How to pass a custom comparator to "sort"?

Question

Class A has the following comparator:

class A
  attr_accessor x

  def my_comparator(a)
    x**2 <=> (a.x)**2
  end
end

I would like to use this comparator to sort an array where each item is of class A:

class B
  def my_method
    items.sort!(<how can I pass my_comparator here ?>)
  end
end

How should I pass my_comparator to sort!?

Literally, you could use `items.sort! { |x,y| x.my_comparator y }`, but if this is the default sorting behavior for the class you should consider something like what the Tin Man has below. — coreyward, Jan 07 '11 at 03:46

the Tin Man · Accepted Answer · 2018-12-16T23:21:10.793

Define your own <=>, and include Comparable. This is from the Comparable doc:

class SizeMatters
  include Comparable
  attr :str
  def <=>(an_other)
    str.size <=> an_other.str.size
  end
  def initialize(str)
    @str = str
  end
  def inspect
    @str
  end
end

s1 = SizeMatters.new("Z")
s2 = SizeMatters.new("YY")
s3 = SizeMatters.new("XXX")
s4 = SizeMatters.new("WWWW")
s5 = SizeMatters.new("VVVVV")

s1 < s2                       #=> true
s4.between?(s1, s3)           #=> false
s4.between?(s3, s5)           #=> true
[ s3, s2, s5, s4, s1 ].sort   #=> [Z, YY, XXX, WWWW, VVVVV]

You don't actually have to include Comparable, but you get extra functionality for free if you do that after having defined <=>.

Otherwise, you can use Enumerable's sort with a block if your objects implement <=> already.

Another way to use several different comparisons is to use lambdas. This uses the new 1.9.2 declaration syntax:

ascending_sort  = ->(a,b) { a <=> b }
descending_sort = ->(a,b) { b <=> a }

[1, 3, 2, 4].sort( & ascending_sort ) # => [1, 2, 3, 4]
[1, 3, 2, 4].sort( & descending_sort ) # => [4, 3, 2, 1]

foo = ascending_sort
[1, 3, 2, 4].sort( & foo ) # => [1, 2, 3, 4]

Or, specific to this question: `alias_method :<=>, :my_comparator` — Phrogz, Jan 07 '11 at 03:29
+1. Good catch @Phrogz, though it would be more Ruby-ish to call the method `<=>` in the first place. — the Tin Man, Jan 07 '11 at 03:34
In my case, I have several comparators, so I don't want to override `<=>` with `my_comparator`. — Misha Moroshko, Jan 07 '11 at 04:32
I added some examples using lambdas which let you easily swap out comparisons. — the Tin Man, Jan 07 '11 at 05:03

score 26 · Answer 2 · answered Jan 07 '11 at 03:18

26

Both of these should work:

items.sort_by! { |a| (a.x)**2 }
items.sort! { |a1,a2| a1.my_comparator(a2) }

answered Jan 07 '11 at 03:18

Sophie Alpert

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This is good and correct, but @theTinMan's answer is better for custom classes. – Phrogz Jan 07 '11 at 03:29

score 5 · Answer 3 · answered Jan 07 '11 at 03:47

items.sort!(&:my_comparator)

This calls the :my_comparator.to_proc internally, which returns a block

proc {|x,y| x.my_comparator(y)}

thus reducing this answer to Ben Alpert's answer.

(But I agree with Phrogz's observation that if this is the natural order for the class, then you should use the Tin Man's answer instead.)

Stefan Rendevski · Answer 4 · 2018-07-26T21:03:56.097

If you want to reuse these comparators in different places, it would be better to define them as a class, instead of rewriting the same lambda expression every time.

This is based on Java's implementation of Comparable interface:

module Comparator
  def compare(a, b)
    raise NotImplementedError, 'must implement this method'
  end

  def to_proc
    ->(a, b) { compare(a, b) }
  end
end

class LengthComparator
  include Comparator

  def compare(a, b)
    a.length <=> b.length
  end
end

class ReverseLengthComparator < LengthComparator
  def compare(a, b)
    -super
  end
end

You implement your comparison logic in the #compare method. You can then use this class like so: array.sort(&MyCustomComparator.new). It essentially boils down to a lambda expression, but supports more reusability in my opinion.

How to pass a custom comparator to "sort"?

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