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

当然，你可以这样做：

odd, even = list.partition &:odd?

Answer 2

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

Answer 3

正如pguardiario所说，partition方法是最直接的方法。您也可以使用Set#divide：

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

Answer 4

你可以尝试下面的内容：

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

输出：

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

ruby array（enumerable）方法，用于在1次操作中选择和拒绝2个数组

4 个答案: