我正在尝试使用 Numeric.AD 和自定义 Expr 类型。我想计算用户输入表达式的符号渐变。 第一次具有常数的试验表达式效果很好:
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拉出:
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
我得到
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 接受这一点?
我的猜测是您忘记应用提升将Expr
转换为AD s Expr
。
如果您有兴趣使用广告包进行符号区分。伦纳特·奥古斯特森(Lennart Augustsson)的跟踪包效果很好。
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。 我不知道这个库的界面。 但是,我的猜测是calcGrad1 :: (Num a, Floating a) => Expr a -> [Expr a]
正确的签名。