Python Pulp-唯一团队约束数

问题描述

我是Pulp的新手，因此在尝试进行条件约束时遇到了问题。我制作了一个幻想足球优化器，可以选择9名球员的最佳选择，我的求解器目前可以在位置限制，薪水限制等条件下完全正常工作。

我需要添加的最后一件事是一个约束，这使得它从它挑选的9名球员中脱颖而出，需要有8个唯一的球员球队名称。例如：在我的代码###Stack QB with 2 teammates中，鉴于此约束，四分卫和WR / TE将在同一个团队中。因此，其他所有人都应该位于不同的团队中，以拥有8个唯一的团队名称。

下面是我试图用来制作此约束的代码，excel文件的头部已优化，并且到目前为止我的代码在没有约束的情况下仍能正常工作，我想在选择的9个球员中添加8个唯一的团队名称

我目前已经尝试过此方法，但是它不起作用！非常感谢您的帮助！

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()

解决方法

以线性编程术语

如果选择了x_i = 1播放器，让i^th，否则选择0，i = 1....I。假设t_i是i^th玩家的团队，这是一个常量。
假设t_j是j^th唯一的团队，也是一个不变的j = 1....T。
如果t_{ij} = 1，则让t_i == t_j，否则为0。这也是一个常数。

那么您可以说，从t_j团队中选出的球员总数为(t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I)，从逻辑上来说，其取值范围为0到I。

现在，如果任何选定的球员来自团队y_j = 1，则可以让二进制变量t_j进入，否则为0，就像这样：

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

这给您以下情况：

如果(t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) = 0，则y_j为0；
如果(t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij}*x_I) > 0，则y_j可以为0或1。

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

以PULP术语（类似这样，无法对其进行测试）

如果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','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

linear-programming optimization optimization pandas pandas pulp python