Question

我目前正在做exercism.io而我正在进行汉明挑战，我现在已经陷入困境了，因为我需要做的是添加从哈希在一起。

挑战如下：

编写一个程序，可以计算两条DNA链之间的汉明差异。

他们开始为您服务的是（不要担心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

我已使用以下代码成功完成了其中四项：

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value的{{1}}数字加在一起创建一个整数，例如：

key

所以我需要做的是将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]]}和G的值加在一起并输出一个整数T

问题是，如何将两个值一起添加并输出一个整数而不输出整个哈希本身？

Answer 1

从你的哈希开始，

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，您需要每个Array#last提供的最后一个条目。

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

您的号码包含在另一个数组中，因此您可以再次使用last，但Array#flatten在语义上很不错。

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

然后用一个简单的方法来总结它们是Enumerable#inject。

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

我不清楚你是否也想要哈希键中的数字，但是如果你这样做，你可以使用Hash#keys，或者因为你想要一切，Hash#to_a。< / p>

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

将哈希键的值一起添加，以输出单个整数

1 个答案: