Folds
editExercises |
---|
|
and [] = True
and (x:xs) = x && and xs
--As a fold
and = foldr (&&) True
or [] = False
or (x:xs) = x || or xs
--As a fold
or = foldr (||) False
maximum :: Ord a => [a] -> a -- omitting type declaration results in "Ambiguous type variable" error
maximum = foldr1 max --foldl1 also works, but not if you want infinite lists. Not that you can ever be sure of the maximum of an infinite list.
minimum :: Ord a => [a] -> a
minimum = foldr1 min --Same as above.
--Using foldl'; you will want to import Data.List in order to do the same.
reverse :: [a] -> [a]
reverse = foldl' flippedCons []
where
flippedCons xs x = x : xs
Scans
editExercises |
---|
Write your own definition of scanr , first using recursion, and then using foldr . Do the same for scanl first using recursion then foldl . |
-- with recursion
scanr2 step zero [] = [zero]
scanr2 step zero (x:xs) = step x (head prev):prev
where prev = scanr2 step zero xs
scanl2 step zero [] = [zero]
scanl2 step zero (x:xs) = zero : scanl2 step (step zero x) xs
-- An alternative scanl with poorer performance. The problem is that
-- last and init, unlike head and tail, must run through the entire
-- list, and (++) does the same with its first argument.
scanl2Slow step zero [] = [zero]
scanl2Slow step zero xs = prev ++ [step (last prev) (last xs)]
where prev = scanl2Slow step zero (init xs)
--with fold
scanr3 step zero xs = foldr step' [zero] xs
where step' x xs = step x (head xs):xs
scanl3 step zero xs = (reverse . foldl step' [zero]) xs
where step' xs x = step (head xs) x:xs
-- This implementation has poor performance due to (++), as explained
-- for scanl2Slow. In general, using (++) to append to a list
-- repeatdely is not a good idea.
scanl3Slow step zero xs = foldl step' [zero] xs
where step' xs x = xs ++ [step (last xs) x]
In the efficient scanr
implementations above, we use head
and tail
as a way to avoid having to break down the list argument with pattern matching in the where clause only to assemble it back in the right-hand side. In the Pattern matching chapter, we will see how as-patterns allow us to reach the same effect without needing head
.
Exercises |
---|
Define the following functions:
|
factList n = scanl1 (*) [1..n]
factList' n = scanl (*) 1 [2..n]
filter and list comprehension
editExercises |
---|
Write a |
returnDivisible :: Int -> [Int] -> [Int]
returnDivisible n xs = [ x | x<-xs , (mod x n) == 0 ]
--or, if you prefer to use filter...
returnDivisible :: Int -> [Int] -> [Int]
returnDivisible n = filter (\x -> (mod x n) == 0)
Exercises |
---|
|
choosingTails :: [[Int]] -> [[Int]]
choosingTails ls = [ tail l | l <- ls, not (null l), head l > 5 ]
The boolean conditions in a list comprehension are evaluated in order, so if you swapped the conditions like this:
choosingTails ls = [ tail l | l <- ls, head l > 5, not (null l) ]
If one of the l turned out to be an empty list the program would try to apply head on it, which would cause an error. Putting not (null l) first causes the program to, when it meets an empty list, to short-circuit the conditional - that is, ignore the second condition since the first one being false is enough for rejecting the list - thus avoiding doing head [] and the error that follows.
Alternatively you may filter by pattern matching. Only non-empty lists match (l:ll)
divided by the first element l
followed by the possibly empty tail ll
:
choosingTails ls = [ ll | (l:ll) <- ls, l > 5 ] -- filter by pattern match
Exercises |
---|
Over this section we've seen how list comprehensions are essentially syntactic sugar for filter and map. Now work in the opposite direction and define alternative versions of the filter and map using the list comprehension syntax. |
alternativeFilter :: (a -> Bool) -> [a] -> [a]
alternativeFilter cond xs = [ x | x<-xs, cond x ]
alternativeMap :: (a -> b) -> [a] -> [b]
alternativeMap f xs = [ f x | x<-xs ]
--no need for any condition, as all x will be accepted
Exercises |
---|
Rewrite doubleOfFirstForEvenSeconds using filter and map instead of list comprehension. |
doubleOfFirstForEvenSeconds' :: [(Int, Int)] -> [Int]
doubleOfFirstForEvenSeconds ps =
let f pair = 2 * (fst pair)
g pair = isEven (snd pair)
in map f (filter g ps)
--isEven, naturally, remains the same.
isEven :: Int -> Bool
isEven n = (mod n 2 == 0)
doubleOfFirstForEvenSeconds' :: [(Int, Int)] -> [Int]
doubleOfFirstForEvenSeconds' ps = map ((2*) . fst) (filter (isEven . snd) ps)