如何重构 Ruby 方法以通过 Rubocop 测试同时也是 DRY

Question

作为 Launch School 课程的一部分，我正在研究二十一游戏（将游戏视为 Blackjack 的轻量级版本）。

我需要适当地重构我自己的代码以使其保持干燥，同时还通过所有 Rubocop 测试。我目前使用的是 Rubocop 0.85.0 版本（课程需要这个版本）。

在我的游戏中，我将所有数据存储在一个名为 Keep Score 的散列中：

keep_score = { 'player_cards' => [], \
               'dealer_cards' => [], \
               'player_card_values' => [], \
               'dealer_card_values' => [], \
               'player_points' => 0, \
               'dealer_points' => 0, \
               'player_move' => '', \
               'dealer_move' => '', \
               'end_game' => false }

到目前为止我的代码可以正常工作，到目前为止我已经做了很多重构，但是我需要进一步重构这两个方法以使代码更干，同时仍然通过 Rubocop 测试。

我有两种方法，一种 convert_face_cards 转换 K、Q 和 J（A 单独处理）。然后有一个单独的方法，add_integer_points 来处理卡片上的整数值：

def convert_face_cards(keep_score)
  keep_score['player_points'] = 0
  keep_score['dealer_points'] = 0

  keep_score['player_card_values'].each do |card|
    if card == "jack" || card == "queen" || card == "king"
      keep_score['player_points'] += 10
    end
  end
  keep_score['dealer_card_values'].each do |card|
    if card == "jack" || card == "queen" || card == "king"
      keep_score['dealer_points'] += 10
    end
  end
end

def add_integer_points(keep_score)
  keep_score['player_card_values'].each do |card|
    if card.is_a? Integer
      keep_score['player_points'] += card
    end
  end
  keep_score['dealer_card_values'].each do |card|
    if card.is_a? Integer
      keep_score['dealer_points'] += card
    end
  end
end

虽然这段代码从技术角度来看是可行的，但我没有通过 Rubocop 测试。 convert_face_cards 方法返回错误：Cyclomatic complexity for convert_face_cards is too high. [7/6].

我的代码显然不够干。例如，在 convert_face_cards 方法中，我本质上是运行 each 方法，在 player_card_values 哈希键和 dealer_card_values 上都采用相同的方式哈希键。

但我不确定如何进一步压缩这些方法。感谢任何帮助或指导！

Answer 1

def convert_face_cards(keep_score)
  players = keep_score['player_card_values'].select { |card| card == "jack" || card == "queen" || card == "king" }
  dealers = keep_score['dealer_card_values'].select { |card| card == "jack" || card == "queen" || card == "king" }

  keep_score['player_points'] = players.length * 10
  keep_score['dealer_points'] = dealers.length * 10
end

Answer 2

由于无论是闲家还是庄家，算人牌的逻辑都是一样的，你可以把它提取到自己的方法中，稍微干点。

def convert_face_cards(keep_score)
  keep_score['player_points'] = count_faces(keep_score['player_card_values'])
  keep_score['dealer_points'] = count_faces(keep_score['dealer_card_values'])
end

def sum_faces(cards)
  faces = cards.select { |card| face?(card) }
  faces.length * 10
end

def face?(card)
  %w[jack queen king].include?(card)
end

Answer 3

您不应该使用散列来存储此数据并编写循环遍历散列的代码。您应该使用包含这些知识的数据结构作为方法。

例如：

Card = Struct.new(:value) do
  def score
    case value
    when :king, :queen, :jack
      10
    when :ace
      1
    else
      value
    end
  end
end

CARDS = [
  :ace,
  :king,
  :queen,
  :jack,
  10,
  ... # other cards here
  2
].map { |card| Card.new(card) }

player_cards = []
dealer_cards = []

def score(cards)
  cards.sum(&:score)
end

如何重构 Ruby 方法以通过 Rubocop 测试同时也是 DRY

How to refactor Ruby methods to pass Rubocop tests while also being DRY

ruby

refactoring