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我正在尝试从 Numeric.AD 编译以下最小示例：

import Numeric.AD 
timeAndGrad f l = grad f l
main = putStrLn "hi"

我遇到了这个错误：

test.hs:3:24:
    Couldn't match expected type ‘f (Numeric.AD.Internal.Reverse.Reverse
                                       s a)
                                  -> Numeric.AD.Internal.Reverse.Reverse s a’
                with actual type ‘t’
      because type variable ‘s’ would escape its scope
    This (rigid, skolem) type variable is bound by
      a type expected by the context:
        Data.Reflection.Reifies s Numeric.AD.Internal.Reverse.Tape =>
        f (Numeric.AD.Internal.Reverse.Reverse s a)
        -> Numeric.AD.Internal.Reverse.Reverse s a
      at test.hs:3:19-26
    Relevant bindings include
      l :: f a (bound at test.hs:3:15)
      f :: t (bound at test.hs:3:13)
      timeAndGrad :: t -> f a -> f a (bound at test.hs:3:1)
    In the first argument of ‘grad’, namely ‘f’
    In the expression: grad f l

关于为什么会发生这种情况的任何线索？通过查看前面的示例，我了解到这是“扁平化”grad的类型：

grad :: (Traversable f, Num a) => (forall s. Reifies s Tape => f (Reverse s a) -> Reverse s a) -> f a -> f a

但我实际上需要在我的代码中做这样的事情。事实上，这是不会编译的最小示例。我想做的更复杂的事情是这样的：

example :: SomeType
example f x args = (do stuff with the gradient and gradient "function")
    where gradient = grad f x
          gradientFn = grad f
          (other where clauses involving gradient and gradient "function")

这是一个稍微复杂的版本，带有可以编译的类型签名。

{-# LANGUAGE RankNTypes #-}

import Numeric.AD 
import Numeric.AD.Internal.Reverse

-- compiles but I can't figure out how to use it in code
grad2 :: (Show a, Num a, Floating a) => (forall s.[Reverse s a] -> Reverse s a) -> [a] -> [a]
grad2 f l = grad f l

-- compiles with the right type, but the resulting gradient is all 0s...
grad2' :: (Show a, Num a, Floating a) => ([a] -> a) -> [a] -> [a]
grad2' f l = grad f' l
       where f' = Lift . f . extractAll
       -- i've tried using the Reverse constructor with Reverse 0 _, Reverse 1 _, and Reverse 2 _, but those don't yield the correct gradient. Not sure how the modes work

extractAll :: [Reverse t a] -> [a]
extractAll xs = map extract xs
           where extract (Lift x) = x -- non-exhaustive pattern match

dist :: (Show a, Num a, Floating a) => [a] -> a
dist [x, y] = sqrt(x^2 + y^2)

-- incorrect output: [0.0, 0.0]
main = putStrLn $ show $ grad2' dist [1,2]

但是，我不知道如何在代码中使用第一个版本 ,grad2因为我不知道如何处理Reverse s a. 第二个版本 ,grad2'具有正确的类型，因为我使用内部构造函数Lift来创建 a Reverse s a，但我一定不了解内部结构（特别是参数s）是如何工作的，因为输出梯度全为 0。使用其他构造函数Reverse（此处未显示）也会产生错误的渐变。

或者，是否有人们使用过ad代码的库/代码示例？我认为我的用例是一个非常常见的用例。

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1 回答 1

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本质上，where f' = Lift . f . extractAll您为自动微分基础类型创建了一个后门，该类型丢弃了所有导数，只保留常量值。如果您随后将其用于grad，那么您得到零结果也就不足为奇了！

明智的方法是按原样使用grad：

dist :: Floating a => [a] -> a
dist [x, y] = sqrt $ x^2 + y^2
-- preferrable is of course `dist = sqrt . sum . map (^2)`

main = print $ grad dist [1,2]
-- output: [0.4472135954999579,0.8944271909999159]

你不需要知道任何更复杂的东西来使用自动微分。只要您只区分Num或多Floating态函数，一切都会按原样工作。如果您需要区分作为参数传入的函数，则需要使该参数具有 rank-2 多态性（另一种方法是切换到ad函数的 rank-1 版本，但我敢说这不太优雅并且不并没有真正让你获得太多）。

{-# LANGUAGE Rank2Types, UnicodeSyntax #-}

mainWith :: (∀n . Floating n => [n] -> n) -> IO ()
mainWith f = print $ grad f [1,2]

main = mainWith dist
于 2016-12-26T19:46:56.280 回答