pulp constraint: exactly N in a category, up to X of a second choice, same category

Question

I have a problem I’m trying to solve for where I want N players from one team, and up to X players from a second team, but I don’t particularly care which team fills those constraints. For example, if N=5 and X=2, I could have 5 from one team and up to 2 from a second, different, team. How would I write such a constraint?

example dataframe:

	team	pos	name	ceil	salary
0	NYY	OF	Aaron Judge	21.6631	6500
1	HOU	OF	Yordan Alvarez	21.6404	6100
2	ATL	OF	Ronald Acuna Jr.	21.5363	5400
3	HOU	OF	Kyle Tucker	20.0992	4700
4	TOR	1B	Vladimir Guerrero Jr.	20.0722	6000
5	LAD	SS	Trea Turner	20.0256	5700
6	LAD	OF	Mookie Betts	19.5231	6300
7	SEA	OF	Julio Rodriguez	19.3694	5200
8	MIN	OF	Byron Buxton	19.3412	5600
9	LAD	1B	Freddie Freeman	19.3393	5600
10	TOR	OF	George Springer	19.1429	5100
11	NYM	OF	Starling Marte	19.0791	5200
12	ATL	1B	Matt Olson	19.009	4800
13	ATL	3B	Austin Riley	18.9091	5200
14	SF	OF	Austin Slater	18.9052	3700
15	NYM	1B	Pete Alonso	18.8921	5700
16	TEX	OF	Adolis Garcia	18.7115	4200
17	TEX	SS	Corey Seager	18.6957	5100
18	TOR	OF	Teoscar Hernandez	18.6834	5200
19	CWS	1B	Jose Abreu	18.497	4600
20	ATL	SS	Dansby Swanson	18.4679	4900
21	TEX	2B/SS	Marcus Semien	18.4389	4100
22	NYY	1B	Anthony Rizzo	18.4383	5300
23	NYY	2B	Gleyber Torres	18.39	4500
24	CHC	C	Willson Contreras	18.3452	5800

existing code snippet:

        #problem definition
        prob = LpProblem(name="DFS", sense=LpMaximize)
        prob += lpSum(player_vars[i] * slate['ceil'].iloc[i] for i in player_ids), "FPTS"
        
        #salary and total player constraints
        prob += lpSum(player_vars[i] * slate['salary'].iloc[i] for i in player_ids) <= 50000, "Salary"
        prob += lpSum(player_vars[i] for i in player_ids) == 10, "Total Players"
        
        #position constraints
        prob += lpSum(player_vars[i] for i in player_ids if slate['pos'].iloc[i] == 'P') == 2, "Pitcher"
        prob += lpSum([player_vars[i] for i in player_ids if slate['name'].iloc[i] in slate['name'][slate['pos'].str.contains('C')].to_list()]) == 1, "Catcher"
        prob += lpSum([player_vars[i] for i in player_ids if slate['name'].iloc[i] in slate['name'][slate['pos'].str.contains('1B')].to_list()]) == 1, "1B"
        prob += lpSum([player_vars[i] for i in player_ids if slate['name'].iloc[i] in slate['name'][slate['pos'].str.contains('2B')].to_list()]) == 1, "2B"
        prob += lpSum([player_vars[i] for i in player_ids if slate['name'].iloc[i] in slate['name'][slate['pos'].str.contains('3B')].to_list()]) == 1, "3B"
        prob += lpSum([player_vars[i] for i in player_ids if slate['name'].iloc[i] in slate['name'][slate['pos'].str.contains('SS')].to_list()]) == 1, "SS"
        prob += lpSum([player_vars[i] for i in player_ids if slate['name'].iloc[i] in slate['name'][slate['pos'].str.contains('OF')].to_list()]) == 3, "OF"
        
       
        #no opposing pitcher constraint
        for pid in player_ids:
            if slate['pos'].iloc[pid] == 'P':
                prob += lpSum([player_vars[i] for i in player_ids if 
                    slate['team'].iloc[i] == slate['opp'].iloc[pid]] + [9 * player_vars[pid]]) <= 9, "P{pid}".format(pid=pid)

        #three team max constraint
        unique_teams = slate['team'].unique()
        player_in_team = slate['team'].str.get_dummies()        
        team_vars = LpVariable.dicts('team', unique_teams, cat = 'Binary')
        
        for team in unique_teams:
            prob += lpSum([player_in_team[team][i] * player_vars[i] for i in player_ids if slate['pos'].iloc[i] != 'P']) >= team_vars[team], "Team{team}Min".format(team=team)
            
        
        prob += lpSum(team_vars[team] for team in unique_teams) == 3, "3 Teams"

Asked By: Rory Burke

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Source

Answer 1

Here’s how I would attack this… In pseudocode….

make a set of teams that you can use to index a couple of new variables.
Make subsets of your players grouped by team, or use pandas data frame filters to limit summations of players to the team of interest.
Make 2 new variables, that are binary "indicator" variables, one call it use5from[team] and one called use[team] to indicate that the team has been used at all.
Make appropriate constraints to link those to the selection variables. Something like:

for team in teams:
    5 * use5from[team] <= pulp.lpSum(x[i] for i in team[i])

And for the other, a constraint to indicate any use…

for team in teams:
    use[team] <= pulp.lpSum(x[i] for i in team[I])

And then make constraints that those two variables sum to over 1 and 3 respectively.

Answered By: AirSquid

pulp constraint: exactly N in a category, up to X of a second choice, same category

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