Remove duplicate elements from array in Ruby

I have a Ruby array which contains duplicate elements.

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

How can I remove all the duplicate elements from this array while retaining all unique elements without using for-loops and iteration?

array = array.uniq

uniq removes all duplicate elements and retains all unique elements in the array.

This is one of many beauties of the Ruby language.

You can return the intersection.

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

This will also delete duplicates.

You can remove the duplicate elements with the uniq method:

array.uniq  # => [1, 2, 4, 5, 6, 7, 8]

What might also be useful to know is that uniq takes a block, so if you have a have an array of keys:

["bucket1:file1", "bucket2:file1", "bucket3:file2", "bucket4:file2"]

and you want to know what the unique files are, you can find it out with:

a.uniq { |f| f[/\d+$/] }.map { |p| p.split(':').last }

If someone was looking for a way to remove all instances of repeated values, see "https://stackoverflow.com/questions/786879/how-can-i-efficiently-extract-repeated-elements-in-a-ruby-array";.

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]

Just another alternative if anyone cares.

You can also use the to_set method of an array which converts the Array into a Set and by definition, set elements are unique.

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

The simplest ways for me are these ones:

array = [1, 2, 2, 3]

Array#to_set

array.to_set.to_a

# [1, 2, 3]

Array#uniq

array.uniq

# [1, 2, 3]

Just to provide some insight:

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

Which, when run, results in:

# >> 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}>)

Note: these returned bad results:

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

Try using the XOR operator, without using built-in functions:

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

With built-in functions:

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

a.uniq