Function, which finds in the list of integers one of the longest ordered increments of subscripts (not necessarily consecutive) numbers. Example:
• Sequence [21,27,15,14,18,16,14,17,22,13] = [14,16,17,22]
I have a problem with the function which takes the initial number from the array, and looks for a sequence:
fstLen:: Int -> [Int] -> [Int]
fstLen a [] = a: []
fstLen x (l:ls) = if x < l then x:(fstLen l ls) else fstLen x ls
I have problems in place, 14,18,16,14,17,22,13
14 < 18 but then 18 > 16 and my algorithm takes the number 16 as the basis and is looking for a new sequence and I need to go back to 14
How can I do it?
(sorry for my english)
You could always just use subsequences from Data.List to get all the possible subsequences in a list. When you get these subsequences, just take the sorted ones with this function and filter:
isSorted :: (Ord a) => [a] -> Bool
isSorted [] = True
isSorted [_] = True
isSorted(x:y:xs) = x <= y && isSorted (y:xs)
Then get the maximum length subsequence with maximumBy(or another method), with the ordering being comparinglength.
Here is what the code could look like:
import Data.Ord (comparing)
import Data.List (subsequences, maximumBy, nub)
isSorted :: (Ord a) => [a] -> Bool
isSorted [] = True
isSorted [_] = True
isSorted(x:y:xs) = x <= y && isSorted (y:xs)
max_sequence :: (Ord a) => [a] -> [a]
max_sequence xs = maximumBy (comparing length) $ map nub $ filter isSorted (subsequences xs)
Which seems to work correctly:
*Main> max_sequence [21,27,15,14,18,16,14,17,22,13]
[14,16,17,22]
Note: used map nub to remove duplicate elements from the sub sequences. If this is not used, then this will return [14,14,17,22] as the maximum sub sequence, which may be fine if you allow this.
A more efficient n log n solution can be done by maintaining a map where
keys are the first element of an increasing sequence.
values are a tuple: (length of the sequence, the actual sequence)
and the map maintains the invariance that for each possible size of an increasing sequence, only the lexicographically largest one is retained.
Extra traceShow bellow to demonstrate how the map changes while folding from the end of the list:
import Debug.Trace (traceShow)
import Data.Map (empty, elems, insert, delete, lookupGT, lookupLT)
-- longest (strictly) increasing sequence
lis :: (Ord k, Show k, Foldable t) => t k -> [k]
lis = snd . maximum . elems . foldr go empty
where
go x m = traceShow m $ case x `lookupLT` m of
Nothing -> m'
Just (k, v) -> if fst a < fst v then m' else k `delete` m'
where
a = case x `lookupGT` m of
Nothing -> (1, [x])
Just (_, (i, r)) -> (i + 1, x:r)
m' = insert x a m
then:
\> lis [21,27,15,14,18,16,14,17,22,13]
fromList []
fromList [(13,(1,[13]))]
fromList [(22,(1,[22]))]
fromList [(17,(2,[17,22])),(22,(1,[22]))]
fromList [(14,(3,[14,17,22])),(17,(2,[17,22])),(22,(1,[22]))]
fromList [(16,(3,[16,17,22])),(17,(2,[17,22])),(22,(1,[22]))]
fromList [(16,(3,[16,17,22])),(18,(2,[18,22])),(22,(1,[22]))]
fromList [(14,(4,[14,16,17,22])),(16,(3,[16,17,22])),(18,(2,[18,22])),(22,(1,[22]))]
fromList [(15,(4,[15,16,17,22])),(16,(3,[16,17,22])),(18,(2,[18,22])),(22,(1,[22]))]
fromList [(15,(4,[15,16,17,22])),(16,(3,[16,17,22])),(18,(2,[18,22])),(27,(1,[27]))]
[15,16,17,22]
It is not necessary to retain the lists within the map. One can reconstruct the longest increasing sequence only using the keys and the length of the sequences (i.e. only the first element of the tuples).
Excellent question! Looking forward to a variety of answers.
Still improving my answer. The answer below folds to build increasing subsequences from the right. It also uses the the list monad to prepend new elements to subsequences if the new element is smaller than the head of the subsequence. (This is my first real application of the list monad.) For example,
λ> [[3], [1]] >>= (prepIfSmaller 2)
[[2,3],[3],[1]]
This solution is about as short as I can make it.
import Data.List (maximumBy)
maxSubsequence :: Ord a => [a] -> [a]
maxSubsequence [] = []
maxSubsequence xs = takeLongest $ go [] xs
where
takeLongest :: Ord a => [[a]] -> [a]
takeLongest = maximumBy (\ x y -> compare (length x) (length y))
go :: Ord a => [[a]] -> [a] -> [[a]]
go = foldr (\x subs -> [x] : (subs >>= (prepIfSmaller x)))
where prepIfSmaller x s#(h:_) = (if x < h then [x:s] else []) ++ [s]
Quick test.
λ> maxSubsequence [21,27,15,14,18,16,14,17,22,13]
[15,16,17,22]
Related
Im new to Haskell.
How to show only the repeated elements ?
Given as input: bbbool, expected output: bo
I found a way to do this from internet:
import qualified Data.Set as Set
dup :: Ord a => [a] -> Maybe a
dup xs = dup' xs Set.empty
where dup' [] _ = Nothing
dup' (x:xs) s = if Set.member x s
then Just x
else dup' xs (Set.insert x s)
dupString :: (Ord a, Show a) => [a] -> [Char]
dupString x = case dup x of
Just x -> "First duplicate: " ++ (show x)
Nothing -> "No duplicates"
But the problem is, it will only show the first repeated element.
for example : bbbool = b
I hope that my question is clear.
A simple way to do this would be, assuming you only look for adjacent repeated elements:
import Data.List (group)
findRepeating :: Eq a => [a] -> [a]
findRepeating = map head . filter ((> 1) . length) . group
You may want to sequence it with nub to deal with repeated elements, since for "bbbXaaaXbbb" it returns "bab".
Edit: if you meant all repeating elements, this should do:
import qualified Data.Map as M
findRepeating :: (Foldable k, Ord a) => t a -> [a]
findRepeating = M.keys . M.filter (> 1) . foldr (\x acc -> M.insertWith (+) x 1 acc) M.empty
This approach counts how many times each element shows up and then returns those that happened to appear more than once. The order in which they are returned is undefined.
I have created a program to remove first smallest element but I dont how to do for second largest:
withoutBiggest (x:xs) =
withoutBiggestImpl (biggest x xs) [] (x:xs)
where
biggest :: (Ord a) => a -> [a] -> a
biggest big [] = big
biggest big (x:xs) =
if x < big then
biggest x xs
else
biggest big xs
withoutBiggestImpl :: (Eq a) => a -> [a] -> [a] -> [a]
withoutBiggestImpl big before (x:xs) =
if big == x then
before ++ xs
else
withoutBiggestImpl big (before ++ [x]) xs
Here is a simple solution.
Prelude> let list = [10,20,100,50,40,80]
Prelude> let secondLargest = maximum $ filter (/= (maximum list)) list
Prelude> let result = filter (/= secondLargest) list
Prelude> result
[10,20,100,50,40]
Prelude>
A possibility, surely not the best one.
import Data.Permute (rank)
x = [4,2,3]
ranks = rank (length x) x -- this gives [2,0,1]; that means 3 (index 1) is the second smallest
Then:
[x !! i | i <- [0 .. length x -1], i /= 1]
Hmm.. not very cool, let me some time to think to something better please and I'll edit my post.
EDIT
Moreover my previous solution was wrong. This one should be correct, but again not the best one:
import Data.Permute (rank, elems, inverse)
ranks = elems $ rank (length x) x
iranks = elems $ inverse $ rank (length x) x
>>> [x !! (iranks !! i) | i <- filter (/=1) ranks]
[4,2]
An advantage is that this preserves the order of the list, I think.
Here is a solution that removes the n smallest elements from your list:
import Data.List
deleteN :: Int -> [a] -> [a]
deleteN _ [] = []
deleteN i (a:as)
| i == 0 = as
| otherwise = a : deleteN (i-1) as
ntails :: Int -> [a] -> [(a, Int)] -> [a]
ntails 0 l _ = l
ntails n l s = ntails (n-1) (deleteN (snd $ head s) l) (tail s)
removeNSmallest :: Ord a => Int -> [a] -> [a]
removeNSmallest n l = ntails n l $ sort $ zip l [0..]
EDIT:
If you just want to remove the 2nd smallest element:
deleteN :: Int -> [a] -> [a]
deleteN _ [] = []
deleteN i (a:as)
| i == 0 = as
| otherwise = a : deleteN (i-1) as
remove2 :: [a] -> [(a, Int)] -> [a]
remove2 [] _ = []
remove2 [a] _ = []
remove2 l s = deleteN (snd $ head $ tail s) l
remove2Smallest :: Ord a => [a] -> [a]
remove2Smallest l = remove2 l $ sort $ zip l [0..]
It was not clear if the OP is looking for the biggest (as the name withoutBiggest implies) or what. In this case, one solution is to combine the filter :: (a->Bool) -> [a] -> [a] and maximum :: Ord a => [a] -> a functions from the Prelude.
withoutBiggest l = filter (/= maximum l) l
You can remove the biggest elements by first finding it and then filtering it:
withoutBiggest :: Ord a => [a] -> [a]
withoutBiggest [] = []
withoutBiggest xs = filter (/= maximum xs) xs
You can then remove the second-biggest element in much the same way:
withoutSecondBiggest :: Ord a => [a] -> [a]
withoutSecondBiggest xs =
case withoutBiggest xs of
[] -> xs
rest -> filter (/= maximum rest) xs
Assumptions made:
You want each occurrence of the second-biggest element removed.
When there is zero/one element in the list, there isn't a second element, so there isn't a second-biggest element. Having the list without an element that isn't there is equivalent to having the list.
When the list contains only values equivalent to maximum xs, there also isn't a second-biggest element even though there may be two or more elements in total.
The Ord type-class instance implies a total ordering. Otherwise you may have multiple maxima that are not equivalent; otherwise which one is picked as the biggest and second-biggest is not well-defined.
wondering how to implement nub over a Seq a
I get that one could do:
nubSeq :: Seq a -> Seq a
nubSeq = fromList . nub . toList
Just wondering is there something standard that does not convert to Lists in order to call nub :: [a]->[a]?
An implementation that occurred to me, based obviously on nub, is:
nubSeq :: (Eq a) => Seq a -> Seq a
nubSeq = Data.Sequence.foldrWithIndex
(\_ x a -> case x `Data.Sequence.elemIndexR` a of
Just _ -> a
Nothing -> a |> x) Data.Sequence.empty
But there must be something more elegant?
thanks.
Not sure whether this qualifies as more elegant but it splits the concerns in independent functions (caveat: you need an Ord constraint on a):
seqToNubMap takes a Seq and outputs a Map associating to each a the smallest index at which it appeared in the sequence
mapToList takes a Map of values and positions and produces a list of values in increasing order according to the specified positions
nubSeq combines these to generate a sequence without duplicates
The whole thing should be O(n*log(n)), I believe:
module NubSeq where
import Data.Map as Map
import Data.List as List
import Data.Sequence as Seq
import Data.Function
seqToNubMap :: Ord a => Seq a -> Map a Int
seqToNubMap = foldlWithIndex (\ m k v -> insertWith min v k m) Map.empty
mapToList :: Ord a => Map a Int -> [a]
mapToList = fmap fst . List.sortBy (compare `on` snd) . Map.toList
nubSeq :: Ord a => Seq a -> Seq a
nubSeq = Seq.fromList . mapToList . seqToNubMap
Or a simpler alternative following #DavidFletcher's comment:
nubSeq' :: forall a. Ord a => Seq a -> Seq a
nubSeq' xs = Fold.foldr cons nil xs Set.empty where
cons :: a -> (Set a -> Seq a) -> (Set a -> Seq a)
cons x xs seen
| x `elem` seen = xs seen
| otherwise = x <| xs (Set.insert x seen)
nil :: Set a -> Seq a
nil _ = Seq.empty
Another way with an Ord constraint - use a scan to make the sets of
elements that appear in each prefix of the list. Then we can filter out
any element that's already been seen.
import Data.Sequence as Seq
import Data.Set as Set
nubSeq :: Ord a => Seq a -> Seq a
nubSeq xs = (fmap fst . Seq.filter (uncurry notElem)) (Seq.zip xs seens)
where
seens = Seq.scanl (flip Set.insert) Set.empty xs
Or roughly the same thing as a mapAccumL:
nubSeq' :: Ord a => Seq a -> Seq a
nubSeq' = fmap fst . Seq.filter snd . snd . mapAccumL f Set.empty
where
f s x = (Set.insert x s, (x, x `notElem` s))
(If I was using lists I would use Maybes instead of the pairs with
Bool, then use catMaybes instead of filtering. There doesn't seem to be catMaybes
for Sequence though.)
I think your code should be pretty efficient. Since Sequences are tree data structures using another tree type data structure like Map or HashMap to store and lookup the previous items doesn't make too much sense to me.
Instead i take the first item and check it's existence in the rest. If exists i drop that item and proceed the same with the rest recursively. If not then construct a new sequence with first element is the unique element and the rest is the result of nubSeq fed by the rest. Should be typical. I use ViewPatterns.
{-# LANGUAGE ViewPatterns #-}
import Data.Sequence as Seq
nubSeq :: Eq a => Seq a -> Seq a
nubSeq (viewl -> EmptyL) = empty
nubSeq (viewl -> (x :< xs)) | elemIndexL x xs == Nothing = x <| nubSeq xs
| otherwise = nubSeq xs
*Main> nubSeq . fromList $ [1,2,3,4,4,2,3,6,7,1,2,3,4]
fromList [6,7,1,2,3,4]
If I am given a list of objects and another list for some indices from this list, is there an easy way to change every object in this list with an index from the list of indices to a different value?
E.g. I am hoping there exists some function f such that
f 0 [4,2,5] [6,5,8,4,3,6,2,7]
would output
[6,5,0,4,0,0,2,7]
Here is a beautiful version that uses lens:
import Control.Lens
f :: a -> [Int] -> [a] -> [a]
f x is = elements (`elem` is) .~ x
Here is an efficient version that doesn't have any dependencies other than base. Basically, we start by sorting (and removing duplicates from the) indices list. That way, we don't need to scan the whole list for every replacement.
import Data.List
f :: a -> [Int] -> [a] -> [a]
f x is xs = snd $ mapAccumR go is' (zip xs [1..])
where
is' = map head . group . sort $ is
go [] (y,_) = ([],y)
go (i:is) (y,j) = if i == j then (is,x) else (i:is,y)
You can define a helper function to replace a single value and then use it to fold over your list.
replaceAll :: a -> [Int] -> [a] -> [a]
replaceAll repVal indices values = foldl (replaceValue repVal) values indices
where replaceValue val vals index = (take index vals) ++ [val] ++ (drop (index + 1) vals)
Sort the indices first. Then you can traverse the two lists in tandem.
{-# LANGUAGE ScopedTypeVariables #-}
import Prelude (Eq, Enum, Num, Ord, snd, (==), (<$>))
import Data.List (head, group, sort, zip)
f :: forall a. (Eq a, Enum a, Num a, Ord a) => a -> [a] -> [a] -> [a]
f replacement indices values =
go (head <$> group (sort indices)) (zip [0..] values)
where
go :: [a] -> [(a, a)] -> [a]
go [] vs = snd <$> vs
go _ [] = []
go (i:is) ((i', v):vs) | i == i' = replacement : go is vs
go is (v:vs) = snd v : go is vs
The sorting incurs an extra log factor on the length of the index list, but the rest is linear.
I want to write a function which takes a list of elements l, a list of indices i, and a list of replacement values v. The function will replace the values in l corresponding to the indices in i with the corresponding value in v.
Example:
If l = [1,2,3,4,5,6], i = [0,2], and v = [166,667], then
replaceValues l i v == [166,2,667,4,5,6]
My function:
--Replace the values in list l at indices in i with the
-- corresponding value in v
replaceValues :: [a] -> [Int] -> [a] -> [a]
replaceValues l [] [] = l
replaceValues l i v = x ++ [head v] ++ (replaceValues (tail y) shiftedIndices (tail v))
where
(x,y) = splitAt (head i) l
--The indices must be shifted because we are changing the list
shiftedIndices = map ((-)((head i) + 1)) (tail i)
This function manages to correctly replace the value at the first index in i, but it misplaces all of the following values. In the example above, it would give the output [166,667,3,4,5,6].
The problem with your implementation is that you aren't keeping track of which index you're currently at.
First of all, you're better off considering using [(Int,a)] rather than [Int] and [a] arguments separate to ensure that the "lists" are equal in length.
An alternate implementation is as follows:
import Data.Maybe(fromMaybe)
import qualified Data.IntMap as M
replaceValues :: [a] -> [(Int,a)] -> [a]
replaceValues as rs = map rep $ zip [0..] as
where
rsM = M.fromList rs
rep (i,a) = fromMaybe a $ M.lookup i rsM
What's happening here:
Tag each value with its index
See if there's a replacement value for that index: if there is, use it; otherwise, use the original value.
The first thing that springs to mind is that you should use a list of tuples to specify the replacement, that is, work with
l = [1,2,3,4,5,6]
r = [(0,166),(2,667)]
... you can use zip to convert your two lists into that format. Then I'm going to stipulate that that list is sorted by the first element of the tuple (sortBy), and that there are no duplicate indices in it (nubBy). The rest is a simple recursion, replacing as you go, with linear complexity and being maximally lazy:
replaceValues :: [a] -> [(Int, a)] -> [a]
replaceValues xs rs = f 0 xs rs
where
f _ xs [] = xs
f _ [] _ = []
f n (x:xs) is#((i,r):is')
| n < i = x:f (n+1) xs is
| n == i = r:f (n+1) xs is'
| otherwise = error "Can't happen"
Beware of the code, though, I have only proven it to be correct, not actually tried it.
Using a map works too, of course, but then you're dealing with an complexity of O(m log m + n log m) (construct map + n times lookup) instead of O(n), or, taking sorting into account, O(n + m log m), as well as lose the capability of being lazy in case your list is already sorted and have the replacements incrementally garbage collected while you traverse.
nubBy from the library has quadratic complexity, but as the list is sorted it can be dealt with in linear time, too, just replace the call to error with a recursive call throwing away superfluous (i,r)s.
as said before, use tuples - but don't forget about pattern matching. Or use Map if you are not dealing with large collections
replaceValues :: [a] -> [Int] -> [a] ->[a]
replaceValues a b i = map fst $ f (zip a [0..]) (zip b i)
where
f [] _ = []
f xs [] = xs
f ((x,i):xs) s2#((j,y):ys) | i == j = (y,i) : f xs ys
| otherwise = (x,i) : f xs s2