Question

我可以轻松地展平嵌套数据结构，但双重操作似乎非常困难（至少对我而言）。

给定类型ContentPlaceHolder的列表，如何创建嵌套结构，即

[(a,b,c)]

这样每个unflatten :: (Ord a, Ord b) => [(a,b,c)] -> [(a, [(b, [c])])]只出现一次，同样只出现a

的每个组合

我不太清楚为什么我在努力解决这个问题，因为双重操作非常简单

(a,b)

更新

根据EricR的建议，我编写了一个功能，它可以实现一个不平坦的功能，但我仍在努力扩展它

flatten :: [(a, [(b, [c])])] -> [(a,b,c)]
flatten xs = [(a,b,c) | (a,bcs) <- xs, (b,cs) <- bcs, c<-cs]

Answer 1

我个人将这些数据放入嵌套地图中。您可以随后将其再次拉出到嵌套关联列表中。

首先将每个项目转换为（小）嵌套地图，其中包含列表[c]中的1个元素：

import qualified Data.Map as M
import Data.List

triplesToMaps :: [(a,b,c)] -> [M.Map a (M.Map b [c])]
triplesToMaps = map (\(a,b,c) -> M.singleton a (M.singleton b [c]))

然后使用unionWith

将它们全部合并

combineMaps :: (Ord a, Ord b) => [M.Map a (M.Map b [c])] -> M.Map a (M.Map b [c])
combineMaps = foldl' (M.unionWith (M.unionWith (++))) M.empty

然后，如果您愿意，请将其带回列表：

flattenToLists :: M.Map a (M.Map b [c]) -> [(a,[(b,[c])])]
flattenToLists = M.assocs . (M.map M.assocs)

示例：

> flattenToLists . combineMaps . triplesToMaps $ [(1,2,3),(1,2,4),(1,3,5),(2,6,8)]
[(1,[(2,[3,4]),(3,[5])]),(2,[(6,[8])])]

在实践中，您可能不会flattenToLists，因为嵌套地图是更有用的结构。

顺便提一下，如果Map拥有＆＃34;权利＆＃34; Monoid结构然后M.unionWith (M.unionWith (++))只是mappend，整个事情可能是foldMap (\(a,b,c) -> M.singleton a (M.singleton b [c]))。

[编辑：在问题中添加了双重]

哦，回到[(a,b,c)]我想到了两种方法。列表理解与您的一样，但使用assocs：

flatten :: M.Map a (M.Map b [c]) -> [(a,b,c)]
flatten m = [(a,b,c) | (a,bcs) <- M.assocs m, (b,cs) <- M.assocs bcs, c <- cs]

或使用foldMap的版本：

flatten' :: M.Map a (M.Map b [c]) -> [(a,b,c)]
flatten' = M.foldMapWithKey (\a -> M.foldMapWithKey (\b -> foldMap (\c -> [(a,b,c)])))

Answer 2

根据ErikR的建议，出现了一种模式，将其扩展为更长和更长的元组。

-- construct some raw material
l a i = map ((a ++) . show) [1..i]
l4 = [(a,b,c,d) | d<-l "d" 3, c <- l "c" 3, b<-l "b" 3, a<-l "a" 2 ]

grp2 :: Ord a => [(a,b)] -> [(a,[b])]
grp2 xs = do
    a <- (nub . sort . map fst) xs 
    let bs = [b' | (a',b') <- xs, a'==a]
    return (a,bs)

grp3 :: (Ord a, Ord b) => [(a,b,c)] -> [(a,[(b,[c])])]
grp3 ys = do
    (a, bcs) <- grp2 (map split3 ys)
    return (a, grp2 bcs)
          where
              split3 (a,b,c) = (a,(b,c))

grp4 :: (Ord a, Ord b, Ord c) => [(a,b,c,d)] -> [(a,[(b,[(c,[d])])])]
grp4 zs = do
    (a, bcds) <- grp2 (map split4 zs)
    return (a, grp3 bcds)
          where
              split4 (a,b,c,d) = (a,(b,c,d))

现在

*Main> pp $ grp4 l4

打印

[ ( "a1"
  , [ ( "b1"
      , [ ( "c1" , [ "d1" , "d2" , "d3" ] )
        , ( "c2" , [ "d1" , "d2" , "d3" ] )
        , ( "c3" , [ "d1" , "d2" , "d3" ] )
        ]
      )
    , ( "b2"
      , [ ( "c1" , [ "d1" , "d2" , "d3" ] )
        , ( "c2" , [ "d1" , "d2" , "d3" ] )
        , ( "c3" , [ "d1" , "d2" , "d3" ] )
        ]
      )
    , ( "b3"
      , [ ( "c1" , [ "d1" , "d2" , "d3" ] )
        , ( "c2" , [ "d1" , "d2" , "d3" ] )
        , ( "c3" , [ "d1" , "d2" , "d3" ] )
        ]
      )
    ]
  )
, ( "a2"
  , [ ( "b1"
      , [ ( "c1" , [ "d1" , "d2" , "d3" ] )
        , ( "c2" , [ "d1" , "d2" , "d3" ] )
        , ( "c3" , [ "d1" , "d2" , "d3" ] )
        ]
      )
    , ( "b2"
      , [ ( "c1" , [ "d1" , "d2" , "d3" ] )
        , ( "c2" , [ "d1" , "d2" , "d3" ] )
        , ( "c3" , [ "d1" , "d2" , "d3" ] )
        ]
      )
    , ( "b3"
      , [ ( "c1" , [ "d1" , "d2" , "d3" ] )
        , ( "c2" , [ "d1" , "d2" , "d3" ] )
        , ( "c3" , [ "d1" , "d2" , "d3" ] )
        ]
      )
    ]
  )
]

如何取消展平列表

2 个答案: