A队赢得了B队

Question

我正在努力计算体育联盟的排名。领带断路器基于头2记录。因此，为了确定在积分达到平局时排名更高的球队，我需要能够从一个字典中拉开团队之间的头部记录。

我的计划是为每个团队创建一个字典，其中包含针对其他团队的所有头部2个头部记录。

问题是，团队名称每年都会发生变化（以及团队数量）。所以我需要遍历团队列表（从数据库中调用）并为每个团队创建字典。这可能吗？有没有更简单的方法来解决这个问题？因为我可能不知道我在做什么。

我的代码目前看起来像这样：

wins = collections.defaultdict(int)
losses = collections.defaultdict(int)
ties = collections.defaultdict(int)
points = collections.defaultdict(int)
games_rem = collections.defaultdict(int)
poss_points = collections.defaultdict(int)

for team in team_dict:
    wins[team.shortname] = team.wins
    losses[team.shortname] = team.losses
    ties[team.shortname] = team.ties
    points[team.shortname] = team.points
    games_rem[team.shortname] = mens_games.count + team.wins + team.losses + team.ties
    poss_points[team.shortname] = team.points + games_rem[team.shortname] * 2


for team in team_dict:
    for other_teams in team_dict:
        if other_teams.shortname == team.shortname:
            continue
        vs_"string{0}".format(other_teams.shortname) = collections.defaultdict(int)
        for game in game_dict:
            if game.team1 == other_teams.shortname and game.team2 == team.shortname and game.team1score < game.team2score:
                vs_"string{0}".format(other_teams.shortname)[team.shortname] += 1
            if game.team1 == other_teams.shortname and game.team2 == team.shortname and game.team1score > game.team2score:
                vs_"string{0}".format(other_teams.shortname)[team.shortname] -= 1
            if game.team2 == other_teams.shortname and game.team1 == team.shortname and game.team2score < game.team1score:
                vs_"string{0}".format(other_teams.shortname)[team.shortname] += 1
            if game.team2 == other_teams.shortname and game.team1 == team.shortname and game.team2score > game.team1score:
                vs_"string{0}".format(other_teams.shortname)[team.shortname] -= 1

问题显然是我尝试根据团队名称分配dictonaries

vs_"string{0}".format(other_teams.shortname) = collections.defaultdict(int)

我认为这绝对是不允许的，但我无法想出一个更好的方法来解决这个问题。有什么想法吗？

编辑：为不良解释道歉。我试图输出的是每个团队的单独字典，如下所示：

vs_teamA = {'teamB': 1, 'teamC': 1, 'teamD': -1, 'teamE': 0}
vs_teamB = {'teamA': -1, 'teamC': 1, 'teamD': -1, 'teamE': 0}
vs_teamC = {'teamA': -1, 'teamB': -1, 'teamD': -1, 'teamE': 0}
vs_teamD = {'teamA': 1, 'teamB': 1, 'teamC': 1, 'teamE': 0}
vs_teamE = {'teamA': 0, 'teamB': 0, 'teamC': 0, 'teamD': 0}

如果一支球队与另一支球队有胜利记录，那么这个数字将是正数，如果他们有失败的记录则为负数。

Answer 1

这里确实没有具体的问题，但我认为你可以从中学习python模块pandas。作为一个小例子，看看如何使用“类字典”界面组织数据：

import pandas as pd

stat_names = "points", "games_rem", "poss_points"
team_names = "A", "B", "C"

df = pd.DataFrame(0, index=stat_names, columns=team_names)

df["A"]["points"] = 30
df["B"]["points"] = 20
df["C"]["poss_points"] = 2

print df

哪个给出了

              A   B  C
points       30  20  0
games_rem     0   0  0
poss_points   0   0  2

Answer 2

为什么你不能将字典视为价值？

{'vs_teamA': {'teamB': 1, 'teamC': 1, 'teamD': -1, 'teamE': 0},
 'vs_teamB': {'teamA': -1, 'teamC': 1, 'teamD': -1, 'teamE': 0},
 'vs_teamC': {'teamA': -1, 'teamB': -1, 'teamD': -1, 'teamE': 0},
 'vs_teamD': {'teamA': 1, 'teamB': 1, 'teamC': 1, 'teamE': 0},
 'vs_teamE': {'teamA': 0, 'teamB': 0, 'teamC': 0, 'teamD': 0}}

Answer 3

我建议您使用嵌套字典

In [371]: from collections import defaultdict

In [380]: match_results = defaultdict(lambda: defaultdict(int))

A队赢得了B队

In [381]: match_results['vs_teamA']['teamB']+=1

A队输给C队

In [382]: match_results['vs_teamA']['teamC']+=-1

C队输给B队

In [383]: match_results['vs_teamC']['teamB']+=-1

In [386]: for team, games in match_results.iteritems():
   .....:     print team
   .....:     print games


vs_teamA
defaultdict(<type 'int'>, {'teamB': 1, 'teamC': -1})
vs_teamC
defaultdict(<type 'int'>, {'teamB': -1})

Python - 使用列表中的变量为字典创建名称

3 个答案:

A队赢得了B队

A队输给C队

C队输给B队