给定一个元组中由 2 个团队组成的元组列表，返回一个按获胜团队到失败团队排序的列表

Question

我有一个元组列表：

matches = [("Team D","Team A"), ("Team E","Team B"), ("Team T","Team B"), ("Team T","Team D"), ("Team F","Team C"), ("Team C","Team L"), ("Team T","Team F")]

以第一个元组为例，因为元组中的 is 之前，胜过 。在2支球队不相互对抗的情况下，我们以这种方式确定获胜顺序：例如，如果我们想找出和之间的获胜顺序，因为获胜和获胜，整体获胜也是如此。("Team D", "Team A")DADATADATDTAT > D > A

定义一个函数，该函数返回团队的排序列表，例如winning_list(matches)["Team T", "Team D", "Team A", ...]

我有一个辅助方法，可以找到2个特定团队之间的获胜顺序

def winner(matches, team_1, team_2):

    size = len(matches)

    lst1 = []

    lst2 = []

    for i in range(0, size): # extract games with team1

        if matches[i][0] == team1 or matches[i][1] == team1:

            lst1.append(matches[i])

        elif matches[i][0] == team2 or matches[i][1] == team2: # extract games with team2

            lst2.append(matches[i])

    lst_partner1 = [] # opponent teams involving team1

    lst_partner2 = [] # opponent teams involving team2

    for i in range(0, len(lst1)):

        if lst1[i][0] != team1:

            lst_partner1.append(lst1[i][0])

        elif lst1[i][1] != team1:

            lst_partner1.append(lst1[i][1])

    for i in range(0, len(lst2)):

        if lst2[i][0] != team2:

            lst_partner2.append(lst2[i][0])

        elif lst2[i][1] != team2:

            lst_partner2.append(lst2[i][1])

    common = [value for value in lst_partner1 if value in lst_partner2] # opponent team that played against team1 and team2

    # print(common)

    opponent_team = common[0]

    # print(opponent_team)

    if len(common) == 0:

        return 0

    else:

        for  i in range(0, len(lst1)):

            if opponent_team in lst1[i]:

                idx_opp1 = lst1[i].index(opponent_team)

        for l in range(0, len(lst2)):

            if opponent_team in lst2[l]:

                idx_opp2 = lst2[l].index(opponent_team)

        if idx_opp1 == idx_opp2:

            return 0

        elif idx_opp1 < idx_opp2:

            return 2

        elif idx_opp1 > idx_opp2:

            return 1

        else:

            return 0

但这种方法似乎无效。此外，只有当他们有一个共同的对手团队时，它才会起作用。

Answer 1

根据提供的有关如何对解决方案进行排名的信息，可以：from collections import defaultdictmatches = [("Team D", "Team A"), ("Team E", "Team B"), ("Team T", "Team B"),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;("Team T", "Team D"), ("Team F", "Team C"), ("Team C", "Team L"),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;("Team T", "Team F")]def winning_list(mathces):&nbsp; &nbsp; scores = defaultdict(int)&nbsp; &nbsp; for fst, snd in matches:&nbsp; &nbsp; &nbsp; &nbsp; scores[fst] += 1&nbsp; &nbsp; &nbsp; &nbsp; scores[snd] -= 1&nbsp; &nbsp; return sorted(scores.items(), key=lambda e: e[1], reverse=True)ranking = winning_list(matches)print(ranking)为了使它更简单，我们可以使用collections.Counterfrom collections import Counterdef winning_list2(mathces):&nbsp; &nbsp; scores = Counter()&nbsp; &nbsp; for fst, snd in matches:&nbsp; &nbsp; &nbsp; &nbsp; scores[fst] += 1&nbsp; &nbsp; &nbsp; &nbsp; scores[snd] -= 1&nbsp; &nbsp; return scores.most_common()