这是我正在使用的数据：
structure(list(Name = c("Jokic", "Butler", "Murray", "Adebayo",
"Porter", "Gordon", "Martin", "Pope", "Vincent", "Lowry", "Brown",
"Strus", "Robinson", "Green", "Highsmith"), Points = c(62.8,
48.8, 45.8, 41.8, 35.3, 30.3, 29.3, 23.8, 23.3, 22.3, 21.8, 19,
16.3, 8.5, 6.8), Cost = c(14000, 13400, 10800, 9200, 7600, 6600,
7400, 5600, 5800, 5200, 6200, 4800, 4200, 2200, 1400)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -15L))
这是我当前的代码：
library(readxl)
library(tidyverse)
library(lpSolveAPI)
# Read the data from Excel
data <- read_excel("C:/Users/M0185JN/Downloads/NBA_1.xlsx")
data$cpt_points <- 1.5*data$Points
data$cpt_cost <- 1.5*data$Cost
#Players
num_players <- rep(1,nrow(data))
num_captain <- rep(1,nrow(data))
# Define the objective function coefficients
obj <- data$Points*num_players + data$cpt_points*num_captain
# Create a new LP model
lprec <- make.lp(nrow(data), nrow(data))
# Set the optimization direction to maximize
lp.control(lprec, sense = "maximize")
# Set the objective function coefficients
set.objfn(lprec, obj)
# Set type of decision variables
set.type(lprec, 1:nrow(data), type = "binary")
# Constraint: Pick exactly 5 players
add.constraint(lprec, num_players, "=", 5)
add.constraint(lprec, num_captain, "=", 1)
# Constraint: Total cost must be less than or equal to 50,000
add.constraint(lprec, data$Cost*num_players + data$cpt_cost*num_captain, "<=", 50000)
# Constraint: No Duplicate Players
add.constraint(lprec, num_players + num_captain, "<=", 1)
# Solve the linear programming problem
solve(lprec)
# Get the solution status
status <- get.solutioncount(lprec)
# Check if a solution was found
if (status > 0) {
# Retrieve the values of the decision variables
player_picked <- get.variables(lprec)
# Create a data frame with the players and their corresponding picked status
result <- data.frame(Name = data$Name, Picked = player_picked)
# Filter the data frame to show only the players that were picked (Picked = 1)
picked_players <- result[result$Picked == 1, ]
# Print the picked players
print(picked_players)
} else {
print("No feasible solution found.")
}
代码继续给我一个“找不到可行的解决方案”。当我知道确实存在的时候。以下是更好解释的限制：
一些可疑的东西，比如“数量”变量，它们可能应该是指派向量，其中队长不被视为玩家。以下成功，但我还没有将其翻译回 R。
import pandas as pd
import pulp
index = pd.Index(name='Name', data=(
'Jokic', 'Butler', 'Murray', 'Adebayo',
'Porter', 'Gordon', 'Martin', 'Pope', 'Vincent', 'Lowry', 'Brown',
'Strus', 'Robinson', 'Green', 'Highsmith',
))
df = pd.DataFrame(
index=index,
data={
'Points': (
62.8, 48.8, 45.8, 41.8, 35.3, 30.3, 29.3,
23.8, 23.3, 22.3, 21.8, 19, 16.3, 8.5, 6.8,
),
'Cost': (
14000, 13400, 10800, 9200, 7600, 6600, 7400,
5600, 5800, 5200, 6200, 4800, 4200, 2200, 1400,
),
'Captain': pulp.LpVariable.matrix(
name='captain', indices=index, cat=pulp.LpBinary),
'Player': pulp.LpVariable.matrix(
name='player', indices=index, cat=pulp.LpBinary),
},
)
prob = pulp.LpProblem(name='nba_team', sense=pulp.LpMaximize)
# person cannot be both a player and a captain; make these mutually exclusive
for name, row in df.iterrows():
prob.addConstraint(name=f'excl_{name}', constraint=row['Captain'] + row['Player'] <= 1)
# the total cost is the matrix product of costs with both the captain and player assignment vars
captain_cost, player_cost = df['Cost'] @ df[['Captain', 'Player']]
captain_coef = 1.5
total_cost = captain_coef*captain_cost + player_cost
prob.addConstraint(name='total_cost', constraint=total_cost <= 50_000)
# there are exactly four non-captain players
prob.addConstraint(name='n_players', constraint=df['Player'].sum() == 4)
# there is exactly one captain
prob.addConstraint(name='n_captains', constraint=df['Captain'].sum() == 1)
# maximize the number of points for the roster, with more weight to the captain
captain_points, player_points = df['Points'] @ df[['Captain', 'Player']]
prob.setObjective(captain_coef*captain_points + player_points)
print(prob)
prob.solve()
assert prob.status == pulp.LpStatusOptimal
captains = df.loc[
df['Captain'].apply(pulp.value) > 0.5,
['Points', 'Cost'],
]
players = df.loc[
df['Player'].apply(pulp.value) > 0.5,
['Points', 'Cost'],
]
print(f'{prob.objective.value():.1f} points for ${total_cost.value():,.2f}\n')
print('Captain:')
print(captains, end='\n\n')
print('Players:')
print(players)
nba_team:
MAXIMIZE
62.699999999999996*Adebayo_captain + 41.8*Adebayo_player + 32.7*Brown_captain + 21.8*Brown_player + 73.19999999999999*Butler_captain + 48.8*Butler_player + 45.45*Gordon_captain + 30.3*Gordon_player + 12.75*Green_captain + 8.5*Green_player + 10.2*Highsmith_captain + 6.8*Highsmith_player + 94.19999999999999*Jokic_captain + 62.8*Jokic_player + 33.45*Lowry_captain + 22.3*Lowry_player + 43.95*Martin_captain + 29.3*Martin_player + 68.69999999999999*Murray_captain + 45.8*Murray_player + 35.7*Pope_captain + 23.8*Pope_player + 52.949999999999996*Porter_captain + 35.3*Porter_player + 24.450000000000003*Robinson_captain + 16.3*Robinson_player + 28.5*Strus_captain + 19.0*Strus_player + 34.95*Vincent_captain + 23.3*Vincent_player + 0.0
SUBJECT TO
Jokic_excl: Jokic_captain + Jokic_player <= 1
Butler_excl: Butler_captain + Butler_player <= 1
Murray_excl: Murray_captain + Murray_player <= 1
Adebayo_excl: Adebayo_captain + Adebayo_player <= 1
Porter_excl: Porter_captain + Porter_player <= 1
Gordon_excl: Gordon_captain + Gordon_player <= 1
Martin_excl: Martin_captain + Martin_player <= 1
Pope_excl: Pope_captain + Pope_player <= 1
Vincent_excl: Vincent_captain + Vincent_player <= 1
Lowry_excl: Lowry_captain + Lowry_player <= 1
Brown_excl: Brown_captain + Brown_player <= 1
Strus_excl: Strus_captain + Strus_player <= 1
Robinson_excl: Robinson_captain + Robinson_player <= 1
Green_excl: Green_captain + Green_player <= 1
Highsmith_excl: Highsmith_captain + Highsmith_player <= 1
total_cost: 13800 Adebayo_captain + 9200 Adebayo_player + 9300 Brown_captain
+ 6200 Brown_player + 20100 Butler_captain + 13400 Butler_player
+ 9900 Gordon_captain + 6600 Gordon_player + 3300 Green_captain
+ 2200 Green_player + 2100 Highsmith_captain + 1400 Highsmith_player
+ 21000 Jokic_captain + 14000 Jokic_player + 7800 Lowry_captain
+ 5200 Lowry_player + 11100 Martin_captain + 7400 Martin_player
+ 16200 Murray_captain + 10800 Murray_player + 8400 Pope_captain
+ 5600 Pope_player + 11400 Porter_captain + 7600 Porter_player
+ 6300 Robinson_captain + 4200 Robinson_player + 7200 Strus_captain
+ 4800 Strus_player + 8700 Vincent_captain + 5800 Vincent_player <= 50000
n_players: Adebayo_player + Brown_player + Butler_player + Gordon_player
+ Green_player + Highsmith_player + Jokic_player + Lowry_player
+ Martin_player + Murray_player + Pope_player + Porter_player
+ Robinson_player + Strus_player + Vincent_player = 4
n_captains: Adebayo_captain + Brown_captain + Butler_captain + Gordon_captain
+ Green_captain + Highsmith_captain + Jokic_captain + Lowry_captain
+ Martin_captain + Murray_captain + Pope_captain + Porter_captain
+ Robinson_captain + Strus_captain + Vincent_captain = 1
VARIABLES
0 <= Adebayo_captain <= 1 Integer
0 <= Adebayo_player <= 1 Integer
0 <= Brown_captain <= 1 Integer
0 <= Brown_player <= 1 Integer
0 <= Butler_captain <= 1 Integer
0 <= Butler_player <= 1 Integer
0 <= Gordon_captain <= 1 Integer
0 <= Gordon_player <= 1 Integer
0 <= Green_captain <= 1 Integer
0 <= Green_player <= 1 Integer
0 <= Highsmith_captain <= 1 Integer
0 <= Highsmith_player <= 1 Integer
0 <= Jokic_captain <= 1 Integer
0 <= Jokic_player <= 1 Integer
0 <= Lowry_captain <= 1 Integer
0 <= Lowry_player <= 1 Integer
0 <= Martin_captain <= 1 Integer
0 <= Martin_player <= 1 Integer
0 <= Murray_captain <= 1 Integer
0 <= Murray_player <= 1 Integer
0 <= Pope_captain <= 1 Integer
0 <= Pope_player <= 1 Integer
0 <= Porter_captain <= 1 Integer
0 <= Porter_player <= 1 Integer
0 <= Robinson_captain <= 1 Integer
0 <= Robinson_player <= 1 Integer
0 <= Strus_captain <= 1 Integer
0 <= Strus_player <= 1 Integer
0 <= Vincent_captain <= 1 Integer
0 <= Vincent_player <= 1 Integer
Result - Optimal solution found
Objective value: 225.40000000
Enumerated nodes: 0
Total iterations: 0
Time (CPU seconds): 0.03
Time (Wallclock seconds): 0.04
Option for printingOptions changed from normal to all
Total time (CPU seconds): 0.04 (Wallclock seconds): 0.04
225.4 points for $50,000.00
Captain:
Points Cost
Name
Jokic 62.8 14000
Players:
Points Cost
Name
Adebayo 41.8 9200
Porter 35.3 7600
Gordon 30.3 6600
Pope 23.8 5600
如果事实证明您确实想要六名玩家，则将
45224.7 points for $50,000.00
Captain:
Points Cost
Name
Jokic 62.8 14000
Players:
Points Cost
Name
Adebayo 41.8 9200
Porter 35.3 7600
Gordon 30.3 6600
Robinson 16.3 4200
Highsmith 6.8 1400