Question

康纳麦克布莱德和罗斯帕特森的经典论文 Applicative programming with effects 显示'matrice'换位示例：

transpose   :: [[a]] -> [[a]]
transpose [] = repeat []
transpose (xs : xss) = zipWith (:) xs (transpose xss)

transpose正在使用列表的“收集视角”：它成对函数（此处为(:)）并按元素输入并生成结果列表输出。

因此，给定

v = [[1,2,3],[4,5,6]]

然后

transpose  v

结果

[[1,4],[2,5],[3,6]]

他们在论文的后面说了

如果我们想为我们的转置示例做同样的事情，我们必须这样做避免图书馆的“成功列表”（Wadler，1985）monad and take 而是一个支持'矢量化'的实例Applicative []，其中pure = repeat和(~) = zapp，屈服于

transpose'' :: [[a]] -> [[a]]
transpose''         [] = pure []
transpose'' (xs : xss) = pure (:) <*> xs <*> transpose'' xss

这里，transpose''正在使用“非确定性计算点列表的视图：它将函数（此处为(:)）应用于输入转动。

因此

 transpose'' v

结果

 [[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]]

我觉得我错过了一些微妙的观点。我可以看到transpose是确实使用收集的角度来转置“向量” 名单。但transpose''（使用非确定性计算列表的观点）似乎与矢量无关转置。

换句话说，transpose和transpose''似乎无关功能 - 不同的例子。我错过了什么吗？

Answer 1

其中pure = repeat和(❄) = zapp，产生......

这是不标准列表实例。要在Haskell中实现这一点，我们需要

newtype Zapp a = Zapp { runZapp : [a] } deriving (Functor)
zcons :: a -> Zapp a -> Zapp a
zcons x (Zapp xs) = Zapp $ x : xs

instance Applicative Zapp where
  pure = Zapp . repeat
  Zapp a <*> Zapp b = Zapp $ zapp a b

然后

transpose'' :: Zapp (Zapp a) -> Zapp (Zapp a)
transpose''         (Zapp []) = pure $ Zapp []
transpose'' (Zapp (xs : xss)) = pure zcons <*> xs <*> transpose'' xss

Answer 2

如果您为第一个示例列出了实例，那么Applicative实例有pure = repeat和<*> = zapp

instance Applicative [] where
    pure = repeat
    (<*>) = zapp

transpose :: [[a]] -> [[a]]
transpose         [] = pure []
transpose (xs : xss) = pure (:) <*> xs <*> transpose xss

main = do
    print . transpose $ [[1,2,3],[4,5,6]]

你从转置获得转置：

[[1,4],[2,5],[3,6]]

相反，如果您使用Applicative

的普通[]实例

instance Applicative [] where
    pure x = [x]
    fs <*> xs = [f x | f <- fs, x <- xs]

transpose :: [[a]] -> [[a]]
transpose         [] = pure []
transpose (xs : xss) = pure (:) <*> xs <*> transpose xss

main = do
    print . transpose $ [[1,2,3],[4,5,6]]

你得到了

[[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]]

这两个例子的样板是：

module Main (
    main
) where

import Prelude hiding (repeat)

infixl 4 <*>
class Applicative f where
    pure :: a -> f a
    (<*>) :: f (a -> b) -> f a -> f b

repeat :: a -> [a]
repeat x = x : repeat x

zapp :: [a -> b] -> [a] -> [b]
zapp (f : fs) (x : xs) = f x : zapp fs xs
zapp _        _        = []

请求澄清McBride / Paterson申请文件中的换位示例

2 个答案: