Rubinfeld (enumerable) Methode auszuwählen und lehnt sich in 2-Arrays in 1 Betrieb

Question

# this code works 
list = (0..20).to_a 
# => [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20] 

odd = list.select { |x| x.odd? } 
# => [1, 3, 5, 7, 9, 11, 13, 15, 17, 19] 

list.reject! { |x| x.odd? } 
# => [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20] 

# but can i emulate this type of functionality with an enumerable method? 
set = [1,5,10] 
# => [1, 5, 10] 
one, five, ten = set 
# => [1, 5, 10] 
one 
# => 1 
five 
# => 5 
ten 
# => 10 

# ------------------------------------------------ 
# method I am looking for ? 
list = (0..20).to_a 
odd, even = list.select_with_reject { |x| x.odd? } 
# put the matching items into the first variable 
# and the non-matching into the second 

Answer 1

Sicher kann man tun:

odd, even = list.partition &:odd? 

Answer 2

odd = [] 
even = [] 
list = [1..20] 
list.each {|x| x.odd? ? odd << x : even << x } 

Answer 3

Als pguardiario gesagt, die partition Methode der direkteste Weg ist, .

require 'set' 
list = (1..10).to_a 
odd, even = Set.new(list).divide(&:odd?).to_a.map{|x| x.to_a} 

Answer 4

Sie könnten versuchen, unten: Sie könnten auch Set#divide verwenden

odd,even = (0..20).group_by(&:odd?).values 
p odd,even 

Ausgang:

[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20] 
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19]