使用Control.Lens获取子列表中的值列表

时间:2016-12-06 21:33:54

标签: haskell lens

我正在尝试从子结构中获取值列表。 我有以下结构

("Value",[(1,"1"),(2,"2"),(3,"3"),(4,"4"),(5,"5")])

我正试图在列表中获取元组的第二个元素。

["1", "2" , "3" , "4" , "5"]

我要说的是:

view (_2 . toListOf . _2) a

我也试过traverse。但似乎traverse在列表上有折叠效果。我需要将结果作为列表。

Prelude Control.Lens> let a = ("Value", [(i, show i)|i<-[1..5]]) :: (String, [(Int, String)])
Prelude Control.Lens> a
("Value",[(1,"1"),(2,"2"),(3,"3"),(4,"4"),(5,"5")])
Prelude Control.Lens> view (_2 . toListOf . _2) a

<interactive>:36:7: error:
    • Couldn't match type ‘[]’ with ‘Const t’
      Expected type: Getting t (String, [(Int, String)]) t
        Actual type: (t -> Const t t)
                     -> (String, [(Int, String)]) -> [(String, [(Int, String)])]
    • In the first argument of ‘view’, namely ‘(_2 . toListOf . _2)’
      In the expression: view (_2 . toListOf . _2) a
      In an equation for ‘it’: it = view (_2 . toListOf . _2) a
    • Relevant bindings include it :: t (bound at <interactive>:36:1)

<interactive>:36:23: error:
    • Couldn't match type ‘Const t t0’
                     with ‘[(Int, String)]
                           -> Const (Data.Monoid.Endo [[(Int, String)]]) [(Int, String)]’
      Expected type: (t -> Const t t)
                     -> Getting
                          (Data.Monoid.Endo [[(Int, String)]])
                          [(Int, String)]
                          [(Int, String)]
        Actual type: (t -> Const t t)
                     -> ([(Int, String)]
                         -> Const (Data.Monoid.Endo [[(Int, String)]]) [(Int, String)])
                     -> Const t t0
    • In the second argument of ‘(.)’, namely ‘_2’
      In the second argument of ‘(.)’, namely ‘toListOf . _2’
      In the first argument of ‘view’, namely ‘(_2 . toListOf . _2)’
    • Relevant bindings include it :: t (bound at <interactive>:36:1)
Prelude Control.Lens>

1 个答案:

答案 0 :(得分:2)

文档说:

  

查看Getter,Iso或Lens指向的值或折叠指向幺半值的折叠或遍历的所有结果的结果。

您需要使用toListOf代替view。例如:

toListOf (_2.traverse._2) a

或者

a ^.. _2 . traverse . _2

所以,toListOf不是镜头,它只是view之类的另一个运算符,但它会提取折叠的目标列表