如果有人正在寻找一种方法来删除重复值的所有实例，请参阅“如何有效地提取Ruby数组中的重复元素?”。

a = [1, 2, 2, 3]
counts = Hash.new(0)
a.each { |v| counts[v] += 1 }
p counts.select { |v, count| count == 1 }.keys # [1, 3]

对我来说，最简单的方法是:

array = [1, 2, 2, 3]

数组# to_set

array.to_set.to_a

# [1, 2, 3]

数组# uniq

array.uniq

# [1, 2, 3]

如果有人在乎的话，这只是另一个选择。

您还可以使用数组的to_set方法将array转换为Set，并且根据定义，Set元素是唯一的。

[1,2,3,4,5,5,5,6].to_set => [1,2,3,4,5,6]

只是为了提供一些见解:

require 'fruity'
require 'set'

array = [1,2,2,1,4,4,5,6,7,8,5,6] * 1_000

def mithun_sasidharan(ary)
  ary.uniq
end

def jaredsmith(ary)
  ary & ary
end

def lri(ary)
  counts = Hash.new(0)
  ary.each { |v| counts[v] += 1 }
  counts.select { |v, count| count == 1 }.keys 
end

def finks(ary)
  ary.to_set
end

def santosh_mohanty(ary)
    result = ary.reject.with_index do |ele,index|
      res = (ary[index+1] ^ ele)
      res == 0
    end
end

SHORT_ARRAY = [1,1,2,2,3,1]
mithun_sasidharan(SHORT_ARRAY) # => [1, 2, 3]
jaredsmith(SHORT_ARRAY) # => [1, 2, 3]
lri(SHORT_ARRAY) # => [3]
finks(SHORT_ARRAY) # => #<Set: {1, 2, 3}>
santosh_mohanty(SHORT_ARRAY) # => [1, 2, 3, 1]

puts 'Ruby v%s' % RUBY_VERSION

compare do
  _mithun_sasidharan { mithun_sasidharan(array) }
  _jaredsmith { jaredsmith(array) }
  _lri { lri(array) }
  _finks { finks(array) }
  _santosh_mohanty { santosh_mohanty(array) }
end

当运行时，结果是:

# >> Ruby v2.7.1
# >> Running each test 16 times. Test will take about 2 seconds.
# >> _mithun_sasidharan is faster than _jaredsmith by 2x ± 0.1
# >> _jaredsmith is faster than _santosh_mohanty by 4x ± 0.1 (results differ: [1, 2, 4, 5, 6, 7, 8] vs [1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, ...
# >> _santosh_mohanty is similar to _lri (results differ: [1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, ...
# >> _lri is similar to _finks (results differ: [] vs #<Set: {1, 2, 4, 5, 6, 7, 8}>)

注意:这些返回的结果不好:

lri(SHORT_ARRAY) # => [3] > . finks(SHORT_ARRAY) # => #<Set: {1,2,3 santosh_mohanty(SHORT_ARRAY) # => [1,2,3,1]

你可以返回交叉点。

a = [1,1,2,3]
a & a

这也将删除重复项。

尝试使用XOR操作符，不使用内置函数:

a = [3,2,3,2,3,5,6,7].sort!

result = a.reject.with_index do |ele,index|
  res = (a[index+1] ^ ele)
  res == 0
end

print result

内置功能:

a = [3,2,3,2,3,5,6,7]

a.uniq