Question

我在python 3中编写了一个程序，它模拟了不同玩家之间的竞争，这些玩家正在抛球。

我有一个包含玩家对象的列表，对象具有属性名称，平均投掷长度，投掷标准偏差，最后：表单列表的属性，其中每个索引包含玩家的多个投掷在eack比赛期间预制。（为了澄清，在索引0上将有例如来自第一场比赛的6次投掷，在索引1处将有来自第二次比赛的6次投掷等等）。当然有多个球员。我现在需要找到一种方法来确定哪个玩家是每个比赛的赢家，并计算每个玩家的获胜次数。我总是发现赢家的事情，但这是我到目前为止所想到的：

def winner(self):
    for player in players:
        winner= max(players, key = attrgetter("self.results"))

通过此类创建Player对象：

class Player:

    def __init__(self, name, mean, standardDeviation):
        self.name = name
        self.mean = mean
        self.standardDeviation = standardDeviation
        self.results = []

我很确定我上面写的内容不会让我到任何地方，所以你们有没有任何提示？

Answer 1

这是一个只有简化Player类的东西的版本，可以帮助您入门。它可以处理不止一个玩家将球投射到另一个玩家最长距离的关系。

class Player:
    def __init__(self, name, results):
        self.name = name
        self.results = results

def find_winners(players, num_contests):
    winners = []  # one (or more) player entries per contest
    for icontest in range(num_contests):
        contest_winners = []  # may hold multiple players if there are ties
        contest_longest_throw = -1
        for player in players:
            player_longest_throw = max(player.results[icontest])
            if player_longest_throw > contest_longest_throw:
                contest_longest_throw = player_longest_throw
                contest_winners = [player]
            elif player_longest_throw == contest_longest_throw:
                contest_winners.append(player)
        winners.append(contest_winners)
    return winners

# test data
players = [Player('Name1', [(7, 8, 9), (2, 3, 4), (3, 4, 5)]),
           Player('Name2', [(4, 5, 6), (8, 9, 10), (6, 7, 8)]),
           Player('Name3', [(1, 2, 3), (5, 6, 7), (9, 10, 11)]),
           # add one that will tie with first player
           Player('Name4', [(7, 8, 9), (2, 3, 4), (3, 4, 5)])]

# assumes length of first player's results is number of contests
winners_found = find_winners(players, len(players[0].results))

print("Contest  Winner(s)")
for icontest, contest_winners in enumerate(winners_found, start=1):
    print('{:^7}  {}'.format(
        icontest, ', '.join(player.name for player in contest_winners)))

输出：

Contest  Winner(s)
   1     Name1, Name4
   2     Name2
   3     Name3

查找列表中的最大值，即列表中对象中的属性

1 个答案: