考虑以下列表中的4个元组:
players_score = [ ('Joe', 100, 34, 38, 90, 67, 3, 10),
('Bob', 90, 38, 4, 100, 60, 4, 11),
('May', 80, 36, 40, 91, 70, 2, 12),
('Anna', 95, 32, 36, 92, 68, 8, 13) ]球员们已经打了7场比赛了。在第一场比赛中,乔以100分获胜。
我想根据以下内容给每个玩家(每一场比赛)打分:
First/best player: 5 points
Second player: 3 points
Third player: 1 point
Fourth player: -1 point -> But 0 points is what I like to assign the fourth player of each game.到目前为止,我的代码如下:
from itertools import zip_longest as y
zipped = list(y(*[game_score for _, *game_score in players_score]))
tup = list(sorted(i, reverse=True) for i in zipped)
score_dict = {}
for player, *game_score in players_score:
tmp = []
for i,j in zip(game_score, tup):
tmp.append(5-2*j.index(i))
tot_score = sum(tmp)
score_dict[player] = tot_score
print("The overall scores are: ", sorted(score_dict.items(), key=lambda kv: (kv[1], kv[0]), reverse=True))因此,我的代码将-1点应用于每个游戏的第四个玩家,但我希望第四个玩家只获得0个零点。我正在努力应用一个范围(0:3)或其他方法,通过这些方法,我可以跳过第四个玩家的得分,因为那个玩家只获得0分。
发布于 2019-12-17 02:51:58
我会以不同的方式设置它。也就是说,有不止一种方法可以做到这一点。我要做的第一件事是把players_score列表变成一个更漂亮的集合,就像字典一样--这样工作起来就更有趣了。
基本上,我们迭代我们的“轮次”(游戏),并根据玩家在本轮中的得分对玩家姓名进行排序(本轮的获胜者排在第一位)。然后,我们用他们应得的奖励(分数)对排序后的球员姓名进行zip,并使用球员姓名作为关键字来更新我们的collections.Counter集合。
from collections import Counter
players_score = [
("Joe", 100, 34, 38, 90, 67, 3, 10),
("Bob", 90, 38, 4, 100, 60, 4, 11),
("May", 80, 36, 40, 91, 70, 2, 12),
("Anna", 95, 32, 36, 92, 68, 8, 13)
]
original_scores = {player: scores for player, *scores in players_score}
def iter_score(player):
for score in original_scores[player]:
yield player, score
score_rewards = [5, 3, 1, 0]
overall_scores = Counter()
for current_round in zip(*map(iter_score, original_scores)):
sorted_players = map(lambda tpl: tpl[0], sorted(current_round, key=lambda tpl: tpl[1], reverse=True))
for player, score in zip(sorted_players, score_rewards):
overall_scores.update({player: score})
print("The overall scores:")
for player, score in overall_scores.most_common():
print(f"{player}: {score}")输出:
The overall scores:
Anna: 20
May: 17
Bob: 15
Joe: 11发布于 2019-12-17 02:22:57
您可以更改评分函数:
tmp.append(5-2*j.index(i) if 5-2*j.index(i) > 0 else 0)因此,如果存在负值,则将分数设置为0
发布于 2019-12-17 02:23:02
一个非常简单的解决方案
for i,j in zip(game_score, tup):
score_var = 5-2*j.index(i)
if (score_var < 0):
score_var = 0
tmp.append(score_var)这将确保没有玩家会获得负分。
https://stackoverflow.com/questions/59362077
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