我是python的新手，并且努力学习。我有一项非常乏味的任务，我希望你能帮助我。
我正在组织220人的活动。全天有4项活动。这些活动由10人（22支队伍）组成。目标是团队改变的每项活动。所以，我想重新安排团队组成4次。这是为了确保人们尽可能多地了解同事。
在excel中，我在列A中有220个名称的列表。我在列B中为每个名称（1 -220）分配了一个数字。我随机导入了模块和pandas。我的第一步是导入excel文件，将带有Numbers的列B转换为列表并对它们进行洗牌。
在那一步之后我迷失了。从洗牌列表中我想创建22支球队。在第一次抽签之后，我想再次洗牌并重新创建22支球队，但条件是在第二组合中没有来自第一队组成的人。总计这个过程重复四次。
理想情况下，我想将结果导出回excel。
希望有人可以帮助我！
再一次，我正在学习，请分享思考过程:)。
感谢您的时间。
亲切的问候，
BAS
答案 0 :(得分：0)
首先，不要使用熊猫。这是工作的错误工具。请改用NumPy。这是一个代码：
In [1]: import numpy as np
In [2]: people = np.arange(1, 221, 1)
people
Out[2]: array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26,
27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39,
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52,
53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65,
66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78,
79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91,
92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104,
105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117,
118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130,
131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143,
144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156,
157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169,
170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182,
183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195,
196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208,
209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220])
In [3]: np.random.choice(people, 220, replace=False).reshape(22,10)
Out[3]: array([[ 20, 48, 164, 176, 135, 190, 67, 147, 203, 130],
[150, 177, 171, 141, 122, 127, 202, 172, 21, 219],
[167, 206, 84, 118, 163, 181, 34, 87, 116, 184],
[ 33, 26, 70, 119, 129, 191, 105, 69, 86, 217],
[ 94, 24, 146, 9, 31, 208, 179, 148, 57, 102],
[211, 170, 65, 195, 220, 61, 88, 187, 32, 7],
[182, 40, 144, 145, 198, 47, 193, 74, 44, 90],
[174, 121, 216, 63, 82, 38, 201, 113, 13, 66],
[180, 137, 214, 73, 75, 51, 80, 23, 71, 18],
[161, 115, 3, 157, 89, 79, 29, 68, 200, 8],
[142, 173, 98, 36, 133, 215, 138, 50, 53, 4],
[152, 101, 139, 54, 30, 108, 49, 213, 124, 83],
[ 76, 17, 64, 10, 56, 22, 128, 153, 158, 140],
[131, 11, 45, 192, 92, 166, 60, 37, 12, 156],
[104, 25, 6, 205, 212, 197, 77, 46, 199, 96],
[ 59, 19, 112, 132, 126, 159, 151, 207, 85, 109],
[ 42, 55, 204, 188, 185, 35, 62, 41, 27, 178],
[ 14, 194, 2, 186, 143, 78, 134, 103, 106, 110],
[100, 91, 99, 111, 72, 58, 15, 120, 136, 97],
[ 39, 81, 123, 149, 165, 169, 209, 175, 1, 117],
[189, 28, 95, 114, 162, 160, 196, 5, 93, 154],
[210, 16, 168, 218, 43, 107, 155, 125, 52, 183]])
请注意，Out[3]是一个包含22行和10列的矩阵（22个团队，每个团队10人）。您只需要在In [3]中运行代码，就像您拥有的任务一样多。所以基本上：
In [4]: tasks = {} # Tasks is a dictionary with task number as key, and teams array as value.
for i in np.range(1, 5, 1):
tasks['task_' + str(i)] = np.random.choice(people, 220, replace=False).reshape(22,10)
现在，您可以将存储在tasks中的结果返回到Excel文件中，然后从那里继续。如果您有任何疑虑，请告诉我。
如果您坚持使用Pandas，并且已经在df中加载了数据，那么您可以使用Pandas为每个任务返回每个团队的玩家姓名。此代码应该足以满足您的需求：
df.set_index('player_id', inplace=True) # Which is actually pointless because
# Pandas will automatically create a 0 indexed index when you read your data in.
# So you don't even need any player_id column to begin with. If that is the case,
# change In [2] above to people = np.arange(0, 220, 1), or simply, np.arange(220)
tasks = {}
for i in np.arange(1, 5, 1):
tasks['task_' + str(i)] = {}
j = 1
for team in np.random.choice(people, 220, replace=False).reshape(22,10):
tasks[i]['team_' + str(j)] = df.loc[team].values
j += 1
tasks将是嵌套的Dictionary，其结构如下：
tasks = {'task_1' : {'team_1' : [list_of_players],
'team_2' : [list_of_players],
...
'team_19' : [list_of_players],
'team_20' : [list_of_players]},
'task_1' : {'team_1' : [list_of_players],
'team_2' : [list_of_players],
...
'team_19' : [list_of_players],
'team_20' : [list_of_players]},
'task_1' : {'team_1' : [list_of_players],
'team_2' : [list_of_players],
...
'team_19' : [list_of_players],
'team_20' : [list_of_players]},
'task_1' : {'team_1' : [list_of_players],
'team_2' : [list_of_players],
...
'team_19' : [list_of_players],
'team_20' : [list_of_players]}}
希望这有帮助！
答案 1 :(得分：0)
这是一个蛮力算法 - 它取决于随机选择，如果失败，它只是再次尝试。
根本没有使用pandas，只是普通的列表和集合。
met变量from random import choice
class Person:
def __init__(self, name):
self.name = name
self.met = set()
def __repr__(self):
return self.name
class TooManyIteration(Exception):
pass
GAMES = 4
TEAMS = 22
TEAMSIZE = 10
def pick_teams(s):
s_ = s.copy()
teams = [set() for _ in range(TEAMS)]
for t in teams:
c = 0
while len(t) < TEAMSIZE:
p = choice(tuple(s_))
s_.remove(p)
if all(p not in x.met for x in t):
t.add(p)
else:
s_.add(p)
c += 1
if c > len(s):
# failed to pick
raise TooManyIteration()
for t in teams:
for x in t:
x.met.update(t)
return teams
s = set(Person(str(i)) for i in range(220))
# s = set(Person(name) for name in LIST_OF_NAMES)
games = []
while len(games) < GAMES:
try:
games.append(pick_teams(s))
except TooManyIteration:
continue
print(games)