我正在尝试将一些 Python 转换为 F#,特别是 numpy.random.randn .
该函数采用可变数量的 int 参数,并根据参数的数量返回不同维度的数组。
我相信这是不可能的,因为一个函数不能返回不同类型( int[]
、 int[][]
、 int[][][]
等),除非它们是受歧视联合的一部分,但在 promise 之前要确定一种解决方法。
健全性检查:
member self.differntarrays ([<ParamArray>] dimensions: Object[]) =
match dimensions with
| [| dim1 |] ->
[|
1
|]
| [| dim1; dim2 |] ->
[|
[| 2 |],
[| 3 |]
|]
| _ -> failwith "error"
导致错误:
This expression was expected to have type
int
but here has type
'a * 'b
与
expression
正在:[| 2 |], [| 3 |]
和 int
指 [| 1 |]
中的 1即
1
的类型与[| 2 |], [| 3 |]
的类型不一样TLDR;
numpy.random.randn
numpy.random.randn(d0, d1, ..., dn)
Return a sample (or samples) from the “standard normal” distribution.
If positive, int_like or int-convertible arguments are provided, randn generates an array of shape (d0, d1, ..., dn), filled with random floats sampled from a univariate “normal” (Gaussian) distribution of mean 0 and variance 1 (if any of the d_i are floats, they are first converted to integers by truncation). A single float randomly sampled from the distribution is returned if no argument is provided.
交互式 python session 的示例:
np.random.randn(1) - array([-0.28613356])
np.random.randn(2) - array([-1.7390449 , 1.03585894])
np.random.randn(1,1)- array([[ 0.04090027]])
np.random.randn(2,3)- array([[-0.16891324, 1.05519898, 0.91673992],
[ 0.86297031, 0.68029926, -1.0323683 ]])
代码用于 Neural Networks and Deep Learning并且由于出于性能原因这些值需要可变,因此不能选择使用不可变列表。
最佳答案
你是对的 - 浮点数组 float[]
与浮点数组的数组类型不同float[][]
或二维浮点数组 float[,]
所以你不能编写一个函数,根据输入参数返回一个或另一个。
如果你想做一些类似 Python 的 rand
,你可以写一个重载的方法:
type Random() =
static let rnd = System.Random()
static member Rand(n) = [| for i in 1 .. n -> rnd.NextDouble() |]
static member Rand(n1, n2) = [| for i in 1 .. n1 -> Random.Rand(n2) |]
static member Rand(n1, n2, n3) = [| for i in 1 .. n1 -> Random.Rand(n2, n3) |]
关于arrays - 从一个函数返回不同维度的数组;在 F# 中可能吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34599909/