Question

我想了解如何在Haskell中使用Numeric.AD（自动区分）。

我定义了一个简单的矩阵类型和一个标量函数，它以一个数组和两个矩阵作为参数。我想使用AD来获得关于两个矩阵的评分函数的梯度，但是我遇到了编译问题。这是代码

{-# LANGUAGE DeriveTraversable, DeriveFunctor, DeriveFoldable #-}
import Numeric.AD.Mode.Reverse as R
import Data.Traversable as T
import Data.Foldable as F

--- Non-linear function on "vectors"
logistic x = 1.0 / (1.0 + exp(-x) )
phi v = map logistic v
phi' (x:xs) = x : (phi xs)

--- dot product
dot u v = foldr (+) 0 $ zipWith (*) u v

--- simple matrix type
data Matrix a = M [[a]] deriving (Eq,Show,Functor,F.Foldable,T.Traversable)

--- action of a matrix on a vector
mv _ [] = []
mv (M []) _ = []
mv ( M m ) v = ( dot (head m)  v ) :  (mv (M (tail m)) v )

--- two matrices
mbW1 = M $ [[1,0,0],[-1,5,1],[1,2,-3]]
mbW2 = M $ [[0,0,0],[1,3,-1],[-2,4,6]]

--- two different scoring functions
sc1 v m = foldr (+) 0 $ (phi' . (mv m) )  v  

sc2 :: Floating a => [a] -> [Matrix a] -> a
sc2 v [m1, m2] = foldr (+) 0 $ (phi' . (mv m2) . phi' . (mv m1) ) v

strToInt = read :: String -> Double
strLToIntL = map strToInt
--- testing
main = do
        putStrLn $ "mbW1:" ++ (show mbW1)
        putStrLn $ "mbW2:" ++ (show mbW2)
        rawInput <-  readFile "/dev/stdin"
        let xin= strLToIntL $ lines rawInput
        putStrLn "sc xin mbW1"
        print $ sc1 xin mbW1  --- ok. = 
        putStrLn "grad (sc1 xin) mbW1"
        print $ grad ( sc1 xin) mbW1   -- yields an error: expects xin [Reverse s Double] instead of [Double]
        putStrLn "grad (sc1 [3,5,7]) mbW1"
        print $ grad ( sc1 [3,5,7]) mbW1   --- ok. =
        putStrLn "sc2 xin [mbW1,mbW2]"
        print $ sc2 xin [mbW1, mbW2]
        putStrLn "grad (sc2 [3,5,7) [mbW1,mbW2]"
        print $ grad ( sc2 [3,5,7]) [mbW1, mbW2]  --- Error: see text

最后一行（sc2上的grad）给出以下错误：

Couldn't match type ‘Reverse s (Matrix Double)’
               with ‘Matrix (Reverse s (Matrix Double))’
Expected type: [Reverse s (Matrix Double)]
               -> Reverse s (Matrix Double)
  Actual type: [Matrix (Reverse s (Matrix Double))]
               -> Reverse s (Matrix Double)
In the first argument of ‘grad’, namely ‘(sc2 [3, 5, 7])’
In the second argument of ‘($)’, namely
  ‘grad (sc2 [3, 5, 7]) [mbW1, mbW2]’

我不明白矩阵＆＃34; Matrix矩阵＆＃34;在实际看到的类型来自。我正在使用咖喱版本的sc2喂grad，使其成为Matrix列表中的一个函数。

注释掉两条违规行没有问题，即第一个梯度正常工作并正确计算（我将[1,2,3]作为程序的输入）：

mbW1:M [[1.0,0.0,0.0],[-1.0,5.0,1.0],[1.0,2.0,-3.0]]
mbW2:M [[0.0,0.0,0.0],[1.0,3.0,-1.0],[-2.0,4.0,6.0]]
sc1 xin mbW1
1
2
3
2.0179800657874893
grad (sc1 [3,5,7]) mbW1
M [[3.0,5.0,7.0],[7.630996942126885e-13,1.2718328236878141e-12,1.7805659531629398e-12],[1.0057130122694228e-3,1.6761883537823711e-3,2.3466636952953197e-3]]
sc2 xin [mbW1,mbW2]
1.8733609463863194

这两个错误都是个问题。我想取任何这样的sc2评分函数的梯度，取决于矩阵数组，在任何给定的＆＃34;点＆＃34;鑫。显然，我还没有充分理解AD库。任何帮助，将不胜感激。

关于Haskell中矩阵列表的自动区分（AD）

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