问题描述
import Numeric.AD
导入限定的Data.Vector如V
newtype Mat a = Mat {unMat :: V.Vector a}
scale'f = Mat。 V.map（* f）。 unMat
add'ab = Mat $ V.zipWith（+）（unMat a）（unMat b）
sub'ab = Mat $ V.zipWith（ - ）（unMat a）（unMat b）
mul'ab = Mat $ V.zipWith（*）（unMat a）（unMat b）
pow'ae = Mat $ V.map（^ e）（unMat a）
sumElems':: Num a =>垫子 - >
sumElems'= V.sum。 unMat
（为了演示目的...我使用hmatrix但认为问题在那里）
和一个错误函数（ eq3 ）：
eq1':: Num a => [a] - > [Mat a] - > Mat a a
eq1'asφs= foldl1 add'$ zipWith scale'asφs
eq3':: Num a =>垫子 - > [a] - > [Mat a] - > a
eq3'img asφs= negate $ sumElems'（errImg`pow'`（2 :: Int））
where errImg = img`sub``（eq1'asφs）
为什么编译器无法推断出正确的类型？
diffTest :: forall a。 （小数a，Ord a）=>垫子 - > [Mat a] - > [a] - > [[a]]
diffTest mφsas0 = gradientdecent go as0
其中go xs = eq3'm xsφs
确切的错误信息是这样的：
src / Stuff.hs：59：37：
无法推断（a〜Numeric.AD.Internal.Reverse.Reverse sa）
来自上下文（小数a，Ord a）
由类型签名绑定
diffTest ::（分数a，Ord a）=>
垫子 - > [Mat a] - > [a] - > [[a]]
在src / Stuff.hs：58：13-69
或from（reflection-1.5.1.2：Data.Reflection.Reifies
s Numeric.AD.Internal.Reverse .Tape）
由上下文期望的类型绑定：
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 sa
at src / Stuff.hs：59：21-42
'a'是一个刚性类型变量，由
绑定，$ b的类型签名$ b diffTest ::（小数a，Ord a）=>
垫子 - > [Mat a] - > [a] - > [[a]]
at src // Stuff.hs：58：13
预期类型：[Numeric.AD.Internal.Reverse.Reverse s a]
- > Numeric.AD.Internal.Reverse.Reverse s a
实际类型：[a] - > a
相关绑定包括
go :: [a] - > a（绑定在src / Stuff.hs：60：9）
as0 :: [a]（绑定在src / Stuff.hs：59：15）
φs:: [Mat a]（bound在src / Stuff.hs：59：12）
m :: Mat a（绑定在src / Stuff.hs：59：10）
diffTest :: Mat a - > [Mat a] - > [a] - >
（在src / Stuff.hs：59：1处绑定）
在'gradientDescent'的第一个参数中，即'go'
在表达式中：gradientDescent go as0
功能具有类型
gradientDescent ::（Traversable f，Fractional a，Ord a）=>
（forall s。Reifies s Tape => f（Reverse s a） - > Reverse s a） - >
f a - > [fa]
它的第一个参数需要一个 fr - >的函数; r 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 s是相同的，但 Reverse sa 与 a 。
类型拥有许多类型类的实例，可以让我们将 a 转换为反向sa 或返回。最明显的是小数a => Fractional（Reverse sa），它允许我们将 a s转换为反向sa s与 realToFrac 。
为此，我们需要能够映射函数 a - >>在 Mat a 上获得 Mat b 。最简单的方法是为 Mat 派生一个 Functor 实例。
{ - ＃LANGUAGE DeriveFunctor＃ - }
newtype Mat a = Mat {unMat :: V.Vector a}
派生Functor
我们可以将 m 和 fs 放入任何小数a'=>使用 fmap realToFrac 。
diffTest m fs as0 = gradientdecent go as0
其中go xs = eq3'（fmap realToFrac m）xs（fmap（fmap realToFrac）fs）
但隐藏在广告包中有更好的方法。 Reverse sa 通用于所有 s ，但 a 与 diffTest 的类型签名中绑定的那个是相同的 a 。我们真的只需要一个函数 a - > （反过来）。此功能是，其中 Reverse sa 有一个实例。 auto 的类型稍微偏向 Mode t =>>标量t - > t 但是类型标量（反向s a）= a 。专为反向 auto 有类型
auto ::（Reifies s Tape，Num a）=> a - >反向sa
这允许我们将 Mat a s，而不会混淆与 Rational 中的转换。
{ - ＃LANGUAGE ScopedTypeVariables＃ - }
{ - ＃LANGUAGE TypeFamilies＃ - }
diffTest： ：forall a。 （小数a，Ord a）=>垫子 - > [Mat a] - > [a] - > [[a]]
diffTest m fs as0 = gradientdecent go as0
where
go :: forall t。 （标量t〜a，模式t）=> [t] - > t
go xs = eq3'（fmap auto m）xs（fmap（fmap auto）fs）
Given a very simple Matrix definition based on Vector:
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
(for demonstration purposes ... I am using hmatrix but thought the problem was there somehow)
And an error function (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)
Why the compiler not able to deduce the right types in this?
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
The exact error message is this:
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
The gradientDescent function from ad has the type
gradientDescent :: (Traversable f, Fractional a, Ord a) =>
(forall s. Reifies s Tape => f (Reverse s a) -> Reverse s a) ->
f a -> [f a]
Its first argument requires a function of the type f r -> r where r is forall s. (Reverse s a). go has the type [a] -> a where a is the type bound in the signature of diffTest. These as are the same, but Reverse s a isn't the same as a.
The Reverse type has instances for a number of type classes that could allow us to convert an a into a Reverse s a or back. The most obvious is Fractional a => Fractional (Reverse s a) which would allow us to convert as into Reverse s as with realToFrac.
To do so, we'll need to be able to map a function a -> b over a Mat a to obtain a Mat b. The easiest way to do this will be to derive a Functor instance for Mat.
{-# LANGUAGE DeriveFunctor #-}
newtype Mat a = Mat { unMat :: V.Vector a }
deriving Functor
We can convert the m and fs into any Fractional a' => Mat a' with fmap realToFrac.
diffTest m fs as0 = gradientDescent go as0 where go xs = eq3' (fmap realToFrac m) xs (fmap (fmap realToFrac) fs)
But there's a better way hiding in the ad package. The Reverse s a is universally qualified over all s but the a is the same a as the one bound in the type signature for diffTest. We really only need a function a -> (forall s. Reverse s a). This function is auto from the Mode class, for which Reverse s a has an instance. auto has the slightly wierd type Mode t => Scalar t -> t but type Scalar (Reverse s a) = a. Specialized for Reverse auto has the type
auto :: (Reifies s Tape, Num a) => a -> Reverse s a
This allows us to convert our Mat as into Mat (Reverse s a)s without messing around with conversions to and from Rational.
{-# 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)
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