random order

What is an efficient way to randomize the order of something, for example characters in a string or places in an array, without repeating anything before you ran trough all characters/arrayplaces?

without repeating anything before you ran trough all characters
wat

i would use an array

>>1
http://en.wikipedia.org/wiki/Fisher-Yates_shuffle

>>1
Damn, you couldn't think of this yourself?

>>4
import Monad
import Random
import Data.Array

randomize array =
   foldM step array $ reverse [lo..hi]
   where step a i = fmap (swap a i) $ randomRIO (lo, i)
     swap a i j = a // [(j, a!i), (i, a!j)]
     (lo, hi) = bounds array

>>6
I am learning haskell :)

>>6
haskell: short, but superexponential?

Too bad that Haskell has no arrays.
How do you feel about this, haskellFags?

>>9
Haskell has no arrays?  How does it store lists of data?

>>10
Terrible!

>>10
Terrible!

Oh wow.

>>11
>>12
haskellmind

>>8
With the right type signature, >>6 is actually O(n) if use it on a DiffArray.

import Monad
import Random
import Data.Array.IArray

randomize :: (IArray a e, Enum i, Random i, Ix i) => a i e -> IO (a i e)
randomize array =
   foldM step array $ reverse [lo..hi]
   where step a i = fmap (swap a i) $ randomRIO (lo, i)
         swap a i j = a // [(j, a!i), (i, a!j)]
         (lo, hi) = bounds array

>>10
In lists, like Scheme.

>>10
In lists, like Scheme.

>>16,17
In lists, like Scheme

>>15
>         swap a i j = a // [(j, a!i), (i, a!j)]
What's that comment? And how does this work anyway?

>>19
You have been caught trolling, or retarding

>>15
Actually, the type signature is unnecessary. Just importing Data.Array.IArray instead of Data.Array will allow >>6 to work on DiffArray.

>>20
Yeah, I am worse because I don't litter my head with languages that are useless. FLAWLESS LOGIC.

>>22
Sunohara?

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