Ruby - Compare two Enumerators elegantly

问题

I've got two long streams of numbers coming from two different sources (binary data) in Ruby (1.9.2).

The two sources are encapsulated in the form of two Enumerators.

I want to check that the two streams are exactly equal.

I've come with a couple solutions, but both seem quite inelegant.

The first one simply transforms both into an array:

def equal_streams?(s1, s2)
  s1.to_a == s2.to_a
end

This works, but it is not very performant, memory-wise, specially if the streams have lots of information.

The other option is... ugh.

def equal_streams?(s1, s2)
  s1.each do |e1|
    begin
      e2 = s2.next
      return false unless e1 == e2 # Different element found
    rescue StopIteration
      return false # s2 has run out of items before s1
    end
  end

  begin
    s2.next
  rescue StopIteration
    # s1 and s2 have run out of elements at the same time; they are equal
    return true
  end

  return false

end

So, is there a simpler, more elegant way of doing this?

回答1:

Just a slight refactoring to your code, assuming that your streams do not include an element :eof.

def equal_streams?(s1, s2)
  loop do
    e1 = s1.next rescue :eof
    e2 = s2.next rescue :eof
    return false unless e1 == e2
    return true if e1 == :eof
  end
end

Using a keyword like loop should be faster than using a method like each.

回答2:

Comparing them one element at a time is probably the best you're going to be able to do but you can do it nicer than your "ugh" solution:

def grab_next(h, k, s)
  h[k] = s.next
rescue StopIteration
end

def equal_streams?(s1, s2)
  loop do
    vals = { }
    grab_next(vals, :s1, s1)
    grab_next(vals, :s2, s2)
    return true  if(vals.keys.length == 0)  # Both of them ran out.
    return false if(vals.keys.length == 1)  # One of them ran out early.
    return false if(vals[:s1] != vals[:s2]) # Found a mismatch.
  end
end

The tricky part is differentiating between just one stream running out and both running out. Pushing the StopIteration exception into a separate function and using the absence of a key in a hash is a fairly convenient way of doing that. Just checking vals[:s1] will cause problems if your stream contains false or nil but checking for the presence of a key solves that problem.

回答3:

Here's a shot of doing it by creating an alternative for Enumerable#zip, which works lazily and doesn't create an entire array. It's combining my implementation of Closure's interleave and other two answers here (using sentinel value to indicate end of the Enumerable has been reached - the fact causing the problem is that next rewinds the Enumerable once it reached the end).

This solution supports multiple parameters, so you can compare n structures at once.

module Enumerable
  # this should be just a unique sentinel value (any ideas for more elegant solution?)
  END_REACHED = Object.new

  def lazy_zip *others
    sources = ([self] + others).map(&:to_enum)
    Enumerator.new do |yielder|
      loop do
        sources, values = sources.map{|s|
          [s, s.next] rescue [nil, END_REACHED]
        }.transpose
        raise StopIteration if values.all?{|v| v == END_REACHED}
        yielder.yield values.map{|v| v == END_REACHED ? nil : v}
      end
    end
  end
end

So, when you have variant of zip which works lazily and doesn't stop iteration when the first enumerable reaches the end, you can use all? or any? to actually check corresponding elements for equality.

# zip would fail here, as it would return just [[1,1],[2,2],[3,3]]:
p [1,2,3].lazy_zip([1,2,3,4]).all?{|l,r| l == r}
#=> false

# this is ok
p [1,2,3,4].lazy_zip([1,2,3,4]).all?{|l,r| l == r}
#=> true

# comparing more than two input streams:
p [1,2,3,4].lazy_zip([1,2,3,4],[1,2,3]).all?{|vals|
  # check for equality by checking length of the uniqued array
  vals.uniq.length == 1
}
#=> false

回答4:

Following the discussion in the comments, here is zip-based solution, first wrapping block version of zip within an Enumerator, then using it to compare corresponding elements.

It works, but there is already mentioned edge case: if the first stream is shorter than the other, remaining elements from the other will be discarded (see the example below).

I have marked this answer as community wiki as other members could improve on it.

def zip_lazy *enums
  Enumerator.new do |yielder|
    head, *tail = enums
    head.zip(*tail) do |values|
      yielder.yield values
    end
  end
end

p zip_lazy(1..3, 1..4).all?{|l,r| l == r}
#=> true
p zip_lazy(1..3, 1..3).all?{|l,r| l == r}
#=> true
p zip_lazy(1..4, 1..3).all?{|l,r| l == r}
#=> false

回答5:

Here is a 2-source example using a fiber/co-routine. It's a bit long-winded, but it's very explicit about its behavior, which is nice.

def zip_verbose(enum1, enum2)
  e2_fiber = Fiber.new do
    enum2.each{|e2| Fiber.yield true, e2 }
    Fiber.yield false, nil
  end
  e2_has_value, e2_val = true, nil
  enum1.each do |e1_val|
    e2_has_value, e2_val = e2_fiber.resume if e2_has_value
    yield [true, e1_val], [e2_has_value, e2_val]
  end
  return unless e2_has_value
  loop do
    e2_has_value, e2_val = e2_fiber.resume
    break unless e2_has_value
    yield [false, nil], [e2_has_value, e2_val]
  end
end

def zip(enum1, enum2)
  zip_verbose(enum1, enum2) {|e1, e2| yield e1[1], e2[1] }
end

def self.equal?(enum1, enum2)
  zip_verbose(enum1, enum2) do |e1,e2|
    return false unless e1 == e2
  end
  return true
end

来源：https://stackoverflow.com/questions/6486107/ruby-compare-two-enumerators-elegantly

标签

ruby

enumerator