Question

我要说的是:

type Key = String
type Score = Int
data Thing = Thing Key Score

And if I have an array of them like this:

[Thing "a" 7, Thing "b" 5, Thing "a" 10]

是否有减少这种情况的标准方法,以便我不会有任何重复的钥匙? 如果两个关键点相匹配,我想取更好的分数。

[Thing "b" 5, Thing "a" 10]

Answer 1

从根本上说,我们必须首先决定问题解决和执行工作的困难。因此,如果我们首先通过<代码>S分进行分类,然后在分类名单上保留第一个与<代码”有关的编号。关键这应当奏效,请看一看执行情况:

import Data.List
import Data.Function

type Key = String
type Score = Int
data Thing = Thing { key :: Key, score :: Score }
  deriving (Show)

myNub  = nubBy  ((==) `on` key)
mySort = sortBy (compare `on` (negate . score))

selectFinest = myNub . mySort

我们现在尝试在<代码>上这样做。缩略语

Prelude> :load Test.hs 
[1 of 1] Compiling Main             ( Test.hs, interpreted )
Ok, modules loaded: Main.
*Main> selectFinest [Thing "a" 7, Thing "b" 5, Thing "a" 10]
[Thing {key = "a", score = 10},Thing {key = "b", score = 5}]

查询http://haskell.org/hoogle/rel=“noreferer”>gle 如果你对我在解决办法中使用的职能不确定。学习如何使用<条码><<>>>>>>和这些功能确实需要一些时间。

Answer 2

一种非常简单的解决办法是使用 ,其中考虑到将多重价值与同一关键数值结合起来的功能,将关键价值清单转换为地图。

Prelude Data.Map> fromListWith max [("a", 7), ("b", 5), ("a", 10)]
fromList [("a",10),("b",5)]

请注意,这预示着les,以便在必要时转换。此外,它并不维护投入内容的顺序。运行时间为O(nlog n)。

Answer 3

I m 寄出一个O(nlog n)解决方案,因为每个人都似乎因O(n^2)而被罚款。

consolidate :: (Ord a, Ord b) => [Thing a b] -> [Thing a b]
consolidate xs = 
    max_from_each_group (sortBy (compare `on` getKey) xs)
    where
       max_from_each_group [] = []
       max_from_each_group (x:xs) = 
           let (same_key, rest) = span (	 -> x == getKey t) xs in
           let group_max = maximumBy (compare `on` getValue) (x:same_key) in
           group_max : max_from_each_group rest

Answer 4

Here is my feeble attempt. There surely is a nicer way but I m not much of a Haskell programmer.

import Data.List

type Key = String
type Score = Int
data Thing = Thing Key Score
           deriving (Show, Ord)

instance Eq Thing where
    (Thing k1 _) == (Thing k2 _) = k1 == k2
    (Thing k1 _) /= (Thing k2 _) = k1 /= k2

thingSort :: [Thing] -> [Thing]
thingSort = Data.List.sortBy (flip compare)

ex = [Thing "a" 7, Thing "b" 5, Thing "a" 10]

filtered = nub (thingSort ex)

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