Question

{:measure-value-sets (
   {:benchmark "EUR.EONIA.6Y", :bvalue "0"}),
 :dimensions (
   {:type "Currency", :dvalue "EUR"} 
   {:type "ValueUnderlying_0", :dvalue "21.59999999999811"})}

{:measure-value-sets (
   {:benchmark "EUR.6M3M.5Y6M", :bvalue "-0"}), 
 :dimensions (
   {:type "Currency", :dvalue "EUR"} 
   {:type "ValueUnderlying_0", :dvalue "13.199999999998733"})}

{:measure-value-sets (
   {:benchmark "EUR.LIBOR.U4", :bvalue "0.03558586"}), 
 :dimensions (
   {:type "Currency", :dvalue "EUR"} 
   {:type "ValueUnderlying_0", :dvalue "99.76045398474537"})}

... snip...

我有一个类似于上面的数据集，我需要得到基准，bvalue，dvalue的值，有人可以建议迭代这些并杀死这些值的最佳方法吗？

Answer 1

假设您的数据集结构为：

(pprint ds)

[{:measure-value-sets
  [{:bvalue "0", :benchmark "EUR.EONIA.6Y"}
   :dimensions                                ; A symbol on its own. Really ?!?
   [{:dvalue "EUR", :type "Currency"}
    {:dvalue "21.59999999999811", :type "ValueUnderlying_0"}]]}
 {:measure-value-sets
  [{:bvalue "-0", :benchmark "EUR.6M3M.5Y6M"}
   :dimensions
   [{:dvalue "EUR", :type "Currency"}
    {:dvalue "13.199999999998733", :type "ValueUnderlying_0"}]]}]

您可以使用以下内容获取所有bvalues：

(map #(-> % :measure-value-sets first :bvalue) ds)

你的第二次评价：

; highly unlikely your structure is correct
(map #(-> % :measure-value-sets (nth 2) second :dvalue) ds)

与：同时：

(map (juxt 
        #(-> % :measure-value-sets first :bvalue) 
        #(-> % :measure-value-sets (nth 2) second :dvalue)) 
  ds)

返回

(["0" "21.59999999999811"] ["-0" "13.199999999998733"])

但是，我认为您提供的结构不正确。

嵌套的地图矢量如何获取值

1 个答案: