2016-08-04 21 views

回答

1

想通了:

(reduce #(+ %1 (%2 :count)) 0 [{:count 3} {:count 5} {:count 8}]) 
6
(def v [{:count 3} {:count 5} {:count 8}]) 

(reduce #(+ %1 (:count %2)) 0 v);; => 16 

或者

(apply + (map :count v));; => 16 

你需要學習的功能如何組合,那麼天空纔是極限。

1

而另一個問題:

(transduce (map :count) + 0 [{:count 2} {:count 12}])