如何在 Python 上创建具有预算约束和多个条件的 N 项列表

Question

I have the following df of Premier League players (ROI_top_players):

    player                 team         position    cost_2223   total_points  ROI   
0   Mohamed Salah          Liverpool    FWD         13.0        259           29.77 
1   Trent Alexander        Liverpool    DEF         8.4         206           24.52 
2   Jarrod Bowen           West Ham     MID         8.5         204           23.56
3   Kevin De Bruyne        Man  City    MID         12.0        190           15.70
4   Virgil van Dijk        Liverpool    DEF         6.5         183           14.91 
... ... ... ... ... ... ... ... ... ...
151 Jamaal Lascelles       Newcastle    DEF         4.5         45            10.22
152 Ben Godfrey            Everton      GKP         4.5         45            9.57  
153 Aaron Wan-Bissaka      Man Utd      DEF         4.5         41            8.03
154 Brandon Williams       Norwich      DEF         4.0         36            7.23  

I want to create a list of 15 players (must be 15 - not more, not less), with the highest ROI possible, and it has to fulfill certain conditions:

• Position constraints : it must have 2 GKP, 5 DEF, 5 MID, and 3 FWD
• Budget constraint : I have a budget of $100, so for each player I add to the list, I must subtract the player's cost (cost_2223) from the budget.
• Team constraint : It can't have more than 3 players per club.

Here's my current code:

def get_ideal_team_ROI(budget = 100, star_player_limit = 3, gk = 2, df = 5, md = 5, fwd = 3):
    money_team = []
    budget = budget
    positions = {'GK': gk, 'DEF': df, 'MID': md, 'FWD': fwd}
    for index, row in ROI_top_players.iterrows():
       if (budget >= row['cost_2223'] and positions[row['position']] > 0):
           money_team.append(row['player'])
           budget -= row['cost_2223']
           positions[row['position']] = positions[row['position']] - 1
    return money_team

This code has two problems:

1. It creates the list BUT, the list does not end up with 15 players.
2. It doesn't fulfill the team constraint (I have more than 3 players per team).

How should I tackle this? I want my code to make sure that I always have enough budget to buy 15 players and that I always have at maximum 3 players per team.

**I do not need all possible combinations. Just ONE team with the highest possible ROI.

Answer 1

This type of question pops up now and again, and it has been solved a couple of times. For example, you have the following (not a duplicate per se, but close enough to get inspired from): Python PuLP Optimization problem for Fantasy Football, how to add Certain Conditional Constraints? And you have this Medium article: https://medium.com/ml-everything/using-python-and-linear-programming-to-optimize-fantasy-football-picks-dc9d1229db81 And this: https://towardsdatascience.com/how-to-build-a-fantasy-premier-league-team-with-data-science-f01283281236

Now, mostly in an attempt to understand it myself, I will attempt a solution (this is just standing on the above solutions, nothing uniquely brilliant here). As OP did not provide the data, I went and scraped the first 'Fantasy Football players list' I could find. There is no ROI in that data, however there are 'Points', which we will try to maximize, so I guess OP can apply this to maximize the ROI in his data.

from selenium import webdriver
from selenium.webdriver.chrome.service import Service
from selenium.webdriver.chrome.options import Options
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
import time as t
import pandas as pd
from pulp import *

## get some data approximating OP's data
chrome_options = Options()
chrome_options.add_argument("--no-sandbox")

webdriver_service = Service("chromedriver/chromedriver") ## path to where you saved chromedriver binary
browser = webdriver.Chrome(service=webdriver_service, options=chrome_options)

big_df = pd.DataFrame()

url = 'https://fantasy.premierleague.com/player-list/'
browser.get(url)
try:
    WebDriverWait(browser, 20).until(EC.element_to_be_clickable((By.XPATH, "//button[text()='Accept All Cookies']"))).click()
    print('cookies accepted')
except Exception as e:
    print('no cookies for you!')
tables_divs = WebDriverWait(browser, 20).until(EC.presence_of_all_elements_located((By.XPATH, "//table/parent::div/parent::div")))
for t in tables_divs:
    category = t.find_element(By.TAG_NAME, 'h3')
    print(category.text)
    WebDriverWait(t, 20).until(EC.presence_of_all_elements_located((By.XPATH, "//table")))
    dfs = pd.read_html(t.get_attribute('outerHTML'))
    for df in dfs:
        df['Type'] = category.text
        big_df = pd.concat([big_df, df], axis=0, ignore_index=True)
big_df.to_json('f_footie.json')
browser.quit()
footie_df = pd.read_json('f_footie.json')
footie_df.columns = ['Player', 'Team', 'Points', 'Cost', 'Position']
footie_df['Player'] = footie_df.apply( lambda row: row.Player.replace(' ', '_').strip(), axis=1)
footie_df['Cost'] = footie_df.apply( lambda row: row.Cost.split('£')[1], axis=1)
footie_df['Cost'] = footie_df['Cost'].astype('float')
footie_df['Points'] = footie_df['Points'].astype('int')
print(footie_df)
## constraining variables
positions = footie_df.Position.unique()
clubs = footie_df.Team.unique()
budget = 100
available_roles = {
    'Goalkeepers': 2,
    'Defenders': 5,
    'Midfielders': 5,
    'Forwards': 3    
}

names = [footie_df.Player[i] for i in footie_df.index]
teams = [footie_df.Team[i] for i in footie_df.index]
roles = [footie_df.Position[i] for i in footie_df.index]
costs = [footie_df.Cost[i] for i in footie_df.index]
points = [footie_df.Points[i] for i in footie_df.index]
players = [LpVariable("player_" + str(i), cat="Binary") for i in footie_df.index]
prob = LpProblem("Secret Fantasy Player Choices", LpMaximize)
## define the objective -> maximize the points
prob += lpSum(players[i] * points[i] for i in range(len(footie_df)))
## define budget constraint
prob += lpSum(players[i] * footie_df.Cost[footie_df.index[i]] for i in range(len(footie_df))) <= budget

for pos in positions:
  prob += lpSum(players[i] for i in range(len(footie_df)) if roles[i] == pos) <= available_roles[pos]
## add max 3 per team constraint
for club in clubs:
  prob += lpSum(players[i] for i in range(len(footie_df)) if teams[i] == club) <= 3
prob.solve()
df_list = []
for variable in prob.variables():
  if variable.varValue != 0:
    name = footie_df.Player[int(variable.name.split("_")[1])]
    club = footie_df.Team[int(variable.name.split("_")[1])]
    role = footie_df.Position[int(variable.name.split("_")[1])]
    points = footie_df.Points[int(variable.name.split("_")[1])]
    cost = footie_df.Cost[int(variable.name.split("_")[1])]
    df_list.append((name, club, role, points, cost))
    

#     print(name, club, position, points, cost)
result_df = pd.DataFrame(df_list, columns = ['Name', 'Club', 'Role', 'Points', 'Cost'])
result_df.to_csv('win_at_fantasy_football.csv')
print(result_df)

This will display some control printouts, the data scraped, the long printout from pulp solver, and the result dataframe in the end, looking like this:

	Name	Club	Role	Points	Cost
0	Alisson	Liverpool	Goalkeepers	176	5.5
1	Lloris	Spurs	Goalkeepers	158	5.5
2	Bowen	West Ham	Midfielders	206	8.5
3	Saka	Arsenal	Midfielders	179	8
4	Maddison	Leicester	Midfielders	181	8
5	Ward-Prowse	Southampton	Midfielders	159	6.5
6	Gallagher	Chelsea	Midfielders	140	6
7	Antonio	West Ham	Forwards	140	7.5
8	Toney	Brentford	Forwards	139	7
9	Mbeumo	Brentford	Forwards	119	6
10	Alexander-Arnold	Liverpool	Defenders	208	7.5
11	Robertson	Liverpool	Defenders	186	7
12	Cancelo	Man City	Defenders	201	7
13	Gabriel	Arsenal	Defenders	146	5
14	Cash	Aston Villa	Defenders	147	5

In case there are better solutions, please, whoever is more seasoned in optimisations problems, chime in with critique.