F＃泛型数学：如何用op_GreaterThan编写函数 [英] F# Generic Math: how to write function with op_GreaterThan
问题描述
一个（Oliver）Heaviside步骤函数是一个函数，如果x是负数，它将返回零，否则它会回退一个。
//尝试1：
let inline stepFct1< ^ T当^ T时：（静态成员op_GreaterThan：^ T * float - > bool）
> （x：^ T）：^ T =
// if（^ T：（static member op_GreaterThan）（x 0.0））那么x // ouch也会失败
if（>）x 0.0 then x
else 0.0
编译器说：错误FS0001：类型参数缺少约束'当^ T：comparison'
//尝试2：
let inline stepFct2 <^ T when ^ T :(静态成员（>）：^ T * ^ T - > bool）> （x：^ T）：^ T =
匹配x和
| x当x> 0.0 - > 1.0
| 0.0
FSC表示：错误FS0010：模式中的意外中缀运算符
动机：
我试图重写Ian的Cumulative-Normal和Black-Scholes函数这里使用自动微分（DiffSharp）。 Ian的Cumulative Normal在浮点工程中工作，我想要一个适用于任何数字类型（包括AutoDiff.DualG）的通用版本。
累积正常函数包含一个大于语句。
编辑：Gustavo，谢谢，我接受了你的答案 - 简单的步骤函数现在编译。
但它似乎没有帮助累积正常情况。
给出以下代码：
//累积正态分布函数 - 试图编写一个通用版本
let （b4，b5）=（ - 1.821255978,1.330274429）（b1，b2，b3）=（0.319381530，-0.356563782，1.781477937）
let（b4，b5）
let（p，c）=（0.2316419，0.39894228）
let（zero，one）=（LanguagePrimitives.GenericZero，LanguagePrimitives.GenericOne）
if x>零然后
让t = 1 /（one + p * x）
（one-c * exp（-x * x / 2.0）* t *（t *（t *（t *（t * b5 + b4）+ b3）+ b2）+ b1））
else
let t = 1.0 /（one-p * x）
（c * exp（-x * x / 2.0）* t *（t *（t *（t *（t * b5 + b4）+ b3）+ b2）+ b1））
FSI说：
C：\stdin（116,32）：警告FS0064 ：这个构造使得代码比通过类型注释指示的通用
更少。
类型变量'T'被限制为'float'类型。
val inline CDF：x：float - >浮动
> CDF 0.1M ;;
CDF 0.1M ;;
---- ^^^^
C：\ stdin（122,5）：错误FS0001：此表达式预计有类型
float
，但这里有类型
十进制
>
有谁知道如何让CDF变得一般？
GenericOne 其余 //尝试1：
let inline stepFct1 x =
let zero = LanguagePrimitives.GenericZero
如果x>零然后x
else零
我看了一下你用函数发送的链接你想实现。 FSharpPlus（F＃+）可以帮助您编写通用的数学代码，因为它包含专用的通用数字模块。或者至少你可以从那里获得一些技巧。
更新
您的更新问题将复杂度提高到更高的水平，这里使用最新版本的 F＃+ project：
$ b $ pre $ let inline CDF（x：^ T）：^ T =
let num x = （b 1，b 2，b 3）=（数字319381530I，数字-356563782I，数字1781477937I）
let（b4，b5）=（数字 - >数字> 1000000000I） （p，c）=（num 0231641900I，num 0398942280I）
let（zero，one，two）= 0G，1G，2G
if x>零然后
让t = 1 /（one + p * x）
（one-c * exp（-x * x / 2）* t *（t *（t *（t *（t * b5 + b4）+ b3）+ b2）+ b1））
else
let t = one /（one-p * x）
（c * exp（-x * x /两个）* t *（t *（t *（t *（t * b5 + b4）+ b3）+ b2）+ b1））
不幸的是，在这个时候，我意识到一些函数在库中被标记为内部函数，因此没有公开，但我在一个工作示例中重新创建了它们 here ，这样你就可以测试你的函数了，它很适合 float 和 FLOAT32 。 b
$ b
新版本将在今年年底之前发布，但同时你可以分支它，删除内部并编译它，或者重新创建函数就像我在链接的例子中所做的那样。
如果你对Generic Maths感兴趣，可以随意使用代码或用例。
in F#, how does one write a generic-math step function?
An (Oliver) Heaviside step function is function that returns zero if x is negative, otherwise it retuns one.
Here is a summary of my attempts so far:
// attempt 1:
let inline stepFct1< ^T when ^T : (static member op_GreaterThan: ^T * float -> bool)
> (x:^T) : ^T =
//if (^T : (static member op_GreaterThan) (x 0.0) ) then x //ouch fails also
if (>) x 0.0 then x
else 0.0
compiler says: error FS0001: A type parameter is missing a constraint 'when ^T : comparison'
// attempt 2:
let inline stepFct2<^T when ^T : (static member (>): ^T * ^T -> bool) > (x:^T) : ^T =
match x with
| x when x > 0.0 -> 1.0
| 0.0
FSC says: error FS0010: Unexpected infix operator in pattern
Motivation:
I am trying to rewrite Ian's Cumulative-Normal and Black-Scholes functions here to use Automatic Differentiation (DiffSharp). Ian's Cumulative Normal works on floats, I would like a generic version that works on any numeric type, including AutoDiff.DualG. The cumulative normal function contains a "greater than" statement.
EDIT: Gustavo, thanks, I have accepted your answer - the simple step function now compiles.
But it doesn't seem to help with the Cumulative Normal case. Given this code:
// Cumulative Normal Distribution Function - attempt to write a generic version
let inline CDF(x:^T) : ^T =
let (b1,b2,b3) = (0.319381530, -0.356563782, 1.781477937)
let (b4,b5) = (-1.821255978, 1.330274429)
let (p , c ) = (0.2316419 , 0.39894228)
let (zero, one) = (LanguagePrimitives.GenericZero, LanguagePrimitives.GenericOne)
if x > zero then
let t = one / (one + p * x)
(one - c * exp( -x * x / 2.0)* t * (t*(t*(t*(t*b5+b4)+b3)+b2)+b1))
else
let t = 1.0 / (one - p * x)
(c * exp( -x * x / 2.0)* t * (t*(t*(t*(t*b5+b4)+b3)+b2)+b1))
FSI says:
C:\stdin(116,32): warning FS0064: This construct causes code to be less generic
than indicated by the type annotations.
The type variable 'T has been constrained to be type 'float'.
val inline CDF : x:float -> float
> CDF 0.1M;;
CDF 0.1M;;
----^^^^
C:\stdin(122,5): error FS0001: This expression was expected to have type
float
but here has type
decimal
>
Does anyone know how to make CDF generic?
Use LanguagePrimitives.GenericZero / GenericOne and let type inference do the rest
// attempt 1:
let inline stepFct1 x =
let zero = LanguagePrimitives.GenericZero
if x > zero then x
else zero
I had a look at the link you sent with the function you want to implement. FSharpPlus (F#+) may help you to write generic math code since it contains a dedicated Generic Numbers module. Or at least you can grab some techniques from there.
UPDATE
Regarding your updated question, which takes the complexity to a higher level, here is a solution using the latest version of the F#+ project:
let inline CDF(x:^T) : ^T =
let num x = fromRational (x </ratio/> 1000000000I)
let (b1,b2,b3) = (num 319381530I , num -356563782I , num 1781477937I)
let (b4,b5) = (num -1821255978I , num 1330274429I)
let (p , c ) = (num 0231641900I , num 0398942280I)
let (zero, one, two) = 0G, 1G, 2G
if x > zero then
let t = one / (one + p * x)
(one - c * exp( -x * x / two)* t * (t*(t*(t*(t*b5+b4)+b3)+b2)+b1))
else
let t = one / (one - p * x)
(c * exp( -x * x / two)* t * (t*(t*(t*(t*b5+b4)+b3)+b2)+b1))
Unfortunately at this time I realised some functions were marked as internal in the library and therefore were not exposed, but I re-created them in a working example here so you can test your function, which works nicely with float and float32.
A new version will be released before end of this year, but in the mean time you can branch it, remove the internals and compile it, or just re-create the functions as I did in the linked example.
If you are interested in Generic Maths feel free to contribute with code or use cases.
这篇关于F＃泛型数学：如何用op_GreaterThan编写函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
