Question

我需要编写一个函数，分别查看包含两个团队及其游戏分数的嵌套列表。该列表包含多个匹配项，我希望输出是一个嵌套列表，其中包含所有团队名称和赢得的游戏数量。该列表如下所示：

L = [['Patriots', 'Giants', '3', '1'], ['Steelers', 'Patriots', '1', 2'], ['Giants', 'Steelers', '3', '5']]

所以在上面的列表中，前两个元素是团队名称，第三和第四个元素是他们在比赛中得分。但是，这个列表比这个要大得多，还有更多的团队。输出看起来像这样：

finalList = [['Patriots', 2], ['Giants', 0], ['Steelers', 1]]

因为爱国者赢了两场比赛，巨人队赢了零比赛，钢人队赢了一场比赛。

我已经尝试了以下代码，但它不起作用而且我被卡住了。

def gamesWon():
    for i in L:
        count = 0
        if i[2]>i[3]:
            count += 1
            i.append(count)

Answer 1

您可以使用defaultdict：

from collections import defaultdict
# initialize the result as a defaultdict with default value of 0
result = defaultdict(lambda : 0)   

for t1,t2,s1,s2 in L:
    if int(s1) > int(s2):
        result[t1] += 1
    elif int(s2) > int(s1):
        result[t2] += 1

result
# defaultdict(<function __main__.<lambda>>, {'Patriots': 2, 'Steelers': 1})

请注意，即使在结果中，得分为零的团队也会丢失，但如果您拨打result[team]，则会给您零。

Answer 2

您可以使用defaultdict

from collections import defaultdict

L = [['Patriots', 'Giants', '3', '1'], ['Steelers', 'Patriots', '1', '2'], ['Giants', 'Steelers', '3', '5']]

D = defaultdict(int)

for match in L:
    team1, team2, score1, score2 = match
    D[team1] # make sure the team exist in the dict even if it never wins a match
    D[team2] # make sure the team exist in the dict even if it never wins a match
    if int(score1) > int(score2):
        D[team1] += 1
    if int(score2) > int(score1):
        D[team2] += 1

如果您绝对需要，可以轻松地将D转换为列表...

Answer 3

或者，您可以使用Counter，类似于dict：

import collections as ct

L = [
    ['Patriots', 'Giants', '3', '1'], 
    ['Steelers', 'Patriots', '1', '2'], 
    ['Giants', 'Steelers', '3', '5'],
    ['Giants', 'Patriots', '1', '1']                       # tie  
]    

def count_wins(games):
    """Return a counter of team wins, given a list of games."""
    c = ct.Counter()                                        
    for team1, team2, s1, s2 in games:
        c[team1] += 0
        c[team2] += 0
        if int(s1) == int(s2):
            continue
        elif int(s1) > int(s2):
            c[team1] += 1
        else:
            c[team2] += 1
    return c

season = count_wins(L)
season
# Counter({'Giants': 0, 'Patriots': 2, 'Steelers': 1})

后一个代码为新条目提供了默认的零增量并处理了关系：

L_tie = [['Cowboys', 'Packers', '3', '3']]
game = count_wins(L_tie)
game
# Counter({'Cowboys': 0, 'Packers': 0})

计数器有一些有用的方法可以找到顶级团队：

season.most_common(2)
# [('Patriots', 2), ('Steelers', 1)]

计数器很灵活。您可以轻松更新计数器：

season.update(game)
season
# Counter({'Cowboys': 0, 'Giants': 0, 'Packers': 0, 'Patriots': 2, 'Steelers': 1})

您还可以add (subtract and perform set operations with)其他计数器：

L_last = [['49ers', 'Raiders', '7', '10'], ['Packers', 'Patriots', '3', '7']] 
last_season = count_wins(L_last)
season + last_season
# Counter({'Patriots': 3, 'Raiders': 1, 'Steelers': 1})

更新：另请参阅this related answer了解Counter / generator表达式变体。

Answer 4

ll = [['Patriots', 'Giants', '3', '1'], ['Steelers', 'Patriots', '1', '2'], ['Giants', 'Steelers', '3', '5']]

teamStatus = {}

for l in ll:
  team1,team2,team1_score,team2_score = l
  if team1 not in teamStatus:
      teamStatus[team1] = 0
  if team2 not in teamStatus:    
      teamStatus[team2] = 0

  if int(team1_score) > int(team2_score):
    teamStatus[team1] += 1 
  else:
    teamStatus[team2] += 1

print(teamStatus)

<强> RESULT

{'Patriots': 2, 'Giants': 0, 'Steelers': 1}

计算橄榄球队使用嵌套列表获胜的次数

4 个答案: