5 6 7 8 9 10)
total: 97
10: (0 1 2 3 4 5 6 7 8 9)
12: (0 1 2 3 4 5 6 7 8 9 10 11)
11: (0 1 2 3 4 5 6 7 8 9 10)
12: (0 1 2 3 4 5 6 7 8 9 10 11)
5: (0 1 2 3 4)
12: (0 1 2 3 4 5 6 7 8 9 10 11)
6: (0 1 2 3 4 5)
11: (0 1 2 3 4 5 6 7 8 9 10)
7: (0 1 2 3 4 5 6)
11: (0 1 2 3 4 5 6 7 8 9 10)
total: 97
请注意,您不必使用原子(此示例比ref简单)。 您可以进行打印和添加诸如print-n-sum
类的单个功能,然后将其馈送给reduce
。
您可以在您的reduce append函数中打印它
(defn print-and-append [acc x]
(println (string/join " " x))
(+ acc (count x)))
(println (reduce print-and-append 0 xs))
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