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I am pretty new to Haskell and I am trying to write a function neighbours :: Int -> Metric a -> Point a -> [Point a] -> [Point a] such that neighbours k d p xs returns a list of the k nearest neighbours, in distance order, according to metric d to point p in the list xs of points. My code is
type Point a = (a, a)
type Metric a = Point a -> Point a -> Double
type Tuple a = (Double, Point a)
create:: Metric a -> Point a -> [Point a] -> [Tuple a] -> [Tuple a]
create d p (x:xs) ys | length xs == 0 = sort(((d p x), x) : ys)
| otherwise = create d p xs (((d p x), x) : ys)
takeP:: Tuple a -> Point a
takeP (_,p) = p
pList:: [Tuple a] ->[Point a]-> [Point a]
pList (x:xs) ys | length xs == 0 = reverse (takeP x : ys)
| otherwise = pList xs (takeP x : ys)
neighbours :: Int -> Metric a -> Point a -> [Point a] -> [Point a]--
neighbours k d p xs = take k (pList (create d p xs []) [])
But I am getting an error on sort which is:
* No instance for (Ord a) arising from a use of `sort'
Possible fix:
add (Ord a) to the context of
the type signature for:
create :: forall a.
Metric a -> Point a -> [Point a] -> [Tuple a] -> [Tuple a]
* In the expression: sort (((d p x), x) : ys)
In an equation for `create':
create d p (x : xs) ys
| length xs == 0 = sort (((d p x), x) : ys)
| otherwise = create d p xs (((d p x), x) : ys)
I used type Point a = (Int, Int) at first and it was working fine but in the specification it is required that Point is type Point a = (a, a) which caused my error. The other problem is that I cannot change the function types so I can't just add (Ord a) as proposed.
Is there a way to sort the Tuples' list by the first variable without encountering errors?
In your create function, you make use of sort :: Ord a => [a] -> [a]:
… = sort (((d p x), x) : ys)
this thus means that type of objects that we are sorting, in this case Tuple a, needs to be an instance of the Ord typeclass. A 2-tuple is an instance of the Ord typeclass, if both the type of the items are instances of Ord as well, so in this case Double and Point a. Since Point a is also a 2-tuple, but of two as, this thus means that Tuple a is an instance of Ord, if a is an instance of Ord. You thus should add a type constraint:
create :: Ord a => Metric a -> Point a -> [Point a] -> [Tuple a] -> [Tuple a]
create d p (x:xs) ys | length xs == 0 = sort(((d p x), x) : ys)
| otherwise = create d p xs (((d p x), x) : ys)
The create function makes use of some anti-patterns like using length, which takes linear time. You can in fact rewrite this to sorting a mapping:
create :: Ord a => Metric a -> Point a -> [Point a] -> [Tuple a]
create d p = sort . map f
where f x = (d p x, x)
This removes the ys parameter, which here only seems to be used as a accumulator.
If you wish to only sort on the first item of the 2-tuple, you can make use of sortOn :: Ord b => (a -> b) -> [a] -> [a]:
create :: Metric a -> Point a -> [Point a] -> [Tuple a]
create d p = sortOn fst . map f
where f x = (d p x, x)
In haskell, how can I generate a balanced partition of a set?
Assuming I have a set {1,3,4,6,9}, a balanced partition of that set would be s1{9,3} and s2{6,4,1}, seeing as s1-s2 is 1.
Well, for brute force, we can generate all partitions recursively by generating partitions for the tail and then putting the head on the left list or the right:
partitions :: [a] -> [([a], [a])]
partitions [] = [([], [])]
partitions (x : xs) = let ps = partitions xs in
[(x : ys, zs) | (ys, zs) <- ps] ++ [(ys, x : zs) | (ys, zs) <- ps]
have a way to compute the unbalance:
unbalance :: Num a => ([a], [a]) -> a
unbalance (ys, zs) = abs (sum ys - sum zs)
and then put it all together:
balancedPartition :: (Num a, Ord a) => [a] -> ([a], [a])
balancedPartition = minimumBy (comparing unbalance) . partitions
Here's the complete module:
module Balance where
import Data.List(minimumBy)
import Data.Ord(comparing)
partitions :: [a] -> [([a], [a])]
partitions [] = [([], [])]
partitions (x : xs) = let ps = partitions xs in
[(x : ys, zs) | (ys, zs) <- ps] ++ [(ys, x : zs) | (ys, zs) <- ps]
unbalance :: Num a => ([a], [a]) -> a
unbalance (ys, zs) = abs (sum ys - sum zs)
balancedPartition :: (Num a, Ord a) => [a] -> ([a], [a])
balancedPartition = minimumBy (comparing unbalance) . partitions
Here's a solution that does a little better:
balancedPartition :: (Num a, Ord a) => [a] -> ([a], [a])
balancedPartition = snd . head . partitionsByBadness . sort
where
-- recursively builds a list of possible partitionings and their badness
-- sorted by their (increasing) badness
partitionsByBadness [] = [(0, ([], []))]
partitionsByBadness (x:xs) = let res = partitionsByBadness xs
withX = map ( (+x) *** first (x:)) res
sansX = map (subtract x *** second (x:)) res
in merge withX $ normalize sansX
-- When items are added to the second list, the difference between the sums
-- decreases - and might become negative
-- We take those cases and swap the lists, so that the first list has always
-- a greater sum and the difference is always positive
-- So that we can sort the list again (with linear complexity)
normalize xs = let (neg, pos) = span ((<0) . fst) xs
in merge pos $ reverse $ map (negate *** swap) neg
-- merge two sorted lists (as known from mergesort, but
-- omits "duplicates" with same badness)
merge :: Ord k => [(k, v)] -> [(k, v)] -> [(k, v)]
merge [] zss = zss
merge yss [] = yss
merge yss#(y:ys) zss#(z:zs) = case comparing fst y z of
LT -> y : merge ys zss
EQ -> merge ys zss
GT -> z : merge yss zs
Bin packing works pretty well:
% stack ghci --package Binpack
λ: import Data.BinPack
λ: let bins numberOfBins items = let totalSize = sum items; binSize = succ (totalSize `div` (max 1 numberOfBins)) in binpack WorstFit Decreasing id (replicate numberOfBins (emptyBin binSize)) items
λ: bins 2 [1,3,4,6,9]
([(0,[3,9]),(1,[1,4,6])],[])
If you know your input will fit into the bins you can extract out the partitions:
λ: map snd . fst . bins 2 $ [1,3,4,6,9]
[[3,9],[1,4,6]]
I have managed to implement insertionsort and quicksort in a couple of lines, but selectionsort and mergesort still give me headaches ;)
selectionsort [] = []
selectionsort (x:xs) =
let (minimum, greater) = extractMinimum x [] xs
in minimum : selectionsort greater
extractMinimum minimumSoFar greater [] = (minimumSoFar, greater)
extractMinimum minimumSoFar greater (x:xs)
| x < minimumSoFar = extractMinimum x (minimumSoFar:greater) xs
| otherwise = extractMinimum minimumSoFar (x:greater) xs
Is something like the extractMinimum function available in the standard library? I tried hoogling for (a -> a -> Bool/Ordering) -> [a] -> (a, [a]) without any luck.
mergesort [ ] = [ ]
mergesort [x] = [x]
mergesort xs =
let (left, right) = splitAt (length xs `div` 2) xs
in merge (mergesort left) (mergesort right)
merge xs [] = xs
merge [] ys = ys
merge xxs#(x:xs) yys#(y:ys)
| x < y = x : merge xs yys
| otherwise = y : merge xxs ys
Again, do I have to write merge myself, or can I reuse existing components? Hoogle gave me no useful results for (a -> a -> Bool/Ordering) -> [a] -> [a] -> [a].
There's nothing in the standard libraries, but at least merge is provided by a package on hackage, although I'm not sure it's worth pulling in a dependency for such a simple function.
However,
merge xxs#(x:xs) yys#(y:ys)
| x < y = x : merge xs yys
| otherwise = y : merge xxs ys
produces a non-stable sort, to get a stable sort, the condition to place x should be x <= y.
For extractMinimum, I haven't found anything either, but I can offer an alternative definition,
extractMinimum :: Ord a => a -> [a] -> (a,[a])
extractMinimum x = foldl' select (x, [])
where
select (mini, greater) y
| y < mini = (y, mini:greater)
| otherwise = (mini, y:greater)
A nice definition of selectionSort would be
import Data.List -- for unfoldr
selectionSort :: Ord a => [a] -> [a]
selectionSort = unfoldr getMin
where
getMin [] = Nothing
getMin (x:xs) = Just $ extractMinimum x xs
My suggestion for selection sort:
import Data.List
selectionsort xs = unfoldr f xs where
f [] = Nothing
f xs = Just $ extractMinimum xs
extractMinimum (x:xs) = foldl' f (x,[]) xs where
f (minimum, greater) x | x < minimum = (x, minimum : greater)
| otherwise = (minimum, x : greater)
How can I sort list with IO Compare function?
sortWith :: [String] -> (String -> String -> IO Ordering) -> IO [String]
Sortby expects (a->a->Ordering) and I don't know, how to deal with it. I am too lazy to implement quick sort myself.
I'm afraid there is no simple way. If it was possible to lift
sortBy :: Ord a => (a -> a -> Ordering) -> [a] -> [a]
to
sortByM :: (Ord a, Monad m) => (a -> a -> m Ordering) -> [a] -> m [a]
you could see the order of comparisons in implementation of sortBy, violating referential transparency.
In general, it's easy to go from xxxM to xxx but not conversely.
Possible options:
Implement a monadic sorting method
Use the monadlist library, which contains insertion sort (as in dflemstr's answer)
Use unsafePerformIO as a hack
Switch to sorting by key and use the Schwartzian transform
sortOnM :: (Monad m, Ord k) => (a -> m k) -> [a] -> m [a]
sortOnM f xs = liftM (map fst . sortBy (comparing snd)) $
mapM (\x -> liftM (x,) (f x)) xs
The sortBy function uses merge sort as the algorithm in GHC, but the Haskell 98 Report dictates that insertion sort should be used.
For simplicity, because I don't have a compiler so I cannot test my code, I will implement insertion sort here:
import Data.Foldable (foldrM)
insertByM :: (a -> a -> IO Ordering) -> a -> [a] -> IO [a]
insertByM _ x [] = return [x]
insertByM cmp x ys#(y:ys') = do
p <- cmp x y
case p of
GT -> do
rest <- insertByM cmp x ys'
return $ y : rest
_ -> return $ x : ys
sortByM :: (a -> a -> IO Ordering) -> [a] -> IO [a]
sortByM cmp = foldrM (insertByM cmp) []
As I said, I haven't tested this code, but it could/should work.
Oh, I've done this one before! Merge sort with monadic comparator:
type MComparator m a = a -> a -> m Ordering
sortByM :: (Monad m, Functor m) => MComparator m a -> [a] -> m [a]
sortByM cmp [] = return []
sortByM cmp [x] = return [x]
sortByM cmp xs = do
let (ys, zs) = partition xs
ys' <- sortByM cmp ys
zs' <- sortByM cmp zs
merge ys' zs'
where merge [] bs = return bs
merge as [] = return as
merge (a:as) (b:bs) = do
comparison <- cmp a b
case comparison of
LT -> (a:) <$> merge as (b:bs)
_ -> (b:) <$> merge (a:as) bs
partition xs = splitAt (length xs `quot` 2) xs
From my blog post: http://unknownparallel.wordpress.com/2012/07/03/using-monadic-effects-to-reverse-a-merge-sort/
Was it Larry Wall who said that laziness is one of the 3 great virtues of a programmer?
It seems you want to transform a function of type (a -> a -> b) into a function of type (a -> a -> c b). Let's plug that into Hoogle. Now, if you know that IO is a Monad, you'll see about the 10th match down in liftM2. Check the type of (liftM2 sortBy), is that what you want?
I'm a newbie to Haskell, and I'm trying to write an elegant function to merge an arbitrary number of sorted lists into a single sorted list... Can anyone provide an elegant and efficient reference implementation?
Thanks!
Something like this should work:
merge2 pred xs [] = xs
merge2 pred [] ys = ys
merge2 pred (x:xs) (y:ys) =
case pred x y of
True -> x: merge2 pred xs (y:ys)
False -> y: merge2 pred (x:xs) ys
merge pred [] = []
merge pred (x:[]) = x
merge pred (x:xs) = merge2 pred x (merge pred xs)
Here, the function merge2 merges 2 lists. The function merge merges a list of lists. The pred is predicate you use for sorting.
Example:
merge (<) [[1, 3, 9], [2, 3, 4], [7, 11, 15, 22]]
should return
[1,2,3,3,4,7,9,11,15,22]
Since I like taking advantage of infix operators and higher-order functions where it makes sense to, I would write
infixr 5 ##
(##) :: (Ord a) => [a] -> [a] -> [a]
-- if one side is empty, the merges can only possibly go one way
[] ## ys = ys
xs ## [] = xs
-- otherwise, take the smaller of the two heads out, and continue with the rest
(x:xs) ## (y:ys) = case x `compare` y of
LT -> x : xs ## (y:ys)
EQ -> x : xs ## ys
GT -> y : (x:xs) ## ys
-- a n-way merge can be implemented by a repeated 2-way merge
merge :: (Ord a) => [[a]] -> [a]
merge = foldr1 (##)
Here, xs ## ys merges two lists by their natural ordering (and drops duplicates), while merge [xs, ys, zs..] merges any number of lists.
This leads to the very natural definition of the Hamming numbers:
hamming :: (Num a, Ord a) => [a]
hamming = 1 : map (2*) hamming ## map (3*) hamming ## map (5*) hamming
hamming = 1 : merge [map (n*) hamming | n <- [2, 3, 5]] -- alternative
-- this generates, in order, all numbers of the form 2^i * 3^j * 5^k
-- hamming = [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36,40,45,48,50,..]
Stealing yairchu's unimplemented idea:
{-# LANGUAGE ViewPatterns #-}
import qualified Data.Map as M
import Data.List (foldl', unfoldr)
import Data.Maybe (mapMaybe)
-- merge any number of ordered lists, dropping duplicate elements
merge :: (Ord a) => [[a]] -> [a]
-- create a map of {n: [tails of lists starting with n]}; then
-- repeatedly take the least n and re-insert the tails
merge = unfoldr ((=<<) step . M.minViewWithKey) . foldl' add M.empty where
add m (x:xs) = M.insertWith' (++) x [xs] m; add m _ = m
step ((x, xss), m) = Just (x, foldl' add m xss)
-- merge any number of ordered lists, preserving duplicate elements
mergeDup :: (Ord a) => [[a]] -> [a]
-- create a map of {(n, i): tail of list number i (which starts with n)}; then
-- repeatedly take the least n and re-insert the tail
-- the index i <- [0..] is used to prevent map from losing duplicates
mergeDup = unfoldr step . M.fromList . mapMaybe swap . zip [0..] where
swap (n, (x:xs)) = Just ((x, n), xs); swap _ = Nothing
step (M.minViewWithKey -> Just (((x, n), xs), m)) =
Just (x, case xs of y:ys -> M.insert (y, n) ys m; _ -> m)
step _ = Nothing
where merge, like my original, eliminates duplicates, while mergeDup preserves them (like Igor's answer).
if efficiency wasn't a concern I'd go with
merge = sort . concat
otherwise:
merge :: Ord a => [[a]] -> [a]
merge [] = []
merge lists =
minVal : merge nextLists
where
heads = map head lists
(minVal, minIdx) = minimum $ zip heads [0..]
(pre, ((_:nextOfMin):post)) = splitAt minIdx lists
nextLists =
if null nextOfMin
then pre ++ post
else pre ++ nextOfMin : post
note however that this implementation always linearly searches for the minimum (while for a large number of list one may wish to maintain a heap etc.)
Unlike the other posts, I would have merge :: [a] -> [a] -> [a]
type SortedList a = [a]
merge :: (Ord a) => SortedList a -> SortedList a -> SortedList a
merge [] ys = ys
merge xs [] = xs
merge (x:xs) (y:ys)
| x < y = x : merge xs (y : ys)
| otherwise = y : merge (x : xs) ys
mergeAll :: (Ord a) => [SortedList a] -> SortedList a
mergeAll = foldr merge []
Just a quick note: if you want to have the optimal log n behavior when merging several lists (such as you'd get with a priority queue), you can do it very easily with a tweak to Igor's beautiful solution above. (I would have put this as a comment on his answer above, but I don't have enough reputation.) In particular, you do:
merge2 pred xs [] = xs
merge2 pred [] ys = ys
merge2 pred (x:xs) (y:ys) =
case pred x y of
True -> x: merge2 pred xs (y:ys)
False -> y: merge2 pred (x:xs) ys
everyother [] = []
everyother e0:[] = e0:[]
everyother (e0:e1:es) = e0:everyother es
merge pred [] = []
merge pred (x:[]) = x
merge pred xs = merge2 pred (merge pred . everyother $ xs)
(merge pred . everyother . tail $ xs)
Note that a real priority queue would be a bit faster/more space efficient, but that this solution is asymptotically just as good and, as I say, has the advantage that it's a very minor tweak to Igor's beautifully clear solution above.
Comments?