CODEWITHSUNDEEP
haskell
Kale Joe
Kale Joe

Reputation: 163

How to set the range of a list?

I have to write a code that that returns all numbers in the input list within the range given by the first two arguments.

I tried to set the head of the list as the lowest input. 

inRange :: Int -> Int -> [Int] -> [Int]
inRange lo hi xs = (x == lo : xs == hi)

I expect having a result such as inRange 5 10 [1..15] == [5,6,7,8,9,10], but since I cannot write the code the modular is not loading.

Upvotes: 3

Views: 294

Answers (3)

developer_hatch
developer_hatch

Reputation: 16224

With pattern matching you can do this like:

inRange :: Int -> Int -> [Int] -> [Int]
inRange _ _ [] = []
inRange lo hi (x:xs)
  | lo > hi             = []
  | lo <= x && x <= hi  = x : (inRange lo hi xs)
  | x > hi              = []
  | otherwise           = inRange lo hi xs

Upvotes: 0

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476493

We can filter :: (a -> Bool) -> [a] -> [a] the list, such that it only retains values x for which lo <= x && x <= hi holds: 

inRange :: Ord a => a -> a -> [a] -> [a]
inRange lo hi = filter (\x -> lo <= x && x <= hi)

We can optimize this for the case where lo > hi, since then, no element can satisfy the condition, and hence we can return an empty list.

inRange :: Ord a => a -> a -> [a] -> [a]
inRange lo hi
    | lo > hi = []
    | otherwise = filter (\x -> lo <= x && x <= hi)

Upvotes: 4

karakfa
karakfa

Reputation: 67467

with list comprehensions

> inRange lo hi xs = [ x | x <- xs, x>=lo && x<=hi ]
> inRange 5 10 [1..15]
[5,6,7,8,9,10]

Upvotes: 1

Related Questions