Question

在得到一些帮助之后，理解我试图编译代码的问题，在这个问题中（Trouble understanding GHC complaint about ambiguity）Ness建议我重新设计我的类型类以避免我不满意的解决方案。

有问题的类型是：

class (Eq a, Show a) => Genome a where
    crossover       :: (Fractional b) => b -> a -> a -> IO (a, a)
    mutate          :: (Fractional b) => b -> a -> IO a
    develop         :: (Phenotype b)  => a -> b

class (Eq a, Show a) => Phenotype a where
    --In case of Coevolution where each phenotype needs to be compared to every other in the population
    fitness         :: [a] -> a -> Int 
    genome          :: (Genome b) => a -> b

我试图在Haskell中创建一个可扩展的进化算法，该算法应支持不同的Genomes和Phenotypes。例如，一个Genome可以是一个位数组，另一个可以是一个整数列表，Phenotypes也将不同于http://en.wikipedia.org/wiki/Colonel_Blotto中表示部队移动的双精度列表，或者它可以代表人工神经网络。

由于Phenotype是从Genome开发的，所使用的方法必须完全互换，并且一个Genome类应该能够通过提供不同的Phenotypes来支持多个-- |Full generational replacement selection protocol fullGenerational :: (Phenotype b) => (Int -> [b] -> IO [(b, b)]) -> --Selection mechanism Int -> --Elitism Int -> --The number of children to create Double -> --Crossover rate Double -> --Mutation rate [b] -> --Population to select from IO [b] --The new population created fullGenerational selection e amount cross mute pop = do parents <- selection (amount - e) pop next <- breed parents cross mute return $ next ++ take e reverseSorted where reverseSorted = reverse $ sortBy (fit pop) pop breed :: (Phenotype b, Genome a) => [(b, b)] -> Double -> Double -> IO [b] breed parents cross mute = do children <- mapM (\ (dad, mom) -> crossover cross (genome dad) (genome mom)) parents let ch1 = map fst children ++ map snd children mutated <- mapM (mutate mute) ch1 return $ map develop mutated开发方法（这可以在代码中静态完成，而不必在运行时动态完成）。

使用这些类型类的代码在很大程度上应该幸福地不知道所使用的特定类型，这就是让我提出上述问题的原因。

我想要适应这些类型类的一些代码是：

Genome

我知道必须更改此代码并且必须添加新的约束，但我想使用类型类来展示我想到的一些代码。例如，上面的完整世代替换不需要知道有关基础Phenotypes正常运行的任何信息;它只需要知道Genome可以产生生成它的fullGenerational，以便它可以将它们一起繁殖并创建新的孩子。 Phenotype的代码应该尽可能通用，这样一旦设计了新的Genome或创建了更好的{{1}}，就不需要更改它。

如何更改上面的类型类以避免我遇到类型类歧义的问题，同时在一般EA代码中保留我想要的属性（应该是可重用的）？

Answer 1

“它只需要知道表型可以产生产生它的基因组”

这意味着Phenotype实际上是两种类型的关系，另一种是用于产生给定表型的基因组类型：

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}

import Data.List (sortBy)

class (Eq a, Show a) => Genome a where
    crossover       :: (Fractional b) => b -> a -> a -> IO (a, a)
    mutate          :: (Fractional b) => b -> a -> IO a
    develop         :: (Phenotype b a) => a -> b

class (Eq a, Show a, Genome b) => Phenotype a b | a -> b where
    --  In case of Coevolution where each phenotype needs to be compared to 
    --  every other in the population
    fitness         :: [a] -> a -> Int 
    genome          :: a -> b

breed :: (Phenotype b a, Genome a) => [(b, b)] -> Double -> Double -> IO [b]
breed parents cross mute = do
    children <- mapM (\(dad, mom)-> crossover cross (genome dad) (genome mom)) 
                     parents
    let ch1 = map fst children ++ map snd children
    mutated <- mapM (mutate mute) ch1
    return $ map develop mutated

-- |Full generational replacement selection protocol
fullGenerational :: (Phenotype b a, Genome a) =>
    (Int -> [b] -> IO [(b, b)]) -> --Selection mechanism
    Int -> --Elitism
    Int -> --The number of children to create
    Double -> --Crossover rate
    Double -> --Mutation rate
    [b] -> --Population to select from
    IO [b] --The new population created
fullGenerational selection e amount cross mute pop = do
    parents <- selection (amount - e) pop
    next <- breed parents cross mute
    return $ next ++ take e reverseSorted
            where reverseSorted = reverse $ sortBy (fit pop) pop

fit pop a b = LT   -- dummy function

这个编译。 genome的每个表型will have to provide exactly one implementation。

重新设计Haskell类型

1 个答案: