Correction: the version in monte-carlo takes time O(n), but it only consumes k random numbers. The streaming algorithm consumes n random numbers. On Jun 14, 2011, at 12:04 PM, Patrick Perry wrote:
There are time/space tradeoffs in sampling without replacement. The version in monte-carlo takes space O(n) and time O(k), for all k. I chose this algorithm instead of a streaming algorithm, with takes space O(k) and time O(n).
If k is much less than n, you can improve a bit. Here is a version taking space O(k) and time O(k^2), *provided k << n*; when k is large relative to n, the time increases to roughly O(n log n).
import Control.Monad( liftM ) import Control.Monad.MC import Data.List( nub )
sampleSmallIntSubset :: MonadMC m => Int -> Int -> m [Int] sampleSmallIntSubset n k = (take k . nub) `liftM` repeatMC (sampleInt n)
sampleSmallSubset :: MonadMC m => [a] -> Int -> m [a] sampleSmallSubset xs k = (map (xs !!)) `liftM` sampleSmallIntSubset (length xs) k
-- strict version of the above sampleSmallSubset' :: MonadMC m => [a] -> Int -> m [a] sampleSmallSubset' xs k = do ys <- sampleSmallSubset xs k (length ys) `seq` return ys -- by forcing the length, we make sure we are done with the RNG when we return
If you're wondering about the strict version, the following example should be instructive:
Let's note the first three numbers generated by the RNG with seed 0:
ghci> (replicateM 3 $ uniform 0 1) `evalMC` mt19937 0 [0.999741748906672,0.16290987539105117,0.28261780529282987]
Now, let's look at the first number we get after a call to sampleSmallSubset, if we don't evaluate the subset:
ghci> sampleSmallSubset [1..10000] 2 >> uniform 0 1) `evalMC` mt19937 0 0.999741748906672
Apparently, the call to sampleSmallSubset doesn't consume any random numbers. Now, let's try the strict version:
ghci> (sampleSmallSubset' [1..10000] 2 >> uniform 0 1) `evalMC` mt19937 0 0.28261780529282987
This time, the call to sampleSmallSubset' consumes two random numbers, even if we throw away the result.
Since you're just starting out, I recommend using the strict versions of the functions. They are less space-efficient than the lazy versions, but they are guaranteed to be safe.
From: Jonas Almström Duregård
Subject: Re: [Haskell-cafe] Acquiring a random set of a specific size (w/o dups) from a range of Ints To: "michael rice" < nowgate at yahoo.com
Cc: "Felipe Almeida Lessa" < felipe.lessa at gmail.com>, haskell-cafe at haskell.org
Date: Tuesday, June 14, 2011, 5:17 AM
Shuffle [1..20], then take 5?
Yes, so simple, I'm embarrassed I didn't think of it.
That works well for small numbers, but I'm guessing it will evaluate the entire list so it should not be used for large inputs. If you have a large interval and use a relatively small part of it, the following function should be significantly faster (it builds a random permutation lazily):