Python Pulp - 独特团队的数量约束

Question

我是 Pulp 的新手，因此在尝试进行条件约束时遇到了问题。我制作了一个梦幻足球优化器，可以选择 9 名球员的最佳选择，我的求解器目前完全适用于职位限制、工资限制等。
我需要添加的最后一件事是一个约束，使它在它选择的 9 名球员中，需要有 8 名球员的唯一球队名称。例如:在我的代码 ###Stack QB with 2 teammates 中，鉴于此限制，有一个四分卫和一个 WR/TE 将在同一个团队中.因此，其他所有人都应该在不同的团队中，才能拥有 8 个唯一的团队名称。
下面是我试图用来做这个约束的代码，正在优化的 excel 文件的头部和我的代码到目前为止没有约束我想在选定的 9 名球员中添加 8 个唯一的球队名称。
我目前已经尝试过这个，但它不起作用!真的很感激任何帮助!

list_of_teams = raw_data['Team'].unique()
team_vars = pulp.LpVariable.dicts('team', list_of_teams, cat = 'Binary')

for team in list_of_teams:
  prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Team'][i] == team] + [-9*team_vars[team]]) <= 0

prob += pulp.lpSum([team_vars[t] for t in list_of_teams]) >= 8

file_name = 'C:/Users/Michael Arena/Desktop/Football/Simulation.csv'
raw_data = pd.read_csv(file_name,engine="python",index_col=False, header=0, delimiter=",", quoting = 3)


player_ids = raw_data.index
player_vars = pulp.LpVariable.dicts('player', player_ids, cat='Binary')

prob = pulp.LpProblem("DFS Optimizer", pulp.LpMaximize)

prob += pulp.lpSum([raw_data['Projection'][i]*player_vars[i] for i in player_ids])

##Total Salary upper:
prob += pulp.lpSum([raw_data['Salary'][i]*player_vars[i] for i in player_ids]) <= 50000

##Total Salary lower:
prob += pulp.lpSum([raw_data['Salary'][i]*player_vars[i] for i in player_ids]) >= 49900

##Exactly 9 players:
prob += pulp.lpSum([player_vars[i] for i in player_ids]) == 9

##2-3 RBs:
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'RB']) >= 2
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'RB']) <= 3

##1 QB:
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'QB']) == 1
##3-4 WRs:
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'WR']) >= 3
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'WR']) <= 4

##1-2 TE's:
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'TE']) >= 1
# prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'TE']) <= 2

##1 DST:
prob += pulp.lpSum([player_vars[i] for i in player_ids if raw_data['Position'][i] == 'DST']) == 1


###Stack QB with 2 teammates
for qbid in player_ids:
    if raw_data['Position'][qbid] == 'QB':
        prob += pulp.lpSum([player_vars[i] for i in player_ids if 
                          (raw_data['Team'][i] == raw_data['Team'][qbid] and 
                            raw_data['Position'][i] in ('WR', 'TE'))] + 
                            [-1*player_vars[qbid]]) >= 0

###Don't stack with opposing DST:
for dstid in player_ids:
    if raw_data['Position'][dstid] == 'DST':
        prob += pulp.lpSum([player_vars[i] for i in player_ids if
                            raw_data['Team'][i] == raw_data['Opponent'][dstid]] +
                            [8*player_vars[dstid]]) <= 8



###Stack QB with 1 opposing player:
for qbid in player_ids:
    if raw_data['Position'][qbid] == 'QB':
        prob += pulp.lpSum([player_vars[i] for i in player_ids if
                            (raw_data['Team'][i] == raw_data['Opponent'][qbid] and 
                            raw_data['Position'][i] in ('WR', 'TE'))]+
                            [-1*player_vars[qbid]]) >= 0


prob.solve()

Answer 1

在线性规划术语中
让 x_i = 1如果i^th玩家被选中，否则为 0，i = 1....I .
让 t_i成为i^th的团队播放器，这是一个常数。
让 t_j是 j^th独特的团队，也是不变的，j = 1....T .
并让 t_{ij} = 1如果 t_i == t_j , 否则为 0。这也是一个常数。
那么你可以说从球队中选出的球员总数t_j是 (t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) ，它在逻辑上取一个介于 0 和 I 之间的值。

现在，您可以让二进制变量 y_j = 1如果任何选定的球员来自团队t_j , 否则为 0，如下所示:

(t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) >= y_j

这为您提供了以下情况:

如 (t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) = 0 ，然后 y_j是 0；

如 (t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) > 0 ，然后 y_j可以是 0 或 1。

现在，如果您添加约束 (y_1 + y_2 + ... + y_T) >= 8 ，这意味着 (t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) > 0至少 8 个不同的团队 t_j .

在 PuLP 方面 (像这样的东西，无法对其进行测试)
如 player_vars是等效于 x_i 的二进制变量

teams = raw_data['Team']  # t_i
unique_teams = teams.unique()  # t_j
player_in_team = teams.str.get_dummies()  # t_{ij}

# Example output for `teams = pd.Series(['A', 'B', 'C', 'D', 'E', 'F', 'A', 'C', 'E'])`:
#    A  B  C  D  E  F
# 0  1  0  0  0  0  0
# 1  0  1  0  0  0  0
# 2  0  0  1  0  0  0
# 3  0  0  0  1  0  0
# 4  0  0  0  0  1  0
# 5  0  0  0  0  0  1
# 6  1  0  0  0  0  0
# 7  0  0  1  0  0  0
# 8  0  0  0  0  1  0

team_vars = pulp.LpVariable.dicts('team', unique_teams, cat='Binary')  # y_j

for team in unique_teams:
  prob += pulp.lpSum(
      [player_in_team[team][i] * player_vars[i] for i in player_ids]
  ) >= team_vars[team]

prob += pulp.lpSum([team_vars[t] for t in unique_teams]) >= 8