Question

给出一个基于Vector的非常简单的矩阵定义：

import Numeric.AD
import qualified Data.Vector as V

newtype Mat a = Mat { unMat :: V.Vector a }

scale' f = Mat . V.map (*f) . unMat
add' a b = Mat $ V.zipWith (+) (unMat a) (unMat b)
sub' a b = Mat $ V.zipWith (-) (unMat a) (unMat b)
mul' a b = Mat $ V.zipWith (*) (unMat a) (unMat b)
pow' a e = Mat $ V.map (^e) (unMat a)

sumElems' :: Num a => Mat a -> a
sumElems' = V.sum . unMat

（出于演示目的......我正在使用hmatrix，但认为问题存在于某种程度上）

错误函数（eq3）：

eq1' :: Num a => [a] -> [Mat a] -> Mat a
eq1' as φs = foldl1 add' $ zipWith scale' as φs

eq3' :: Num a => Mat a -> [a] -> [Mat a] -> a
eq3' img as φs = negate $ sumElems' (errImg `pow'` (2::Int))
  where errImg = img `sub'` (eq1' as φs)

为什么编译器无法在此推断出正确的类型？

diffTest :: forall a . (Fractional a, Ord a) => Mat a -> [Mat a] -> [a] -> [[a]]
diffTest m φs as0 = gradientDescent go as0
  where go xs = eq3' m xs φs

确切的错误信息是：

src/Stuff.hs:59:37:
    Could not deduce (a ~ Numeric.AD.Internal.Reverse.Reverse s a)
    from the context (Fractional a, Ord a)
      bound by the type signature for
                 diffTest :: (Fractional a, Ord a) =>
                             Mat a -> [Mat a] -> [a] -> [[a]]
      at src/Stuff.hs:58:13-69
    or from (reflection-1.5.1.2:Data.Reflection.Reifies
               s Numeric.AD.Internal.Reverse.Tape)
      bound by a type expected by the context:
                 reflection-1.5.1.2:Data.Reflection.Reifies
                   s Numeric.AD.Internal.Reverse.Tape =>
                 [Numeric.AD.Internal.Reverse.Reverse s a]
                 -> Numeric.AD.Internal.Reverse.Reverse s a
      at src/Stuff.hs:59:21-42
      ‘a’ is a rigid type variable bound by
          the type signature for
            diffTest :: (Fractional a, Ord a) =>
                        Mat a -> [Mat a] -> [a] -> [[a]]
          at src//Stuff.hs:58:13
    Expected type: [Numeric.AD.Internal.Reverse.Reverse s a]
                   -> Numeric.AD.Internal.Reverse.Reverse s a
      Actual type: [a] -> a
    Relevant bindings include
      go :: [a] -> a (bound at src/Stuff.hs:60:9)
      as0 :: [a] (bound at src/Stuff.hs:59:15)
      φs :: [Mat a] (bound at src/Stuff.hs:59:12)
      m :: Mat a (bound at src/Stuff.hs:59:10)
      diffTest :: Mat a -> [Mat a] -> [a] -> [[a]]
        (bound at src/Stuff.hs:59:1)
    In the first argument of ‘gradientDescent’, namely ‘go’
    In the expression: gradientDescent go as0

Answer 1

gradientDescent中的ad函数的类型为

gradientDescent :: (Traversable f, Fractional a, Ord a) =>
                   (forall s. Reifies s Tape => f (Reverse s a) -> Reverse s a) ->
                   f a -> [f a]

它的第一个参数需要f r -> r类型的函数，其中r是forall s. (Reverse s a)。 go的类型为[a] -> a，其中a是diffTest签名中绑定的类型。这些a是相同的，但Reverse s a与a不同。

Reverse类型包含许多类型类的实例，可以让我们将a转换为Reverse s a或返回。最明显的是Fractional a => Fractional (Reverse s a)，这样我们就可以a将Reverse s a转换为realToFrac。{/ p>

为此，我们需要能够在a -> b上映射函数Mat a以获得Mat b。最简单的方法是为Functor派生Mat个实例。

{-# LANGUAGE DeriveFunctor #-}

newtype Mat a = Mat { unMat :: V.Vector a }
    deriving Functor

我们可以将m和fs转换为Fractional a' => Mat a'的任何fmap realToFrac。

diffTest m fs as0 = gradientDescent go as0
  where go xs = eq3' (fmap realToFrac m) xs (fmap (fmap realToFrac) fs)

但隐藏在广告包中的方式更好。 Reverse s a普遍适用于所有s，但a与a的类型签名中绑定的diffTest相同。a -> (forall s. Reverse s a)。我们真的只需要一个函数Mode。此函数来自Reverse s a类auto，auto有一个实例。 Mode t => Scalar t -> t有一个稍微奇怪的类型type Scalar (Reverse s a) = a，但Reverse。专为auto auto :: (Reifies s Tape, Num a) => a -> Reverse s a而设的类型为

Mat a

这样，我们就可以将Mat (Reverse s a)转换为Rational，而无需与{-# LANGUAGE ScopedTypeVariables #-} {-# LANGUAGE TypeFamilies #-} diffTest :: forall a . (Fractional a, Ord a) => Mat a -> [Mat a] -> [a] -> [[a]] diffTest m fs as0 = gradientDescent go as0 where go :: forall t. (Scalar t ~ a, Mode t) => [t] -> t go xs = eq3' (fmap auto m) xs (fmap (fmap auto) fs)进行转换。

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如何对复杂数据类型进行自动区分？

1 个答案: