Parameterized Array#uniq (i.e., uniq_by)

Question

What if instead of removing duplicate elements from an array, I want to remove elements that have a specific property in common?

Specifically, I want to remove all strings from an array with duplicate "essences", where essence is defined like this:

class String
  def essence
    downcase.gsub('&', 'and').gsub(/[^a-z0-9]/, '')
  end
end

I want something like this:

['a', 'A', 'b'].uniq_by(&:essence)
# => ['a', 'b'] (or ['A', 'b']; I don't really care)

What's the best way to accomplish this?

score 10 · Answer 1 · edited Feb 06 '13 at 21:38

10

Since 1.9.2, Array#uniq (and uniq!) takes a block, so no more need for uniq_by.

edited Feb 06 '13 at 21:38

answered Mar 15 '12 at 17:36

steenslag

2

http://ruby-doc.org/core-1.9.2/Array.html#method-i-uniq confirms this. – Andrew Grimm Mar 15 '12 at 22:06
Just in case anyone else runs into this problem, ruby 1.9.2p180 unique with block doesn't seem to work as reliably as uniq_by, though I haven't done extensive testing to figure out what is up. – Kevin Nov 12 '12 at 07:06
And available to earlier Rubies with `require 'backports'`. – Marc-André Lafortune Feb 06 '13 at 21:39

tokland · Accepted Answer · 2011-03-09T21:17:28.580

4

Activesupport has a Array#uniq_by, and this is the code:

class Array
  def uniq_by
    hash, array = {}, []
    each { |i| hash[yield(i)] ||= (array << i) }
    array
  end
end

Facets also has a Enumerable#uniq_by:

module Enumerable    
  def uniq_by
    h = {}
    inject([]) {|a,x| h[yield(x)] ||= a << x}
  end    
end

edited Mar 09 '11 at 21:17

answered Mar 09 '11 at 21:02

tokland

score 3 · Answer 3 · answered Mar 09 '11 at 20:52

3

Quick and dirty way:

['a', 'A', 'b'].group_by {|i| i.essence}.values.map(&:first)

And some monkey patching:

class Array
  def uniq_by(&block)
    group_by(&block).values.map(&:first)
  end
end

answered Mar 09 '11 at 20:52

Tomas

Hey, just wanted to comment that you can actually do `group_by(&:essence)` so that if you wish to update your answer you can! – Jordon Bedwell Apr 13 '16 at 22:06

3 Answers3