如何使用random-fu(Haskell)从正态分布生成n个随机数？

Question

这个问题对我来说似乎很简单,但我找不到答案.

假设您要从正态分布中获取大小为n的样本,其均值为10,方差为1,然后得到它们的平均值和方差,以验证样本是否来自正态分布.

我想这会是这样的:

values = take n $ sample (normal 10 1)
(avg values, variance values)

我正在尝试使用库random-fu,所以如果你能用lib提供答案,我真的很感激.

Answer 1

这里的尴尬总是存在于Haskell中的随机性 - 因为Haskell是纯粹的,你需要有某种"随机"源.random-fu使用RandomSource这个目的,那么你这个单子里面,当你想操纵随机值工作.

import Data.Random
import Control.Monad (replicateM)

average :: [Double] -> Double
average xs = sum xs / fromIntegral (length xs)

variance :: [Double] -> Double
variance xs = average [ (x - m)^2  | x <- xs ] 
  where m = average xs

main :: IO ()
main = do
  sample <- runRVar (replicateM 10 (normal 10 1)) StdRandom :: IO [Double]
  putStrLn $ "Average: " ++ show (average sample)
  putStrLn $ "Variance: " ++ show (variance sample)

试运行似乎给了我合理的输出:

ghci> main
Average: 10.294887142436771
Variance: 0.7129578122237161
ghci> main
Average: 9.677325092160597
Variance: 0.9894150286175698
ghci> main
Average: 9.714089727813253
Variance: 1.0279068711054316
ghci> main
Average: 10.32028785267642
Variance: 0.8574243439019995
ghci> main
Average: 9.696843993234065
Variance: 0.45301180269725994