clojure update value during doseq

Question

what is the best way to update / calculate a variable during doseq?

Obviously I could just do the following:

(doseq [x xs]
        (println (string/join " " x)))
(println "total:" (reduce + 0 (map count xs)))

But I want to avoid having to map over the entire list at the end to calculate a value... It makes more sense to just update a count as we iterate.

I found that this works , but it seems like a bit of a cludge.

(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))))

I know that doseq allows for :let but I need the value after doseq is finished.

Or

(println "total:" (reduce (fn [m x] (do (println x) (+ m (count x)))) 0 xs))

Answer 1

You should not worry about the at most miniscule overhead of looping. It is much more important to separate different tasks. I quite like your first version.

Your second version could look better, but I'd still prefer the first.

(defn display-xs [xs]
  (let [n (atom 0)]
    (doseq [x xs]
      (swap! n + (count x))
      (println (string/join " " x)))
    (println "total: " @n)))

Answer 2

I rewrote your code a little bit so show how I would do it:

(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)

with result

> 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

Note that you don't have to use an atom (simpler than a ref for this example). You could do the printing and adding in a single function like print-n-sum that you feed to reduce .

Answer 3

You can print it in your reduce append function

(defn print-and-append [acc x]
      (println (string/join " " x))
      (+ acc (count x)))

(println (reduce print-and-append 0 xs))