问题
I need small lists of gaussian random numbers for a simulation and so I tried the following:
import System.Random
seed = 10101
gen = mkStdGen seed
boxMuller mu sigma (r1,r2) = mu + sigma * sqrt (-2 * log r1) * cos (2 * pi * r2)
This is just the Box-Muller algorithm - given r1, r2 uniform random numbers in the [0,1] interval it returns a gaussian random number.
normals 0 g = []
normals n g = take n $ map (boxMuller 0 1) $ pairs $ randoms g
where pairs (x:y:zs) = (x,y):(pairs zs)
So I used this normals function everytime I needed my list of random numbers.
The problem with that must be apparent: it generates always the same sequence cause I'm using always the same seed! I'm not getting new sequences, I'm only getting the first n values of the sequence all the time.
What I was pretending clearly was that, when I type:
x = normal 10
y = normal 50
I'd have x to be the first 10 values of map (boxMuller 0 1) $ pairs $ randoms g and y to be the next 50 values on this list, and so on.
Of course this is impossible, cause a function must always return the same values given the same input. How do I escape this trap?
回答1:
I think that doing your computations that require random numbers inside a monad that abstracts away the generator is the cleanest thing. Here is what that code would look like:
We are going to put the StdGen instance in a state monad, then provide some sugar over the state monad's get and set method to give us random numbers.
First, load the modules:
import Control.Monad.State (State, evalState, get, put)
import System.Random (StdGen, mkStdGen, random)
import Control.Applicative ((<$>))
(Normally I would probably not specify the imports, but this makes it easy to understand where each function is coming from.)
Then we will define our "computations requiring random numbers" monad; in this case, an alias for State StdGen called R. (Because "Random" and "Rand" already mean something else.)
type R a = State StdGen a
The way R works is: you define a computation that requires a stream of random numbers (the monadic "action"), and then you "run it" with runRandom:
runRandom :: R a -> Int -> a
runRandom action seed = evalState action $ mkStdGen seed
This takes an action and a seed, and returns the results of the action. Just like the usual evalState, runReader, etc.
Now we just need sugar around the State monad. We use get to get the StdGen, and we use put to install a new state. This means, to get one random number, we would write:
rand :: R Double
rand = do
gen <- get
let (r, gen') = random gen
put gen'
return r
We get the current state of the random number generator, use it to get a new random number and a new generator, save the random number, install the new generator state, and return the random number.
This is an "action" that can be run with runRandom, so let's try it:
ghci> runRandom rand 42
0.11040701265689151
it :: Double
This is a pure function, so if you run it again with the same arguments, you will get the same result. The impurity stays inside the "action" that you pass to runRandom.
Anyway, your function wants pairs of random numbers, so let's write an action to produce a pair of random numbers:
randPair :: R (Double, Double)
randPair = do
x <- rand
y <- rand
return (x,y)
Run this with runRandom, and you'll see two different numbers in the pair, as you'd expect. But notice that you didn't have to supply "rand" with an argument; that's because functions are pure, but "rand" is an action, which need not be pure.
Now we can implement your normals function. boxMuller is as you defined it above, I just added a type signature so I can understand what's going on without having to read the whole function:
boxMuller :: Double -> Double -> (Double, Double) -> Double
boxMuller mu sigma (r1,r2) = mu + sigma * sqrt (-2 * log r1) * cos (2 * pi * r2)
With all the helper functions/actions implemented, we can finally write normals, an action of 0 arguments that returns a (lazily-generated) infinite list of normally-distributed Doubles:
normals :: R [Double]
normals = mapM (\_ -> boxMuller 0 1 <$> randPair) $ repeat ()
You could also write this less concisely if you want:
oneNormal :: R Double
oneNormal = do
pair <- randPair
return $ boxMuller 0 1 pair
normals :: R [Double]
normals = mapM (\_ -> oneNormal) $ repeat ()
repeat () gives the monadic action an infinite stream of nothing to invoke the function with (and is what makes the result of normals infinitely long). I had initially written [1..], but I rewrote it to remove the meaningless 1 from the program text. We are not operating on integers, we just want an infinite list.
Anyway, that's it. To use this in a real program, you just do your work requiring random normals inside an R action:
someNormals :: Int -> R [Double]
someNormals x = liftM (take x) normals
myAlgorithm :: R [Bool]
myAlgorithm = do
xs <- someNormals 10
ys <- someNormals 10
let xys = zip xs ys
return $ uncurry (<) <$> xys
runRandom myAlgorithm 42
The usual techniques for programming monadic actions apply; keep as little of your application in R as possible, and things won't be too messy.
Oh, and on other thing: laziness can "leak" outside of the monad boundary cleanly. This means you can write:
take 10 $ runRandom normals 42
and it will work.
回答2:
The list you get from randoms is infinite, and when you use the finite prefix, you need not to throw away the infinite tail. You can pass the random numbers in as an additional parameter and return the unused random numbers as an additional result, or you can park an infinite sequence of random numbers in a state monad.
A similar problem occurs for compilers and other codes that want a supply of unique symbols. This is just a real pain in Haskell, because you are threading state (of the random-number generator or of the symbol generator) throughout the code.
I've done randomized algorithms both with explicit parameters and with a monad, and neither one is really satisfying. If you grok monads I probably have a slight recommendation to use a state monad containing the infinite list of random numbers that have not yet been used.
回答3:
You could also sidestep the problem by using newStdGen and then you'll get a different seed (virtually) every time.
来源：https://stackoverflow.com/questions/2110535/sampling-sequences-of-random-numbers-in-haskell