Numeric.AD和打字问题 [英] Numeric.AD and typing problem

问题描述

我正在尝试使用Numeric.AD和一个自定义Expr类型。我希望计算
用户输入表达式的符号梯度。具有恒定
表达式的第一次试验效果很好：

  calcGrad0 :: [Expr Double] 
 calcGrad0 = grad df vars 
其中
 df [x，y] = eval（env [x，y]）（EVarx* EVary）
 env vs = zip varNames vs 
 varNames = [x，y] 
 vars =地图EVar varNames 
  

 > calcGrad0 
 [Const 0.0：+（Const 0.0：+（EVar （const 1.0）：const Const 0.0：+（const 0.0：+（EVarx：* Const 1.0））] 
  

但是，如果我将表达式作为参数提取：

  calcGrad1 :: [Expr Double] 
 calcGrad1 = calcGrad1'（EVarx* EVary）
 calcGrad1'e = grad df vars 
其中
 df [x，y] = eval（ env [x，y]）e 
 env vs = zip varNames vs 
 varNames = [x，y] 
 vars =地图EVar varNames 
  
 
 
 
 $ b pre $ code $无法推断（a〜AD s（Expr a1））
上下文（Num a1，Floating a）
被
 calcGrad1'的推断类型约束:: ::（Num a1，Floating a）=> Expr a  - > Symbolics.hs中的[Expr a1] 
：（60,1） - （65,29）
或from（Mode s）
由上下文期望的类型限制：
模式s => [AD s（Expr a1）]  - > AD s（Expr a1）
在Symbolics.hs：60：16-27 
`a'是一个刚性类型变量，由
绑定
 calcGrad1'的推断类型:: （数字a1，浮动a）=> Expr a  - > [Expr a1] 
在Symbolics.hs：60：1 
预期类型：[AD s（Expr a1）]  - > AD s（Expr a1）
实际类型：[a]  - > a 
在`grad'的第一个参数中，即`df'
在表达式中：grad df vars 
  

我如何得到ghc接受这个？

解决方案

我的猜测是你忘记申请 lift 将 Expr 转换为 AD s Expr 。

如果您有兴趣使用广告包进行符号区分。 Lennart Augustsson的追踪套餐运作良好。

I'm trying to work with Numeric.AD and a custom Expr type. I wish to calculate the symbolic gradient of user inputted expression. The first trial with a constant expression works nicely:

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

This works:

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

However, if I pull the expression out as a parameter:

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

I get

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

How do I get ghc to accept this?

解决方案

My guess is you are forgetting to apply lift to convert an Expr to an AD s Expr.

If you are interested in using the ad package for symbolic differentiation. Lennart Augustsson's traced package works well.

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