Elixir: Split list into odd and even elements as t

问题:

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

回答1:

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]}

回答2:

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]}