Python Pulp - 独特团队数量限制

Question

我是 Pulp 的新手，因此在尝试设置条件约束时遇到了问题。我制作了一个 Fantasy Football 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可以处理位置限制、工资限制等。

我需要添加的最后一件事是一个约束，使得在它选择的 9 名球员中，需要有 8 个唯一的球员团队名称。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，其他人应该属于不同的团队，以便拥有 8 个唯一的团队名称。

下面是我尝试用来进行此约束的代码，正在优化的 Excel 文件的头部和我的代码，到目前为止，没有我想在选定的 9 名球员中添加 8 个唯一团队名称的约束.

我目前已经尝试过这个，但它不起作用!非常感谢任何帮助!

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

用线性规划术语

如果选择了i^th个玩家，则令x_i = 1，否则为0，i = 1....I。
令t_i为第i^th个玩家的队伍，这是一个常量。
令 t_j 为第 j^th 个唯一团队，也是一个常数，j = 1....T。
如果t_i == t_j，则令t_{ij} = 1，否则为0。这也是一个常数。

那么你可以说从球队t_j中选出的球员总数为(t_{1j}*x_1 + t_{1j}*x_2 + ... + t_{Ij }*x_I)，逻辑上取 0 到 I 之间的值。

---

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

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

这会给你带来以下情况:

• 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 对于至少 8 个不同的团队t_j。

---

用 PULP 术语(类似这样的东西，无法测试它)

如果player_vars是一个相当于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