Random Integer in Haskell [duplicate] - haskell

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Random number in Haskell [duplicate]
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I am in the process of learning Haskell and to learn I want to generate a random Int type. I am confused because the following code works. Basically, I want an Int not an IO Int.
In ghci this works:
Prelude> import System.Random
Prelude System.Random> foo <- getStdRandom (randomR (1,1000000))
Prelude System.Random> fromIntegral foo :: Int
734077
Prelude System.Random> let bar = fromIntegral foo :: Int
Prelude System.Random> bar
734077
Prelude System.Random> :t bar
bar :: Int
So when I try to wrap this up with do it fails and I don't understand why.
randomInt = do
tmp <- getStdRandom (randomR (1,1000000))
fromIntegral tmp :: Int
The compiler produces the following:
Couldn't match expected type `IO b0' with actual type `Int'
In a stmt of a 'do' block: fromIntegral tmp :: Int
In the expression:
do { tmp <- getStdRandom (randomR (1, 1000000));
fromIntegral tmp :: Int }
In an equation for `randomInt':
randomInt
= do { tmp <- getStdRandom (randomR (1, 1000000));
fromIntegral tmp :: Int }
Failed, modules loaded: none.
I am new to Haskell, so if there is a better way to generate a random Int without do that would be preferred.
So my question is, why does my function not work and is there a better way to get a random Int.

The simple answer is that you can't generate random numbers without invoking some amount of IO. Whenever you get the standard generator, you have to interact with the host operating system and it makes a new seed. Because of this, there is non-determinism with any function that generates random numbers (the function returns different values for the same inputs). It would be like wanting to be able to get input from STDIN from the user without it being in the IO monad.
Instead, you have a few options. You can write all your code that depends on a randomly generated value as pure functions and only perform the IO to get the standard generator in main or some similar function, or you can use the MonadRandom package that gives you a pre-built monad for managing random values. Since most every pure function in System.Random takes a generator and returns a tuple containing the random value and a new generator, the Rand monad abstracts this pattern out so that you don't have to worry about it. You can end up writing code like
import Control.Monad.Random hiding (Random)
type Random a = Rand StdGen a
rollDie :: Int -> Random Int
rollDie n = getRandomR (1, n)
d6 :: Random Int
d6 = rollDie 6
d20 :: Random Int
d20 = rollDie 20
magicMissile :: Random (Maybe Int)
magicMissile = do
roll <- d20
if roll > 15
then do
damage1 <- d6
damage2 <- d6
return $ Just (damage1 + damage2)
else return Nothing
main :: IO ()
main = do
putStrLn "I'm going to cast Magic Missile!"
result <- evalRandIO magicMissile
case result of
Nothing -> putStrLn "Spell fizzled"
Just d -> putStrLn $ "You did " ++ show d ++ " damage!"
There's also an accompanying monad transformer, but I'd hold off on that until you have a good grasp on monads themselves. Compare this code to using System.Random:
rollDie :: Int -> StdGen -> (Int, StdGen)
rollDie n g = randomR (1, n) g
d6 :: StdGen -> (Int, StdGen)
d6 = rollDie 6
d20 :: StdGen -> (Int, StdGen)
d20 = rollDie 20
magicMissile :: StdGen -> (Maybe Int, StdGen)
magicMissile g =
let (roll, g1) = d20 g
(damage1, g2) = d6 g1
(damage2, g3) = d6 g2
in if roll > 15
then (Just $ damage1 + damage2, g3)
else Nothing
main :: IO ()
main = do
putStrLn "I'm going to case Magic Missile!"
g <- getStdGen
let (result, g1) = magicMissile g
case result of
Nothing -> putStrLn "Spell fizzled"
Just d -> putStrLn $ "You did " ++ show d ++ " damage!"
Here we have to manually have to manage the state of the generator and we don't get the handy do-notation that makes our order of execution more clear (laziness helps in the second case, but it makes it more confusing). Manually managing this state is boring, tedious, and error prone. The Rand monad makes everything much easier, more clear, and reduces the chance for bugs. This is usually the preferred way to do random number generation in Haskell.
It is worth mentioning that you can actually "unwrap" an IO a value to just an a value, but you should not use this unless you are 100% sure you know what you're doing. There is a function called unsafePerformIO, and as the name suggests it is not safe to use. It exists in Haskell mainly for when you are interfacing with the FFI, such as with a C DLL. Any foreign function is presumed to perform IO by default, but if you know with absolute certainty that the function you're calling has no side effects, then it's safe to use unsafePerformIO. Any other time is just a bad idea, it can lead to some really strange behaviors in your code that are virtually impossible to track down.

I think the above is slightly misleading. If you use the state monad then you can do something like this:
acceptOrRejects :: Int -> Int -> [Double]
acceptOrRejects seed nIters =
evalState (replicateM nIters (sample stdUniform))
(pureMT $ fromIntegral seed)
See here for an extended example of usage:Markov Chain Monte Carlo

Related

Couldn't match type ‘IO’ with ‘[]’

I wrote a function to generate two random numbers, which I then pass to a different function to use them there. The code for this is:
randomIntInRange :: (Int, Int, Int, Int) -> Board
randomIntInRange (min,max,min2,max2) = do r <- randomRIO (min, max)
r2 <- randomRIO (min2, max2)
randomCherryPosition (r, r2)
And the function this function calls in its 'do' block is:
randomCherryPosition :: (Int, Int) -> Board
randomCherryPosition (x, y) = initialBoard & element y . element x .~ C
Where initialBoard is a list of lists and C is a predefined data type. I am using lens to change values inside the list. Running this gives me the error:
Couldn't match type ‘IO’ with ‘[]’
Expected type: [Int]
Actual type: IO Int
for both r and r2 lines. I have absolutely no idea what is going on here, or what i'm doing wrong, so I would greatly appreciate any help.
randomRIO has type IO Int, not Int. As long as you use any IO functions, your surrounding function must also be in IO:
randomIntInRange :: (Int, Int, Int, Int) -> IO Board
randomIntInRange (min,max,min2,max2) = do r <- randomRIO (min, max)
r2 <- randomRIO (min2, max2)
pure $ randomCherryPosition (r, r2)
randomRIO is not a pure function. It returns a different value every time. Haskell bans such functions. There are numerous benefits that come from banning such functions, which I'm going to go into here. But you can still have such function if you wrap it in IO. The type IO Int means "it's a program that, when executed, will produce an Int". So when you call randomRIO (min, max), it returns you not an Int, but a program, which you can then execute to get an Int. You do the execution via the do notation with left arrow, but the result of that would also be a similar program.
Unfortunately there is no perfect solution to this problem. It has already been discussed on Stackoverflow, for example here.
The above solution provided by Fyodor involves IO. It works. The main drawback is that IO will propagate into your type system.
However, it is not strictly necessary to involve IO just because you want to use random numbers. An in-depth discussion of the pros and cons involved is available there.
There is no perfect solution, because something has to take care of updating the state of the random number generator every time you pick a random value. In imperative languages such as C/C++/Fortran, we use side effects for this. But Haskell functions have no side effects. So that something can be:
the Haskell IO subsystem (as in randomRIO)
yourself as the programmer - see code sample #1 below
a more specialized Haskell subsystem, for which you need to have: import Control.Monad.Random - see code sample #2 below
You can solve the problem without involving IO by creating your own random number generator, using library function mkStdGen, and then passing the updated states of this generator manually around your computations. In your problem, this gives something like this:
-- code sample #1
import System.Random
-- just for type check:
data Board = Board [(Int, Int)] deriving Show
initialBoard :: Board
initialBoard = Board [(0, 0)]
randomCherryPosition :: (Int, Int) -> Board
randomCherryPosition (x, y) = -- just for type check
let ls0 = (\(Board ls) -> ls) initialBoard
ls1 = (x, y) : ls0
in Board ls1
-- initial version with IO:
randomIntInRange :: (Int, Int, Int, Int) -> IO Board
randomIntInRange (min,max, min2,max2) = do r1 <- randomRIO (min, max)
r2 <- randomRIO (min2, max2)
return $ randomCherryPosition (r1, r2)
-- version with manual passing of state:
randomIntInRangeA :: RandomGen tg => (Int, Int, Int, Int) -> tg -> (Board, tg)
randomIntInRangeA (min1,max1, min2,max2) rng0 =
let (r1, rng1) = randomR (min1, max1) rng0
(r2, rng2) = randomR (min2, max2) rng1 -- pass the newer RNG
board = randomCherryPosition (r1, r2)
in (board, rng2)
main = do
-- get a random number generator:
let mySeed = 54321 -- actually better to pass seed from the command line.
let rng0 = mkStdGen mySeed
let (board1, rng) = randomIntInRangeA (0,10, 0,100) rng0
putStrLn $ show board1
This is cumbersome but can be made to work.
A more elegant alternative consists in using MonadRandom.
The idea is to define a monadic action representing the randomness-involving computation, and then to run this action using the aptly named runRand function.
This gives this code instead:
-- code sample #2
import System.Random
import Control.Monad.Random
-- just for type check:
data Board = Board [(Int, Int)] deriving Show
initialBoard :: Board
initialBoard = Board [(0, 0)]
-- just for type check:
randomCherryPosition :: (Int, Int) -> Board
randomCherryPosition (x, y) =
let ls0 = (\(Board ls) -> ls) initialBoard
ls1 = (x, y) : ls0
in Board ls1
-- monadic version of randomIntInRange:
randomIntInRangeB :: RandomGen tg => (Int, Int, Int, Int) -> Rand tg Board
randomIntInRangeB (min1,max1, min2,max2) =
do
r1 <- getRandomR (min1,max1)
r2 <- getRandomR (min2,max2)
return $ randomCherryPosition (r1, r2)
main = do
-- get a random number generator:
let mySeed = 54321 -- actually better to pass seed from the command line.
let rng0 = mkStdGen mySeed
-- create and run the monadic action:
let action = randomIntInRangeB (0,10, 0,100) -- of type: Rand tg Board
let (board1, rng) = runRand action rng0
putStrLn $ show board1
This is definitely less error prone than code sample #1, so you would typically prefer this solution as soon as your computations become complex enough. All the functions involved are ordinary pure Haskell functions, which the compiler can fully optimize using its usual techniques.

Haskell random numbers

I am trying to generate a sample of random numbers in Haskell
import System.Random
getSample n = take n $ randoms g where
g = newStdGen
but it seems I am not quite using newStdGen the right way. What am I missing?
First off, you probably don't want to use newStdGen. The biggest problem is that you'll get a different seed every time you run your program, so no results will be reproducible. In my opinion, mkStdGen is a better choice as it requires you to give it a seed. This means you will get the same sequence of (pseudo)random numbers every time. If you want a different sequence, just change the seed.
The second problem with newStdGen is that since it's impure, you'll end up in the IO monad which can be a bit inconvenient.
sample :: Int -> IO [Int]
sample n = do
gen <- newStdGen
return $ take n $ randoms gen
You can use do-notation to 'extract' the values and then sum them:
main :: IO ()
main = do
xs <- sample 10
s = sum xs
print s
Or you could 'fmap' the function over the result (but notice that at some point you will probably need to extract the value):
main :: IO ()
main = do
s <- fmap sum $ sample 10
print s
The fmap function is a generalized version of map. Just like map applies a function to the values inside a list, fmap can apply a function to values inside IO.
Another problem with this sample function is that if we call it again, it starts with a fresh seed instead of continuing the previous (pseudo)random sequence. Again, this make reproducing results impossible. In order to fix this problem, we need to pass in the seed and return a new seed. Unfortunately, randoms does not return the next seed for us, so we'll have to write this from scratch using random.
sample :: Int -> StdGen -> ([Int],StdGen)
sample n seed1 = case n of
0 -> ([],seed1)
k -> let (rs,seed2) = sample (k-1) seed1
(r, seed3) = random seed2
in ((r:rs),seed3)
Our main function is now
main :: IO ()
main = do
let seed1 = mkStdGen 123456
(xs,seed2) = sample 10 seed1
s = sum xs
(ys,seed3) = sample 10 seed2
t = sum ys
print s
print t
I know this seems like an awful lot of work just to to use random numbers, but the advantages are worth it. We can generate all of our randomness with a single seed which guarantees that the results can be reproduced.
Of course, this being Haskell, we can take advantage of Monads to get rid of all the manual threading of the seed values. This is a slightly more advanced method, but well worth learning since monads are ubiquitous in Haskell code.
We need these imports:
import System.Random
import Control.Monad
import Control.Applicative
Then we'll create a newtype which represents the action of turning a seed into a value and the next seed.
newtype Rand a = Rand { runRand :: StdGen -> (a,StdGen) }
We need Functor and Applicative instances or GHC will complain, but we can avoid implementing them for this example.
instance Functor Rand
instance Applicative Rand
And now for the Monad instance. This is where the magic happens. The >>= function (called bind) is the one place where we specify how to thread the seed value through the computation.
instance Monad Rand where
return x = Rand ( \seed -> (x,seed) )
ra >>= f = Rand ( \s1 -> let (a,s2) = runRand ra s1
in runRand (f a) s2 )
newRand :: Rand Int
newRand = Rand ( \seed -> random seed )
Now our sample function is extremely simple! We can take advantage of replicateM from Control.Monad which repeats a given action and accumulates the results in a list. All that funny business with the seed values is taken care of behind the scenes
sample :: Int -> Rand [Int]
sample n = replicateM n newRand
main :: IO ()
main = do
let seed1 = mkStdGen 124567
(xs,seed2) = runRand (sample 10) seed1
s = sum xs
print s
We can even stay inside the Rand monad if we need to generate random values multiple times.
main :: IO ()
main = do
let seed1 = mkStdGen 124567
(xs,seed2) = flip runRand seed1 $ do
x <- newRand
bs <- sample 5
cs <- sample 10
return $ x : (bs ++ cs)
s = sum xs
print s
I hope this helps!

Write list of random numbers to file. No Instance for (Show (IO a0))

I am trying to write to file a list of random Integers in a file. There seems to be a problem with writeFile here. When I use my function randomFile it says no instance for (Show (IO a0)). I see writeFile doesn't print anything to screen but IO(), so when I call the function randomFile 1 2 3 it says no Instance for Show (IO a0) but actually I just want to execute the function and not have to print anything but how can I avoid this problem. I might be making a lot of errors here. Any help.
import Control.Monad
import Control.Applicative
import System.Random
randNo mind maxd = randomRIO (mind,maxd)
randomFile mind maxd noe = do
let l=(replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd))
writeFile "RFile.txt" (show l)
I think you have a misunderstanding of what IO is. If you haven't done it, I strongly recommend going through the Input and Output section of Learn You a Haskell.
IO doesn't necessarily have anything to do with print. In Haskell every entry in memory that was made by your own code is considered "pure" while any entry that touches the rest of the computer lives in IO (with some exceptions you will learn about over time).
We model IO using something called a Monad. Which you will learn more about the longer you do Haskell. To understand this, let's look at an example of some code that does and doesn't use IO:
noIOused :: Int -> Int
noIOused x = x + 5
usesIO :: Int -> IO Int
usesIO x = print x >> return (x + 5)
usesIO2 :: Int -> IO Int
usesIO2 x = do
print x
return (x + 5)
The first function is "pure". The second and third functions have an IO "effect" that comes in the form of printing to the screen. usesIO and usesIO2 are just 2 different ways of doing the same thing (it's the same code but with different syntax). I'll use the second format, called do notation from here.
Here are some other ways you could have had IO effects:
add5WithFile :: Int -> IO Int
add5WithFile x = do
writeFile "someFile.txt" (show x)
return (x + 5)
Notice that in that function we didn't print anything, we wrote a file. But writing a file has a side effect and interacts with the rest of the system. So any value we return has to get wrapped in IO.
addRandom :: Int -> IO Int
addRandom x = do
y <- randomRIO (1,10)
return (x + y)
In addRandom we called randomRIO (1,10). But the problem is that randomRIO doesn't return an Int. It returns an IO Int. Why? Because in order to get true randomness we need to interact with the system in some way. To get around that, we have to temporarily strip away the IO. That's where this line comes in:
y <- randomRIO (1,10)
That <- arrow tells us that we want a y value outside of IO. For as long as we remain inside the do syntax that y value is going to be "pure". Now we can use it just like any other value.
So for example we couldn't do this:
let w = x + (randomRIO (1,10))
Because that would be trying to add Int to IO Int. And unfortunately our + function doesn't know how to do that. So first we have to "bind" the result of randomRIO to y before we can add it to x.
Now let's look at your code:
let l=(replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd))
writeFile "RFile.txt" (show l)
The type of l is actually IO a0. It's a0 because you haven't told the compiler what kind of number you want. So it doesn't know if you want a fraction, a double, a big integer or whatever.
So the first problem is to let the compiler know a little bit more about what kind of random number you want. We do this by adding a type annotation:
randNo :: Int -> Int -> IO Int
randNo mind maxd = randomRIO (mind,maxd)
Now both you and the compiler knows what kind of value randNo is.
Now we need to "bind" that value inside of the do notation to temporarily escape IO. You might think that would be simple, like this:
randomFile mind maxd noe = do
l <- replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd)
writeFile "RFile.txt" (show l)
Surely that will "bind" the IO Int to l right? Unfortunately not. The problem here is that replicate is a function of the form Int -> a -> [a]. That is, given a number and a type, it will give you a list of that type.
If you give replicate an IO Int it's going to make [IO Int]. That actually looks more like this: List (IO Int) except we use [] as syntactic sugar for lists. Unfortunately if we want to "bind" an IO value to something with <- it has to be the out-most type.
So what you need is a way to turn an [IO Int] into an IO [Int]. There are two ways to do that. If we put \[IO a\] -> IO \[a\] into Hoogle we get this:
sequence :: Monad m => [m a] -> m [a]
As I mentioned before, we generalise IO to something called a Monad. Which isn't really that big a deal, we could pretend that sequence has this signature: sequence :: [IO a] -> IO [a] and it would be the same thing just specialised to IO.
Now your function would be done like this:
randomFile mind maxd noe = do
l <- sequence (replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd))
writeFile "RFile.txt" (show l)
But a sequence followed by replicate is something people have to do all the time. So someone went and made a function called replicateM:
replicateM :: Monad m => Int -> m a -> m [a]
Now we can write your function like this:
randomFile mind maxd noe = do
l <- replicateM (fromInteger(noe ^ noe)) ( mind `randNo` maxd)
writeFile "RFile.txt" (show l)
And for some real Haskell magic, you can write all 3 lines of code in a single line, like this:
randomFile mind maxd noe = randomRIO >>= writeFile "RFile.txt" . replicateM (fromInteger(noe ^ noe))
If that looks like gibberish to you, then there's a lot you need to learn. Here is the suggested path:
If you haven't already, start from the beginning with Learn You a Haskell
Then learn about how You could have invented Monads
Then learn more about how to use randomness in Haskell
Finally see if you can complete the 20 intermediate Haskell exercises

How can I unpack an arbitrary length list of IO Bool

I'm writing a program that should be able to simulate many instances of trying the martingale betting system with roulette. I would like main to take an argument giving the number of tests to perform, perform the test that many times, and then print the number of wins divided by the total number of tests. My problem is that instead of ending up with a list of Bool that I could filter over to count successes, I have a list of IO Bool and I don't understand how I can filter over that.
Here's the source code:
-- file: Martingale.hs
-- a program to simulate the martingale doubling system
import System.Random (randomR, newStdGen, StdGen)
import System.Environment (getArgs)
red = [1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36]
martingale :: IO StdGen -> IO Bool
martingale ioGen = do
gen <- ioGen
return $ martingale' 1 0 gen
martingale' :: Real a => a -> a -> StdGen -> Bool
martingale' bet acc gen
| acc >= 5 = True
| acc <= -100 = False
| otherwise = do
let (randNumber, newGen) = randomR (0,37) gen :: (Int, StdGen)
if randNumber `elem` red
then martingale' 1 (acc + bet) newGen
else martingale' (bet * 2) (acc - bet) newGen
main :: IO ()
main = do
args <- getArgs
let iters = read $ head args
gens = replicate iters newStdGen
results = map martingale gens
--results = map (<-) results
print "THIS IS A STUB"
Like I have in my comments, I basically want to map (<-) over my list of IO Bool, but as I understand it, (<-) isn't actually a function but a keyword. Any help would be greatly appreciated.
map martingale gens will give you something of type [IO Bool]. You can then use sequence to unpack it:
sequence :: Monad m => [m a] -> m [a]
A more natural alternative is to use mapM directly:
mapM :: Monad m => (a -> m b) -> [a] -> m [b]
i.e. you can write
results <- mapM martingale gens
Note - even after doing it this way, your code feels a bit unnatural. I can see some advantages to the structure, in particular because martingale' is a pure function. However having something of type IO StdGen -> IO Bool seems a bit odd.
I can see a couple of ways to improve it:
make martingale' return an IO type itself and push the newStdGen call all the way down into it
make gens use replicateM rather than replicate
You may want to head over to http://codereview.stackexchange.com for more comprehensive feedback.

Using a random number to encrypt a message

I'm currently trying to encrypt a message (String) with the help of a random generated number in Haskell. The idea is to get the message, generate a random String of numbers with the same length (or more and then to take the length I need).
Then i want to perform some actions based on the ASCII representation and then return the encrypted String.
Unfortunately I'm not very versed with monads in Haskell, so it might be a very simple problem to solve, which I can't comprehend yet.
generateMyKey string = newStdGen >>= \x -> print $ concatMap show $ map abs $ rs x
where rs x = randomlist (length string) x
randomlist :: Int -> StdGen -> [Int]
randomlist n = take n . unfoldr (Just . random)
So the problem is I get an IO() out of getMyKey, but I want to have a String, or atleast a IO(String) to perform the encrypting mechanism.
Right now I'm getting a big list of positive (hence the abs + map) random numbers, but I can't access them.
There are two basic ways to go about this (and one more complicated but easier). If you're just using System.Random, you can generate random numbers in two ways, either by accepting a StdGen and staying pure, or using the OS's random generator and staying in IO. At some point, you'll have to make a call to the OS's random functionality to get a seed or value, but this can happen in main far away from your actual code.
To keep your functions pure, you'll need to pass around a StdGen and use the functions
random :: Random a => StdGen -> (a, StdGen)
randoms :: Random a => StdGen -> [a]
(Note: I've substituted RandomGen g => g for StdGen, there's no need to write a custom RandomGen instance for your case)
You can then write your function generateMyKey as
randomList :: Int -> StdGen -> [Int]
randomList n = take n . randoms
generateMyKey :: String -> StdGen -> String
generateMyKey text g
= concatMap show
$ map abs
$ randomList (length text) g
And this entirely avoids having to live in IO. Be wary, though, if you re-use the same g, you'll generate the same random list each time. We can avoid this by using IO and its related functions
randomList :: Int -> IO [Int]
randomList 0 = return []
randomList n = do
first <- randomIO
rest <- randomList (n - 1) -- Recursively generate the rest
return $ first : rest
generateMyKey :: String -> IO String
generateMyKey text = do
key <- randomList (length text)
return $ concatMap show $ map abs $ key
This will come with a performance hit, and now we've lost the ability to generate the same key repeatedly, making it difficult to test our functions reliably! How can we reconcile these two approaches?
Enter the package MonadRandom. This package provides a monad (and monad transformer, but you don't need to worry about that right now) that lets you abstract away how you generate random numbers so that you can choose how you want to run your code in different circumstances. If you want IO, you can use IO. If you want to supply a seed, you can supply a seed. It's very handy. You can install it with cabal install MonadRandom and use it as
import Control.Monad.Random
randomList :: Int -> Rand StdGen [Int]
randomList n = fmap (take n) getRandoms
generateMyKey :: String -> Rand StdGen String
generateMyKey text = do
key <- randomList (length text)
return $ concatMap show $ map abs $ key
Our generateMyKey code is even the same as the IO version other than the type signature!
Now to run it.
main :: IO ()
main = do
-- Entirely impure, have it automatically grab a StdGen from IO for us
ioVersion <- evalRandIO $ generateMyKey "password"
-- Make a StdGen that stays the same every time we run the program, useful for testing
let pureStdGen = mkStdGen 12345
pureVersion = evalRand (generateMyKey "password") pureStdGen
-- Get a StdGen from the system, but still evaluate it purely
ioStdGen <- getStdGen
let pureVersion2 = evalRand (generateMyKey "password") ioStdGen
-- Print out all three versions
putStrLn ioVersion
putStrLn pureVersion
putStrLn pureVersion2
There are a number of solutions to this problem, but at first glance it might seem that you need to have your entire program operate in the IO monad, but you don't! The entry (/exit) point of your program is the only place that needs to see IO -- you can factor out any transformations on your random list into pure functions, i.e:
import Data.List
import System.Random
generateMyKey :: String -> IO String
generateMyKey string = do
x <- newStdGen
let rs = randomlist (length string)
return $ concatMap show $ map abs $ rs x
randomlist :: Int -> StdGen -> [Int]
randomlist n = take n . unfoldr (Just . random)
change :: String -> String
change = reverse -- for example
main :: IO ()
main = do
key <- generateMyKey "what"
putStrLn $ change key
generateMyKey is identical to what you had before, except that it's written in do notation now and is returning the string instead of just printing it. This allows us to "pull out" a random key from inside the IO monad and transform it with regular pure functions, like change, for example. This allows you to reason about the pure functions as normal, while still pulling in your values from IO.

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