99 questions/Solutions/3: Difference between revisions
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(two more solutions using prelude function `take') |
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<haskell> | <haskell> | ||
elementAt'' [1..] 0 | elementAt'' [1..] 0 | ||
</haskell> | |||
A few more solutions using prelude functions: | |||
<haskell> | |||
elementAt''' xs n = last . take n $ xs | |||
</haskell> | |||
<haskell> | |||
elementAt'''' xs n = head . reverse . take n $ xs | |||
</haskell> | </haskell> |
Revision as of 02:06, 1 December 2011
(*) Find the K'th element of a list. The first element in the list is number 1.
This is (almost) the infix operator !! in Prelude, which is defined as:
(!!) :: [a] -> Int -> a
(x:_) !! 0 = x
(_:xs) !! n = xs !! (n-1)
Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:
elementAt :: [a] -> Int -> a
elementAt list i = list !! (i-1)
Or without using the infix operator:
elementAt' :: [a] -> Int -> a
elementAt' (x:_) 1 = x
elementAt' [] _ = error "Index out of bounds"
elementAt' (_:xs) k
| k < 1 = error "Index out of bounds"
| otherwise = elementAt' xs (k - 1)
Alternative version:
elementAt'' :: [a] -> Int -> a
elementAt'' (x:_) 1 = x
elementAt'' (_:xs) i = elementAt xs (i - 1)
elementAt'' _ _ = error "Index out of bounds"
This does not work correctly on invalid indexes and infinite lists, e.g.:
elementAt'' [1..] 0
A few more solutions using prelude functions:
elementAt''' xs n = last . take n $ xs
elementAt'''' xs n = head . reverse . take n $ xs