[Haskell-cafe] Idiomatically using lists
Dan Weston
westondan at imageworks.com
Tue Jun 5 15:18:12 EDT 2007
Here is my list-based version. There are redundant calls to get the
length of the same list, but I didn't feel like factoring them out (call
it an exercise for the reader). The key to its simplicity is that
shifting an element is a similarity transform of shifting the first
element, with pre- and post list rotation. The shift just maps
(h:xs++ys) to xs++(h:ys). Rotating a list is easy with the drop . cycle
pattern, and shares the list up to the point of rotation (when brought
in from cycle 2 to end the list).
Dan
module RotateList where
import Control.Arrow((&&&))
rotateList :: Int -> [a] -> [a]
rotateList offset = uncurry take
. (length &&& uncurry drop .
(mod offset . length &&& cycle))
shiftElem :: Int -> [a] -> [a]
shiftElem _ [] = []
shiftElem offset (h:t) = a ++ (h:b)
where (a,b) = splitAt ((offset-1) `mod` (length t) + 1) t
-- rotateElem is a similarity transform of shiftElem
rotateElem :: Int -> Int -> [a] -> [a]
rotateElem start offset = rotateList (negate start)
. shiftElem offset
. rotateList start
kevin birch wrote:
> On 火, 2007-6月-05, at 02:54, Greg Fitzgerald wrote:
>
>> > rotating the fourth element 2 positions would result in: [1, 2, 4, 3, 5]
>> Seems odd. Should that be [4,1,2,3,5]?
>>
> Yes, I meant to use the 5 element in my second example. Sorry for the
> confusion.
>
>> > Is there an idomatic way to handle both of these cases in a function?
>> Generally people like to see your attempt at a solution before giving
>> the idomatic one so that they are sure it's not a homework question.
>> What do you have so far?
>>
> Yeah, I only wish I had gone to a school that would be forward thinking
> enough to each FP. ;-)
>
> Here is my version:
>
> rotate :: Array Integer Card -> Integer -> Integer -> Array Integer Card
> rotate a i n
> | i <= u - n = a // [(i, a ! (i + 1)), (i + 1, a ! (i + 2)), (i + 2,
> a ! i)]
> | otherwise = a // zip [l..u] (h ++ [a ! i] ++ filter (not . (== (a
> ! i))) t)
> where (l, u) = bounds a
> (h, t) = splitAt (fromInteger ((i - u) + n)) $ elems a
>
> This function is part of my implementation of the Solitaire encryption
> algorithm, so that is why I have the reference to a Card data type.
> This does what I want, and seems basically idiomatic, but perhaps it
> could be better.
>
> Thanks,
> Kevin
>
>
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