Menu
I have a list, I want to break the list into two, one with elements in the even indexes, and the other from the odd indexes.
breakByIndexes :: [a] -> ([a], [a])
For example:
> breakByIndexes ["A", "B", "C", "D"]
(["A", "C"], ["B", "D"]
> breakByIndexes ["A", "B", "C", "D", "E"]
(["A", "C", "E"], ["B", "D"]
I got a solution like this
breakByIndexes [] = ([], [])
breakByIndexes [e] = ([e], [])
breakByIndexes (e:o:xs) =
let (es, os) = breakByIndexes xs
in (e : es, o : os)
But I'm curious is it possible to implement without using recursion?
And is it possible to implement by composing existing functions from Data.List?
988
One way is using the range() function, such that it starts at the first odd index, 1 , and has a step value of 2 , so that it returns the odd indices 1, 3, 5, ... . Another way to obtain the odd indices of a list is using list comprehension, in which you can use a condition that will only include the odd indices.
If a number is evenly divisible by 2 with no remainder, then it is even. You can calculate the remainder with the modulo operator % like this num % 2 == 0 . If a number divided by 2 leaves a remainder of 1, then the number is odd.
Yes you are right, using the partition function from Data.List.
Prelude Data.List> (s, u) = partition (even . fst) (zip [0 .. ] "ABCD")
Prelude Data.List> (_, s2) = unzip s
Prelude Data.List> (_, u2) = unzip u
Prelude Data.List> (s2, u2)
("AC","BD")
How I found this? Go to Hoogle and fill in [a] -> ([a], [a]).
62
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
