在Clojure中将2元组(x，y)的列表排序为集合的排序映射

Order a list of 2-tuple (x,y) into a sorted-map of sets in Clojure

我正在尝试使用功能list-of-xy->sorted-map-of-sets创建sorted-set的sorted-map：

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(def in
'([1 9] [1 8] [1 7]
[2 1] [2 2] [2 3]
[2 1] [2 2] [2 3]
[2 1] [2 2] [2 3]))

(def out
(into (sorted-map)
{1 (sorted-set 9 8 7)
2 (sorted-set 1 2 3)}))

(defn list-of-xy->sorted-map-of-sorted-sets [list-of-xy]
"Take a list (or lazy-seq) of 2-tuple and return a sorted-map of sorted-sets"
(reduce ????? list-of-xy))

; should return true
(= (list-of-xy->sorted-map-of-sorted-sets in) out)

到目前为止，我尝试分两个步骤创建out：

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(def int1
(group-by #(first %) in))
;=> { 1 [[1 9] [1 8] [1 7]],
; 2 [[2 1] [2 2] [2 3] [2 1] [2 2] [2 3] [2 1] [2 2] [2 3]]}

(def int2
(flatten
(map
#(let [[x xys] %]
(list x (sorted-set (map last xys))))
int1)))
;=> (1 #{7 8 9} 2 #{1 2 3}) ; <-- this is not a sorted-map (yet!)

有什么更好的方法来转换以性能为优先的in --> out？

BTW

@Ankur答案被接受。到目前为止，这是更快的解决方案。

对于我的实际问题，@ amalloy解决方案(+1)中的(update-in acc [x] conj y)为通过get-in通往reduced的方式打开了道路。我正在使用的归约函数是：

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(fn [a [x y]]
(if-not (get-in a [x y])
(update-in a [x] conj y)
(reduced a)))

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(= out (into (sorted-map)
(map (fn [[k v]]
[k (apply sorted-set (map second v))])
(group-by first in))))

让我知道这是否通过了您的性能测试:)。