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假设我有一个球员名单

listOfPlayers = [
                 ("Player1","PG",Cost,projectedPoints),
                 ("Player2","PG",Cost,projectedPoints),
                 ("Player3","SG",Cost,projectedPoints),
                 ("Player4","SG",Cost,projectedPoints),
                 ("Player5","SF",Cost,projectedPoints),
                 ("Player6","SF",Cost,projectedPoints),
                 ("Player7","PF",Cost,projectedPoints),
                 ("Player8","PF",Cost,projectedPoints),
                 ("Player9","C",Cost,projectedPoints),
                 ("Player10","C",Cost,projectedPoints) 
                ]

然后是一个空列表

selectedList = []

我需要按位置对球员进行排序，并将最佳成本/预计分数比拉到所选列表中，但问题是我每个位置只能有 1 个，例如 (pg, sg, sf, pf, c) . 我该怎么做？

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4 回答 4

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首先按成本/预计点数比排序：

listOfPlayers.sort(key=lambda (name, pos, cost, pts): float(cost)/pts, reverse=True)

然后遍历列表并将它们拉出：

results = {}
for name, pos, cost, pts in listOfPlayers:
    if pos in results:
        #position already filled
        continue 
    results[pos] = (name, pos, cost, pts)

您可能希望提前停止，具体取决于您是否提前知道需要哪些职位，例如：

needed_positions = set(["PG", "SG", "SF", "C"])
results = {}
for name, pos, cost, pts in listOfPlayers:
    if not needed_positions:
        #all spots filled
        break
    if pos in results:
        #this position already filled
        continue 
    results[pos] = (name, pos, cost, pts)
    needed_positions.remove(pos)
于 2013-10-15T20:45:47.263 回答
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players.sort(operator.itemgetter(1)) # sort by position
for pos, players in itertools.groupby(players, key=operator.itemgetter(1)):
  best = max(players, key=lambda p:float(p[2])/p[3])
  print "The best player for position %s is %s" %(best[1], best[0])
于 2013-10-15T20:48:17.727 回答
0 
import itertools

players = [
    # name          position    cost    projected_points
    ("Player1",     "PG",       1,      13),
    ("Player2",     "PG",       2,      54),
    ("Player3",     "SG",       4,      44),
    ("Player4",     "SG",       8,      34),
    ("Player5",     "SF",       3,      14),
    ("Player6",     "SF",       2,      91),
    ("Player7",     "PF",       10,     20),
    ("Player8",     "PF",       4,      32),
    ("Player9",     "C",        9,      12),
    ("Player10",    "C",        3,      82)
]

def cost(p):
    return float(p[3]) / p[2]

players.sort(reverse=True, key=lambda x: (x[1], cost(x)))

for k, g in itertools.groupby(players, lambda x: x[1]):
    best = list(g)[0]
    print(best)
于 2013-10-15T20:57:38.130 回答
0

如果您有兴趣使用Pandas DataFrame对象执行此操作，则此代码可以工作。如果您最终需要经常或对较大的数据执行此类排序/聚合操作，这将很有帮助——它使用高效的 NumPy 数组操作来快速执行计算。

df = pandas.DataFrame(listOfPlayers, 
                      columns=["Player", "Pos", "Cost", "ProjectedPoints"])

df['Ratio'] = df.Cost / df.ProjectedPoints
best_players = df.groupby("Pos").apply(lambda x: x.Player.values[x.Ratio.argmax()])
于 2013-10-15T20:58:43.980 回答