我有一個元組列表：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")]以第一個元組為例，因為元組中的 is 之前，勝過 。在2支球隊不相互對抗的情況下，我們以這種方式確定獲勝順序：例如，如果我們想找出和之間的獲勝順序，因為獲勝和獲勝，整體獲勝也是如此。("Team D", "Team A")DADATADATDTAT > D > A定義一個函數(shù)，該函數(shù)返回團(tuán)隊的排序列表，例如winning_list(matches)["Team T", "Team D", "Team A", ...]我有一個輔助方法，可以找到2個特定團(tuán)隊之間的獲勝順序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但這種方法似乎無效。此外，只有當(dāng)他們有一個共同的對手團(tuán)隊時，它才會起作用。