[英]“No instance for .. ”, Haskell'98 code on GHC7.8.3
I'm trying out the attached code for linear regression with automatic differentiation . 我正在尝试使用自动微分进行线性回归的附加代码。 It specifies a datatype [Dual][2] made of two Floats, and declares it to be instance of Num, Fractional and Floating.
它指定由两个Float组成的数据类型[Dual] [2],并将其声明为Num,Fractional和Floating的实例。 As in all fitting/regression tasks, there's a scalar cost function, parametrized by the fitting parameters c and m, and an optimizer which improves on estimates of these two parameters by gradient descent.
像在所有拟合/回归任务中一样,存在一个由拟合参数c和m参数化的标量成本函数,以及一个优化器,该优化器通过梯度下降改进了这两个参数的估计。
Question I'm using GHC 7.8.3, and the authors explicitly mention that this is H98 code (I mentioned it in the title because it's the only substantial difference I can think of between my setup and the Author's, however plz correct if wrong). 问题我使用的是GHC 7.8.3,作者明确提到这是H98代码(我在标题中提到了它,因为这是我可以想到的设置与作者设置之间的唯一实质性区别,但是如果出错,请plz正确) 。 Why does it choke within the definition of the cost function?
为什么它在成本函数的定义之内cho塞? My understanding is: the functions idD and constD map Floats to Duals, g is polymorphic (it can perform algebraic operations on Dual inputs because Dual inherits from Num, Fractional and Floating), and deriv maps Duals to Doubles.
我的理解是:函数idD和constD将Floats映射为Duals,g是多态的(由于Dual从Num,Fractional和Floating继承,因此它可以对Dual输入执行代数运算),并且将Duals映射为Doubles。 The type signature for g (the eta-reduced cost function wrt the data) was inferred.
推断出g的类型签名(数据减少了eta的成本函数)。 I tried omitting it, and making it more general by substituting a Floating type constraint to the Fractional one.
我尝试省略它,并通过将Floating类型约束替换为Fractional约束使它更通用。 Moreover, I tried converting the numeric types of c and m inline with (fromIntegral c :: Double), to no avail.
而且,我尝试将(fromIntegral c :: Double)内联的c和m的数字类型转换为无效。
Specifically this code gives this error: 具体来说,这段代码给出了以下错误:
No instance for (Integral Dual) arising from a use of ‘g’
In the first argument of ‘flip’, namely ‘g’
In the expression: flip g (constD c)
In the second argument of ‘($)’, namely ‘flip g (constD c) $ idD m’
Any hints, please? 有什么提示吗? I'm sure it's a very noob question, but I just don't get it.
我敢肯定这是一个非常菜鸟的问题,但我只是不明白。
The complete code is the following: 完整的代码如下:
{-# LANGUAGE NoMonomorphismRestriction #-}
module ADfw (Dual(..), f, idD, cost) where
data Dual = Dual Double Double deriving (Eq, Show)
constD :: Double -> Dual
constD x = Dual x 0
idD :: Double -> Dual
idD x = Dual x 1.0
instance Num Dual where
fromInteger n = constD $ fromInteger n
(Dual x x') + (Dual y y') = Dual (x+y) (x' + y')
(Dual x x') * (Dual y y') = Dual (x*y) (x*y' + y*x')
negate (Dual x x') = Dual (negate x) (negate x')
signum _ = undefined
abs _ = undefined
instance Fractional Dual where
fromRational p = constD $ fromRational p
recip (Dual x x') = Dual (1.0 / x) (- x' / (x*x))
instance Floating Dual where
pi = constD pi
exp (Dual x x') = Dual (exp x) (x' * exp x)
log (Dual x x') = Dual (log x) (x' / x)
sqrt (Dual x x') = Dual (sqrt x) (x' / (2 * sqrt x))
sin (Dual x x') = Dual (sin x) (x' * cos x)
cos (Dual x x') = Dual (cos x) (x' * (- sin x))
sinh (Dual x x') = Dual (sinh x) (x' * cosh x)
cosh (Dual x x') = Dual (cosh x) (x' * sinh x)
asin (Dual x x') = Dual (asin x) (x' / sqrt (1 - x*x))
acos (Dual x x') = Dual (acos x) (x' / (-sqrt (1 - x*x)))
atan (Dual x x') = Dual (atan x) (x' / (1 + x*x))
asinh (Dual x x') = Dual (asinh x) (x' / sqrt (1 + x*x))
acosh (Dual x x') = Dual (acosh x) (x' / (sqrt (x*x - 1)))
atanh (Dual x x') = Dual (atanh x) (x' / (1 - x*x))
-- example
-- f = sqrt . (* 3) . sin
-- f' x = 3 * cos x / (2 * sqrt (3 * sin x))
-- linear fit sum-of-squares cost
-- cost :: Fractional s => s -> s -> [s] -> [s] -> s
cost m c x y = (/ (2 * (fromIntegral $ length x))) $
sum $ zipWith errSq x y
where
errSq xi yi = zi * zi
where
zi = yi - (m * xi + c)
-- test data
x_ = [1..10]
y_ = [a | a <- [1..20], a `mod` 2 /= 0]
-- learning rate
gamma = 0.04
g :: (Integral s, Fractional s) => s -> s -> s
g m c = cost m c x_ y_
deriv (Dual _ x') = x'
z_ = (0.1, 0.1) : map h z_
h (c, m) = (c - gamma * cd, m - gamma * md) where
cd = deriv $ g (constD m) $ idD c
md = deriv $ flip g (constD c) $ idD m
-- check for convergence
main = do
take 2 $ drop 1000 $ map (\(c, m) -> cost m c x_ y_) z_
take 2 $ drop 1000 $ z_
where the test data x_ and y_ are arrays and the learning rate gamma a scalar. 其中测试数据x_和y_是数组,学习率gamma是标量。
[2]: The two fields of a Dual object are in fact adjoint one with the other, if we see the derivative as an operator [2]:如果我们将导数视为运算符,则对偶对象的两个字段实际上彼此相邻
In the original code, I don't see a type signature for g
. 在原始代码中,我没有看到
g
的类型签名。 In your code, you have specifically written 在您的代码中,您专门编写了
g :: (Integral s, Fractional s) => s -> s -> s
The error message says there's no Integral
instance for Dual
. 错误消息指出
Dual
没有Integral
实例。 The code manually defines instances for Num
and Fractional
, but not Integral
. 该代码手动定义了
Num
和Fractional
实例,但没有Integral
实例。
I'm not actually sure why g
needs to be Integral
. 我实际上不确定为什么
g
必须是Integral
。 If you remove that constraint, the code may even work... 如果您删除该约束,则该代码甚至可以工作...
EDIT: It seems the Integral
instance is necessary because of your use of mod
to generate test data. 编辑:由于您使用
mod
生成测试数据,因此似乎需要Integral
实例。 I'm not really sure what this huge block of code does, but I suspect if you apply fromIntegral
to convert everything to (say) Double
, then it may work. 我不确定这块巨大的代码是做什么的,但是我怀疑如果您应用
fromIntegral
将所有内容都转换为(例如) Double
,那么它可能会起作用。
(I suspect making Dual
an instance of Integral
is probably not what the original authors intended. Then again, I don't really understand the code, so...) (我怀疑使
Dual
成为Integral
的实例可能不是原始作者想要的。然后,我也不是很了解代码,所以...)
First, (Integral s, Fractional s)
makes no sense; 首先,
(Integral s, Fractional s)
没有意义; Integral
is for Euclidean domains (ones with div
and mod
), while Fractional
is for fields (ones with /
). Integral
适用于欧几里得域(带有div
和mod
),而Fractional
适用于字段(带/
)。 If you have true division all your remainders are going to be zero... . 如果您有真正的除法,那么您所有的余数都将为零...。
I think the problem is y_
's attempt to filter to odd numbers. 我认为问题是
y_
尝试过滤为奇数。 Haskell 98 defines a 'stepped' range form for numbers, so you could write y_
as [1,3..19]
. Haskell 98为数字定义了“步进”范围形式,因此您可以将
y_
写为[1,3..19]
。 That should allow y_
to be used at the type [Dual]
, which should allow g
to use it without needing the Integral
constraint. 那应该允许
y_
在[Dual]
类型上使用,这应该允许g
在不需要Integral
约束的情况下使用它。
Edit: Ørjan Johansen points out that you need an Enum
instance for Dual
as well, which is actually fairly easy to implement (this is pretty standard for numeric types; I basically copied GHC's instance for Double
(which is identical to its instance for Float
, for example)): 编辑:ØrjanJohansen指出,您还需要
Dual
的Enum
实例,这实际上很容易实现(对于数字类型,这是相当标准的;我基本上为Double
复制了GHC的实例(这与Float
实例相同,例如)):
instance Enum Dual where
succ x = x + 1
pred x = x - 1
toEnum = fromIntegral
fromEnum (Dual x _) = fromEnum x
enumFrom = numericEnumFrom
enumFromTo = numericEnumFromTo
enumFromThen = numericEnumFromThen
enumFromThenTo = numericEnumFromThenTo
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