python - 循环使几个列表的每个组合

Question

我有一个有点复杂，或者可能不复杂，但很长的问题，我将尝试将其提炼为基本部分，然后也许我可以从那里弄清楚。

所以我要做的是填写一份棒球名册。我有一个球员名单，然后根据位置分成几个其他名单。从这一点来看，我想用每一种可能的球员组合来填补名单。我已经编写了一个完成此操作的基本脚本，但由于我有一个很大的列表，简单地遍历每个组合并不理想。我只是用了一部分玩家来测试它，它仍然花了大约一个半小时。

一些球员有资格被安排在名单中的多个位置，因此，一些名字将出现在我从中提取名字的两个或更多列表中。我的第一个节省时间的尝试是检查每次迭代是否包含重复项，如果是，则跳过该播放器。

import os, csv, time

player_pool_csv = open('Available Player Pool.csv', 'r') player_pool = csv.reader(player_pool_csv) player_pool = list(player_pool)

roster = ['a','b','c','d','e','f','g','h']
# roster with characters is my solution to the TypeError: unhashable type: 'list' 
# when I check len(roster) against len(set(roster)) in the first loop

catcher_pool = []
first_pool = []
second_pool = []
third_pool = []
short_pool = []
of_pool = []

for player in player_pool:
    if 'C' in player[2]:
        catcher_pool.append(player)
    if '1B' in player[2]:
        first_pool.append(player)
    if '2B' in player[2]:
        second_pool.append(player)
    if '3B' in player[2]:
        third_pool.append(player)
    if 'SS' in player[2]:
        short_pool.append(player)
    if 'OF' in player[2]:
        of_pool.append(player)

start = time.time()

for catcher in catcher_pool:
roster[0] = catcher[0]
if len(roster) != len(set(roster)):
    continue
for first_baseman in first_pool:
    roster[1] = first_baseman[0]
    if len(roster) != len(set(roster)):
        continue
    for second_baseman in second_pool:
        roster[2]= second_baseman[0]
        if len(roster) != len(set(roster)):
            continue
        for third_baseman in third_pool:
            roster[3] = third_baseman[0]
            if len(roster) != len(set(roster)):
                continue
            for shortstop in short_pool:
                roster[4] = shortstop[0]
                if len(roster) != len(set(roster)):
                    continue
                for outfielder1 in of_pool:
                    roster[5] = outfielder1[0]
                    if len(roster) != len(set(roster)):
                        continue
                    for outfielder2 in of_pool:
                        roster[6] = outfielder2[0]
                        if len(roster) != len(set(roster)):
                            continue
                        for outfielder3 in of_pool:
                            roster[7] = outfielder3[0]
                            if len(roster) != len(set(roster)):
                                continue
                            print(roster)

end = time.time()
elapsed = end - start
print(elapsed)

当我运行这段代码时，它似乎在“for outfielder2”循环的某个时刻自行停止。我似乎无法确切地知道它停止时为什么会停止，以及如何修复它。

我知道您可能需要比这更多的信息，例如它从中拉出玩家的池中的内容，但如果事实证明没有必要，我不想一开始就过载这个问题。如果需要，我会把这些东西放进去。

知道我做错了什么，以及如何提高效率吗？谢谢。

编辑

好的，我将我的 player_pool 缩减为

['Ender Inciarte', '0.283', 'OF', '3900']
['A.J. Pollock', '0.304', 'OF', '4900']
['Jamie Romak', '0.349', 'OF', '2000']
['Adam Jones', '0.258', 'OF', '3700']
['Paul Goldschmidt', '0.343', '1B', '5600']
['Chris Davis', '0.306', '1B', '4300']
['Aaron Hill', '0.245', '2B/3B', '2400']
['Jimmy Paredes', '0.276', '1B/2B', '3000']
['Jake Lamb', '0.283', '3B', '3700']
['Manny Machado', '0.315', '3B', '4400']
['Welington Castillo', '0.31', 'C', '3800']
['Caleb Joseph', '0.266', 'C', '3200']
['Xander Bogaerts', '0.318', '3B/SS', '4300']
['Eugenio Suarez', '0.294', 'SS', '3000']

当我在所有 if len(roster) != len(set(roster)) 注释掉的情况下运行代码时，它会正确返回每个组合(我编写的每个计时器需要 48.3 秒进去)。但是如果我把 if len(roster) != len(set(roster)) 放回去，这是输出:

['Welington Castillo', 'Paul Goldschmidt', 'Aaron Hill', 'Jake Lamb', 'Xander Bogaerts', 'Ender Inciarte', 'A.J. Pollock', 'Jamie Romak']
['Welington Castillo', 'Paul Goldschmidt', 'Aaron Hill', 'Jake Lamb', 'Xander Bogaerts', 'Ender Inciarte', 'A.J. Pollock', 'Adam Jones']
['Welington Castillo', 'Paul Goldschmidt', 'Aaron Hill', 'Jake Lamb', 'Xander Bogaerts', 'Ender Inciarte', 'Jamie Romak', 'A.J. Pollock']
['Welington Castillo', 'Paul Goldschmidt', 'Aaron Hill', 'Jake Lamb', 'Xander Bogaerts', 'Ender Inciarte', 'Jamie Romak', 'Adam Jones']
0.03500199317932129

Answer 1

您可以通过在外循环中已经选择的玩家过滤内循环池来进行大量优化。这对于集合来说尤其容易。
您还可以为外场球员使用 itertools 中的组合，以避免另外三个循环。
结果如下所示:

import itertools as it
import time

player_pool = [['Ender Inciarte', '0.283', 'OF', '3900'],
               ['A.J. Pollock', '0.304', 'OF', '4900'],
               ['Jamie Romak', '0.349', 'OF', '2000'],
               ['Adam Jones', '0.258', 'OF', '3700'],
               ['Paul Goldschmidt', '0.343', '1B', '5600'],
               ['Chris Davis', '0.306', '1B', '4300'],
               ['Aaron Hill', '0.245', '2B/3B', '2400'],
               ['Jimmy Paredes', '0.276', '1B/2B', '3000'],
               ['Jake Lamb', '0.283', '3B', '3700'],
               ['Manny Machado', '0.315', '3B', '4400'],
               ['Welington Castillo', '0.31', 'C', '3800'],
               ['Caleb Joseph', '0.266', 'C', '3200'],
               ['Xander Bogaerts', '0.318', '3B/SS', '4300'],
               ['Eugenio Suarez', '0.294', 'SS', '3000']]

# create a player hash for later use, I hope player names are unique
player_hash = {p[0]: p for p in player_pool}

# use sets for the pools so that we can use set difference later on
catcher_pool = set()
first_pool = set()
second_pool = set()
third_pool = set()
short_pool = set()
of_pool = set()

# fill the pools only with player names
for player, stats in player_hash.items():
    if 'C' in stats[2]:
        catcher_pool.add(player)
    if '1B' in stats[2]:
        first_pool.add(player)
    if '2B' in stats[2]:
        second_pool.add(player)
    if '3B' in stats[2]:
        third_pool.add(player)
    if 'SS' in stats[2]:
        short_pool.add(player)
    if 'OF' in stats[2]:
        of_pool.add(player)

start = time.time()
playing = set()
all_rosters = []
roster = [None]*8

# create a little generator that only yields players from a pool
# which are currently not playing
def filtered(pool):
    for player in pool-playing:
        playing.add(player)
        yield player
        playing.remove(player)

with open('rosters.txt', 'w') as f:
    for catcher in filtered(catcher_pool):
        roster[0] = catcher
        for first_baseman in filtered(first_pool):
            roster[1] = first_baseman
            for second_baseman in filtered(second_pool):
                roster[2] = second_baseman
                for third_baseman in filtered(third_pool):
                    roster[3] = third_baseman
                    for shortstop in filtered(short_pool):
                        roster[4] = shortstop
                        for outfielders in it.combinations(of_pool-playing, 3):
                            roster[5:8] = outfielders
                            # append result to a list instead of printing
                            # this is a lot faster
                            all_rosters.append(roster[:])
                            # if our list is bigger than 1e6, write it to file
                            if len(all_rosters) > 10**6:
                                f.writelines(repr(roster)+'\n'
                                             for roster
                                             in all_rosters)
                                all_rosters = []
    # don't forget to write the last batch which did not reach 1e6
    f.writelines(repr(roster)+'\n' for roster in all_rosters)

end = time.time()
elapsed = end - start

# make sure no roster contains double names
assert all(len(set(roster)) == len(roster) == 8 for roster in all_rosters)

# make sure all rosters are unique
assert len(all_rosters) == len(set(tuple(roster) for roster in all_rosters))

print(len(all_rosters))
print(elapsed)

打印:

[a lot of rosters]
232
0.00205016136169

但结果是未排序的。我想这不是问题。如果要对它们进行排序，只需在它们生成后对 all_rosters 进行排序即可。