7

我正在尝试使用 Numeric.AD 和自定义 Expr 类型。我希望计算用户输入表达式的符号梯度。使用常量表达式的第一个试验效果很好：

calcGrad0 :: [Expr Double]
calcGrad0 = grad df vars
  where
   df [x,y] = eval (env [x,y]) (EVar "x"*EVar "y")
   env vs = zip varNames vs
   varNames = ["x","y"]
   vars = map EVar varNames

这有效：

>calcGrad0
[Const 0.0 :+ (Const 0.0 :+ (EVar "y" :* Const 1.0)),Const 0.0 :+ (Const 0.0 :+ (EVar "x" :* Const 1.0))]

但是，如果我将表达式作为参数提取出来：

calcGrad1 :: [Expr Double]
calcGrad1 = calcGrad1' (EVar "x"*EVar "y")
calcGrad1' e = grad df vars
  where
   df [x,y] = eval (env [x,y]) e
   env vs = zip varNames vs
   varNames = ["x","y"]
   vars = map EVar varNames

我明白了

Could not deduce (a ~ AD s (Expr a1))
from the context (Num a1, Floating a)
  bound by the inferred type of
           calcGrad1' :: (Num a1, Floating a) => Expr a -> [Expr a1]
  at Symbolics.hs:(60,1)-(65,29)
or from (Mode s)
  bound by a type expected by the context:
             Mode s => [AD s (Expr a1)] -> AD s (Expr a1)
  at Symbolics.hs:60:16-27
  `a' is a rigid type variable bound by
      the inferred type of
      calcGrad1' :: (Num a1, Floating a) => Expr a -> [Expr a1]
      at Symbolics.hs:60:1
Expected type: [AD s (Expr a1)] -> AD s (Expr a1)
  Actual type: [a] -> a
In the first argument of `grad', namely `df'
In the expression: grad df vars

我如何让 ghc 接受这个？

4

2 回答 2

5

我的猜测是您忘记应用lift将 an 转换Expr为AD s Expr.

如果您有兴趣使用广告包进行符号区分。Lennart Augustsson 的跟踪包运行良好。

于 2011-05-09T20:18:03.110 回答
1

当 GHC 无法推断出有效函数的类型相等签名时，如您的情况，解决方案是为函数提供类型签名。我不知道这个库的接口。但是，我的猜测是正确的签名是calcGrad1 :: (Num a, Floating a) => Expr a -> [Expr a].

于 2011-05-09T14:49:30.527 回答