How do I compare two hashes?

Question

I am trying to compare two Ruby Hashes using the following code:

#!/usr/bin/env ruby

require \"yaml\"
require \"active_support\"

file1 = YAML::load(File.op         


        

Answer 1

Here is algorithm to deeply compare two Hashes, which also will compare nested Arrays:

    HashDiff.new(
      {val: 1, nested: [{a:1}, {b: [1, 2]}] },
      {val: 2, nested: [{a:1}, {b: [1]}] }
    ).report

# Output:
val:
- 1
+ 2
nested > 1 > b > 1:
- 2

Implementation:

class HashDiff

  attr_reader :left, :right

  def initialize(left, right, config = {}, path = nil)
    @left  = left
    @right = right
    @config = config
    @path = path
    @conformity = 0
  end

  def conformity
    find_differences
    @conformity
  end

  def report
    @config[:report] = true
    find_differences
  end

  def find_differences
    if hash?(left) && hash?(right)
      compare_hashes_keys
    elsif left.is_a?(Array) && right.is_a?(Array)
      compare_arrays
    else
      report_diff
    end
  end

  def compare_hashes_keys
    combined_keys.each do |key|
      l = value_with_default(left, key)
      r = value_with_default(right, key)
      if l == r
        @conformity += 100
      else
        compare_sub_items l, r, key
      end
    end
  end

  private

  def compare_sub_items(l, r, key)
    diff = self.class.new(l, r, @config, path(key))
    @conformity += diff.conformity
  end

  def report_diff
    return unless @config[:report]

    puts "#{@path}:"
    puts "- #{left}" unless left == NO_VALUE
    puts "+ #{right}" unless right == NO_VALUE
  end

  def combined_keys
    (left.keys + right.keys).uniq
  end

  def hash?(value)
    value.is_a?(Hash)
  end

  def compare_arrays
    l, r = left.clone, right.clone
    l.each_with_index do |l_item, l_index|
      max_item_index = nil
      max_conformity = 0
      r.each_with_index do |r_item, i|
        if l_item == r_item
          @conformity += 1
          r[i] = TAKEN
          break
        end

        diff = self.class.new(l_item, r_item, {})
        c = diff.conformity
        if c > max_conformity
          max_conformity = c
          max_item_index = i
        end
      end or next

      if max_item_index
        key = l_index == max_item_index ? l_index : "#{l_index}/#{max_item_index}"
        compare_sub_items l_item, r[max_item_index], key
        r[max_item_index] = TAKEN
      else
        compare_sub_items l_item, NO_VALUE, l_index
      end
    end

    r.each_with_index do |item, index|
      compare_sub_items NO_VALUE, item, index unless item == TAKEN
    end
  end

  def path(key)
    p = "#{@path} > " if @path
    "#{p}#{key}"
  end

  def value_with_default(obj, key)
    obj.fetch(key, NO_VALUE)
  end

  module NO_VALUE; end
  module TAKEN; end

end

Answer 2

... and now in module form to be applied to a variety of collection classes (Hash among them). It's not a deep inspection, but it's simple.

# Enable "diffing" and two-way transformations between collection objects
module Diffable
  # Calculates the changes required to transform self to the given collection.
  # @param b [Enumerable] The other collection object
  # @return [Array] The Diff: A two-element change set representing items to exclude and items to include
  def diff( b )
    a, b = to_a, b.to_a
    [a - b, b - a]
  end

  # Consume return value of Diffable#diff to produce a collection equal to the one used to produce the given diff.
  # @param to_drop [Enumerable] items to exclude from the target collection
  # @param to_add  [Enumerable] items to include in the target collection
  # @return [Array] New transformed collection equal to the one used to create the given change set
  def apply_diff( to_drop, to_add )
    to_a - to_drop + to_add
  end
end

if __FILE__ == $0
  # Demo: Hashes with overlapping keys and somewhat random values.
  Hash.send :include, Diffable
  rng = Random.new
  a = (:a..:q).to_a.reduce(Hash[]){|h,k| h.merge! Hash[k, rng.rand(2)] }
  b = (:i..:z).to_a.reduce(Hash[]){|h,k| h.merge! Hash[k, rng.rand(2)] }
  raise unless a == Hash[ b.apply_diff(*b.diff(a)) ] # change b to a
  raise unless b == Hash[ a.apply_diff(*a.diff(b)) ] # change a to b
  raise unless a == Hash[ a.apply_diff(*a.diff(a)) ] # change a to a
  raise unless b == Hash[ b.apply_diff(*b.diff(b)) ] # change b to b
end

Answer 3

I had the same problem and sent a pull request to rails

• Works with deeply nested hash
• Works with arrays of hashes

https://github.com/elfassy/rails/commit/5f3410e04fe8c4d4745397db866c9633b80ccec6

Answer 4

This was answered in "Comparing ruby hashes". Rails adds a diff method to hashes. It works well.

Answer 5

You can compare hashes directly for equality:

hash1 = {'a' => 1, 'b' => 2}
hash2 = {'a' => 1, 'b' => 2}
hash3 = {'a' => 1, 'b' => 2, 'c' => 3}

hash1 == hash2 # => true
hash1 == hash3 # => false

hash1.to_a == hash2.to_a # => true
hash1.to_a == hash3.to_a # => false

---

---

You can convert the hashes to arrays, then get their difference:

hash3.to_a - hash1.to_a # => [["c", 3]]

if (hash3.size > hash1.size)
  difference = hash3.to_a - hash1.to_a
else
  difference = hash1.to_a - hash3.to_a
end
Hash[*difference.flatten] # => {"c"=>3}

---

Simplifying further:

Assigning difference via a ternary structure:

  difference = (hash3.size > hash1.size) \
                ? hash3.to_a - hash1.to_a \
                : hash1.to_a - hash3.to_a
=> [["c", 3]]
  Hash[*difference.flatten] 
=> {"c"=>3}

Doing it all in one operation and getting rid of the difference variable:

  Hash[*(
  (hash3.size > hash1.size)    \
      ? hash3.to_a - hash1.to_a \
      : hash1.to_a - hash3.to_a
  ).flatten] 
=> {"c"=>3}

Answer 6

what about convert both hash to_json and compare as string? but keeping in mind that

require "json"
h1 = {a: 20}
h2 = {a: "20"}

h1.to_json==h1.to_json
=> true
h1.to_json==h2.to_json
=> false