问题
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?
回答1:
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.
回答2:
When GHC cannot deduce a type equality signature for a valid function, as in your case, the solution is to give the function a type signature. I do not know the interface to this library. However, my guess is that the correct signature is calcGrad1 :: (Num a, Floating a) => Expr a -> [Expr a]
.
来源:https://stackoverflow.com/questions/5938337/numeric-ad-and-typing-problem