如何在Ruby中对世界杯组表进行排序
我正在编写一个算法来根据匹配数据创建和排序世界杯组表。 因此，给出以下匹配数据：
[ { id: 1, home_team: "Honduras", away_team: "Chile", home_score: 0, away_score: 1 }, { id: 2, home_team: "Spain", away_team: "Switzerland", home_score: 0, away_score: 1 }, { id: 3, home_team: "Chile", away_team: "Switzerland", home_score: 1, away_score: 0 }, { id: 4, home_team: "Spain", away_team: "Honduras", home_score: 2, away_score: 0 }, { id: 5, home_team: "Chile", away_team: "Spain", home_score: 1, away_score: 2 }, { id: 6, home_team: "Honduras", away_team: "Switzerland", home_score: 0, away_score: 0 } ] 我的程序会产生这个（订单很重要）：
[{ goals_for: 4, goals_against: 2, goal_diff: 2, points: 6, name: "Spain" }, { goals_for: 3, goals_against: 2, goal_diff: 1, points: 6, name: "Chile" }, { goals_for: 1, goals_against: 1, goal_diff: 0, points: 4, name: "Switzerland" }, { goals_for: 0, goals_against: 3, goal_diff: -3, points: 1, name: "Honduras" }]
这很棒，除非有双向或三向平局。 然后标准变得复杂。 这是按优先顺序排列的：
- 最多点数
- 最大的目标差异
- 最伟大的目标
- 如果存在平局，则使用以下内容
- 绑定球队之间的比赛得分最多
- 绑定球队之间比赛的最大目标差异
- 最佳目标是从平局球队之间的比赛得分
- 抽签
题
我的排序function满足前三个标准。 如何在有双向或三向平局的情况下更改它？
def sort teams.sort_by! do |team| [ team[:points], team[:goal_diff], team[:goals_for] ] end.reverse! end
三向领带的例子
[ { id: 1, home_team: "Algeria", away_team: "Slovenia", home_score: 2, away_score: 1 }, { id: 2, home_team: "USA", away_team: "Slovenia", home_score: 5, away_score: 1 }, { id: 3, home_team: "England", away_team: "Slovenia", home_score: 4, away_score: 0 }, { id: 4, home_team: "Algeria", away_team: "USA", home_score: 3, away_score: 0 }, { id: 5, home_team: "USA", away_team: "England", home_score: 2, away_score: 0 }, { id: 6, home_team: "England", away_team: "Algeria", home_score: 3, away_score: 2 } ]
这个例子将根据标准1（点）消除斯洛文尼亚。
然后基于匹配数据的子集计算剩余三个队的等级。 该子集应仅包括绑定团队之间的匹配。 在这种情况下，我们将使用包括阿尔及利亚，英格兰和美国在内的所有比赛重建表格。 我们排除涉及斯洛文尼亚的比赛
该表应如下所示：
| POS | TEAM | GF | GA | GD | POINTS | | 1 | Algeria | 5 | 3 | 2 | 3 | | 3 | England | 3 | 4 | -1 | 3 | | 2 | USA | 2 | 3 | -1 | 3 |
阿尔及利亚队取得了净胜球（标准5）。 英格兰排名第二，因为其goals for大于美国（标准6）。
我的程序实际上输出了这个，这是不正确的，因为它对关系没有任何作用，并且在标准3处停止。
[ { goals_for: 7, goals_against: 4, goal_diff: 3, points: 6, name: "England" }, { goals_for: 7, goals_against: 4, goal_diff: 3, points: 6, name: "Algeria" }, { goals_for: 7, goals_against: 4, goal_diff: 3, points: 6, name: "USA" }, { goals_for: 2, goals_against: 11, goal_diff: -9, points: 0, name: "Slovenia" }]
这是完整的程序：
class Calculator attr_reader :games, :teams def initialize(games) defaults = { goals_for: 0, goals_against: 0, goal_diff: 0, points: 0 } @games = games @teams = games.each_with_object([]) do |game, arr| arr.push({ name: game[:home_team] }.merge!(defaults)) arr.push({ name: game[:away_team] }.merge!(defaults)) end.uniq end def build_table build sort return teams end private def build games.each do |game| if game[:home_score].present? && game[:away_score].present? home_team = teams.detect { |team| team[:name] == game[:home_team] } away_team = teams.detect { |team| team[:name] == game[:away_team] } home_team[:goals_for] += game[:home_score] home_team[:goals_against] += game[:away_score] away_team[:goals_for] += game[:away_score] away_team[:goals_against] += game[:home_score] home_team[:goal_diff] = home_team[:goals_for] - home_team[:goals_against] away_team[:goal_diff] = away_team[:goals_for] - away_team[:goals_against] if game[:home_score] > game[:away_score] home_team[:points] += 3 elsif game[:home_score] < game[:away_score] away_team[:points] += 3 else home_team[:points] += 1 away_team[:points] += 1 end end end end def sort teams.sort_by! { |team| [ team[:points], team[:goal_diff], team[:goals_for] ] }.reverse! end end
您有一套明确定义的规则来确定如何订购团队。 一种方法是编写一个一次一个地实现这些规则的排序例程，并在找到胜利者时进行短路：
def compare_points(a, b) a[:points] <=> b[:points] end def compare_goal_diff(a, b) a[:goal_diff] <=> b[:goal_diff] end def compare_teams(a, b) comparison = compare_points(a, b) return comparison unless comparison.zero? comparison = compare_goal_diff(a, b) return comparison unless comparison.zero? # Repeat for each type of comparison # ... comparison.zero? ? flip_coin : comparison end teams.sort! { |a, b| compare_teams(a, b) }.reverse!
比较单个值（如点）时，比较运算符<=>就足够了。 对于更复杂的比较，您需要深入了解@games数组以确定获胜者，例如：
def compare_points_from_matches_between(a, b) # Hand-waving follows # case # when team A has fewer points than team B in their meetings # -1 # when team B has fewer points than team A in their meetings # 1 # else # 0 # end end
根据您的规则应用每个比较。 在每一步，如果比较非零，则返回该值; 否则你继续下一步。 最后，如果比较仍为零，则翻转一枚硬币。