将散列键的值加在一起，得到一个整数

Question

我目前正在做exercism.io，正在接受Hamming挑战，我有点陷入了困境，因为我需要做的是将哈希中出现的数字相加。

挑战如下：

编写一个程序，可以计算两条DNA链之间的汉明差。

他们让你开始的是（不要担心说skip的方法）：

#!/usr/bin/env ruby
gem 'minitest', '>= 5.0.0'
require 'minitest/autorun'
require_relative 'hamming'
# Test data version:
# ab84334 Merge pull request #106 from bennn/grep-meta
class HammingTest < Minitest::Test
  def test_identical_strands
    assert_equal 0, Hamming.compute('A', 'A')
  end
  def test_long_identical_strands
    assert_equal 0, Hamming.compute('GGACTGA', 'GGACTGA')
  end
  def test_complete_distance_in_single_nucleotide_strands
    assert_equal 1, Hamming.compute('A', 'G')
  end
  def test_complete_distance_in_small_strands
    assert_equal 2, Hamming.compute('AG', 'CT')
  end
  def test_small_distance_in_small_strands
    skip
    assert_equal 1, Hamming.compute('AT', 'CT')
  end
  def test_small_distance
    skip
    assert_equal 1, Hamming.compute('GGACG', 'GGTCG')
  end
  def test_small_distance_in_long_strands
    skip
    assert_equal 2, Hamming.compute('ACCAGGG', 'ACTATGG')
  end
  def test_non_unique_character_in_first_strand
    skip
    assert_equal 1, Hamming.compute('AGA', 'AGG')
  end
  def test_non_unique_character_in_second_strand
    skip
    assert_equal 1, Hamming.compute('AGG', 'AGA')
  end
  def test_large_distance
    skip
    assert_equal 4, Hamming.compute('GATACA', 'GCATAA')
  end
  def test_large_distance_in_off_by_one_strand
    skip
    assert_equal 9, Hamming.compute('GGACGGATTCTG', 'AGGACGGATTCT')
  end
  def test_empty_strands
    skip
    assert_equal 0, Hamming.compute('', '')
  end
  def test_disallow_first_strand_longer
    skip
    assert_raises(ArgumentError) { Hamming.compute('AATG', 'AAA') }
  end
  def test_disallow_second_strand_longer
    skip
    assert_raises(ArgumentError) { Hamming.compute('ATA', 'AGTG') }
  end
  # Problems in exercism evolve over time,
  # as we find better ways to ask questions.
  # The version number refers to the version of the problem you solved,
  # not your solution.
  #
  # Define a constant named VERSION inside of Hamming.
  # If you are curious, read more about constants on RubyDoc:
  # http://ruby-doc.org/docs/ruby-doc-bundle/UsersGuide/rg/constants.html
  def test_bookkeeping
    skip
    assert_equal 1, Hamming::VERSION
  end
end

我已经用以下代码成功地完成了其中的四个：

class Hamming
  def self.compute(x, y)
    if x == y
      0
    else
      strings = x, y
      joined = strings.join
      positions = (0...joined.length).group_by{|i| joined[i]}
      length = strings.first.length
      n = strings.length
      diff = Hash[*positions.map{|k, v| 
      [k, v.group_by{|i| i % length}.reject{|i, is| is.length == n}.keys]}]
      diff
    end
  end
end

我现在需要做的是将key的value数字相加，以创建一个整数，例如：

 10) Failure:
HammingTest#test_complete_distance_in_small_strands [hamming_test.rb:23]:
Expected: 2
  Actual: nil
  #<= {["A", [0]]=>["G", [1]], ["C", [0]]=>["T", [1]]}

所以我需要做的是把G和T的值加在一起，输出一个整数2

问题是，如何将两个值相加并输出一个整数，而不输出整个哈希本身？

Answer 1

从您的哈希开始，

h = {["A", [0]]=>["G", [1]], ["C", [0]]=>["T", [1]]}

首先使用Hash#values来获取值。

> h.values
 => [["G", [1]], ["T", [1]]]

这些是数组，您需要Array#last提供的每个数组的最后一个条目。

> h.values.map(&:last)
 => [[1], [1]]

你的数字被包装在另一个数组中，所以你可以再次使用last，但Array#flatten在语义上很好。

> h.values.map(&:last).flatten
 => [1, 1]

然后用Enumerable#inject求和它们的一个简单方法。

> h.values.map(&:last).flatten.inject(:+)
 => 2

我不清楚你是否也想从你的哈希密钥中提取数字，但如果你想，你可以用哈希#密钥来获得它们，或者因为你想要一切，Hash#to_a。

> h.keys + h.values
 => [["A", [0]], ["C", [0]], ["G", [1]], ["T", [1]]]
> h.to_a
 => [[["A", [0]], ["G", [1]]], [["C", [0]], ["T", [1]]]]