CODEWITHSUNDEEP
erlangelixir
ANIL MANE
ANIL MANE

Reputation: 1405

Elixir: Split list into odd and even elements as two items in tuple

I am quiet new to Elixir programming and stuck badly at splitting into two elements tuple.

Given a list of integers, return a two element tuple. The first element is a list of the even numbers from the list. The second is a list of the odd numbers.

Input : [ 1, 2, 3, 4, 5 ]
Output  { [ 2, 4],  [ 1, 3, 5 ] }

I have reached to identify the odd or even but not sure how do I proceed.

defmodule OddOrEven do
import Integer

  def task(list) do

Enum.reduce(list, [],  fn(x, acc) -> 
         case Integer.is_odd(x) do
            :true -> # how do I get this odd value listed as a tuple element
            :false -> # how do I get this even value listed as a tuple element
         end
        #IO.puts(x)
         end
    )     
end

Upvotes: 8

Views: 4497

Answers (3)

Aashish Chakravarty
Aashish Chakravarty

Reputation: 393

You can use Enum.split_while/2

iex> Enum.split_with([1, 2, 3, 4, 5, 6], &rem(&1, 2) == 1)
{[1, 3, 5], [2, 4, 6]}

Another solution

iex> {Enum.take_every([1, 2, 3, 4, 5, 6], 2), Enum.drop_every([1, 2, 3, 4, 5, 6], 2)}
{[1, 3, 5], [2, 4, 6]}

Upvotes: 0

rvirding
rvirding

Reputation: 20916

Or you can use the Erlang :lists module:

iex> :lists.partition(fn (n) -> rem(n, 2) == 1 end, [1,2,3,4,5])
{[1,3,5],[2,4]}

Upvotes: 2

Dogbert
Dogbert

Reputation: 222158

You can use Enum.partition/2:

iex(1)> require Integer
iex(2)> [1, 2, 3, 4, 5] |> Enum.partition(&Integer.is_even/1)
{[2, 4], [1, 3, 5]}

If you really want to use Enum.reduce/2, you can do this:

iex(3)> {evens, odds} = [1, 2, 3, 4, 5] |> Enum.reduce({[], []}, fn n, {evens, odds} ->
...(3)>   if Integer.is_even(n), do: {[n | evens], odds}, else: {evens, [n | odds]}
...(3)> end)
{[4, 2], [5, 3, 1]}
iex(4)> {Enum.reverse(evens), Enum.reverse(odds)}
{[2, 4], [1, 3, 5]}

Upvotes: 14

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