LPsolve混合约束

时间:2016-09-14 15:19:11

标签: r lpsolve

我使用以下R代码来优化我的梦幻体育联盟的足球阵容。到目前为止它一直很好用,但是我想要解决的约束列表中添加了一个新的皱纹。

阵容由8名球员组成。 1GK,2D,2M,2F,& 1 Util。

在创建模型矩阵时,我现在必须考虑混合玩家位置,例如M / F或D / M

在R中,如果球员位置是M / F,那么在M栏中为F添加1和在F栏中添加1的正确方法是什么?这是解决这个问题的正确方法,还是应该考虑其他想法。

具有GK D M F位置的工作解算器代码但不是D / M或M / F

df <- read.csv("players.csv",encoding = "UTF-8")
mm <- cbind(model.matrix(as.formula("FP~Pos+0"), df))
mm <- cbind(mm, mm, 1, df$Salary, df$Salary, df$FP)
colnames(mm) <- c("D", "F", "GK", "M", "D", "F", "GK", "M", "tot", "salary", "minSal", "FP")

mm <- t(mm)
obj <- df$FP
dir <- c("<=", "<=", "<=", "<=", ">=", ">=", ">=", ">=", "==", "<=", ">=", "<=")

x <- 20000
vals <- c()
ptm <- proc.time()
for(i in 1:5){
  rhs <- c(3, 3, 1, 3, 2, 2, 1, 2, 8, 50000, 49500, x)
  lp <- lp(direction = 'max',
           objective.in = obj,
           all.bin = T,
           const.rhs = rhs,
           const.dir = dir,
           const.mat = mm)
  vals <- c(vals, lp$objval)
  x <- lp$objval - 0.00001
  df$selected <- lp$solution
  lineup <- df[df$selected == 1, ]
  lineup = subset(lineup, select = -c(selected))
  lineup <- lineup %>%
    arrange(Pos)
  print("---- Start ----")
  print(i)
  print(lineup)
  print(sum(lineup$FP))
  print(mean(lineup$own, na.rm = TRUE))
  print(sum(lineup$Salary))
  print(sum(lineup$S))
  print("---- END ----")
}
proc.time() - ptm

这是一个约100名玩家的样本库,其中包括一些混合玩家。

structure(list(Name = structure(c(104L, 105L, 92L, 16L, 84L, 
53L, 85L, 37L, 21L, 34L, 100L, 101L, 83L, 31L, 14L, 35L, 98L, 
59L, 60L, 5L, 6L, 78L, 57L, 89L, 26L, 17L, 74L, 63L, 33L, 71L, 
75L, 41L, 9L, 39L, 12L, 1L, 29L, 7L, 2L, 68L, 73L, 90L, 46L, 
72L, 79L, 50L, 88L, 20L, 97L, 64L, 67L, 3L, 94L, 4L, 22L, 103L, 
52L, 47L, 30L, 58L, 10L, 44L, 28L, 38L, 23L, 15L, 49L, 69L, 81L, 
43L, 99L, 93L, 32L, 56L, 82L, 91L, 62L, 36L, 70L, 48L, 11L, 77L, 
27L, 51L, 25L, 24L, 65L, 96L, 42L, 18L, 102L, 86L, 76L, 87L, 
45L, 61L, 40L, 95L, 8L, 55L, 13L, 66L, 80L, 19L, 54L), .Label = c(" Bojan", 
" Oscar", " Willian", "Aaron Ramsey", "Abel Hernandez", "Adam Smith", 
"Adama Diomande", "Adlene Guedioura", "Adnan Januzaj", "Ahmed Elmohamady", 
"Alex Iwobi", "Alex Oxlade-Chamberlain", "Alexis Sanchez", "Andre Gray", 
"Andrew Robertson", "Andros Townsend", "Anthony Martial", "Antonio Valencia", 
"Ben Mee", "Branislav Ivanovic", "Calum Chambers", "Cedric Soares", 
"Cesc Fabregas", "Charlie Daniels", "Christian Fuchs", "Curtis Davies", 
"Daley Blind", "Daniel Drinkwater", "David de Gea", "Demarai Gray", 
"Diego Costa", "Donald Love", "Dusan Tadic", "Eden Hazard", "Eldin Jakupovic", 
"Erik Pieters", "Etienne Capoue", "Fernando Llorente", "Gareth Barry", 
"Glenn Whelan", "Gylfi Sigurdsson", "Hector Bellerin", "Idrissa Gueye", 
"Jack Cork", "Jack Rodwell", "Jason Puncheon", "Jefferson Montero", 
"Jeremain Lens", "Jeremy Pied", "Jermain Defoe", "Joe Allen", 
"Joel Ward", "John Obi Mikel", "Jordi Amat", "Jordon Ibe", "Joshua King", 
"Juan Mata", "Kasper Schmeichel", "Kevin Mirallas", "Kyle Naughton", 
"Laurent Koscielny", "Leighton Baines", "Leroy Fer", "Lukasz Fabianski", 
"Maarten Stekelenburg", "Marc Albrighton", "Mason Holgate", "Matt Targett", 
"Matthew Lowton", "Max Gradel", "Michy Batshuayi", "Modou Barrow", 
"Nacho Monreal", "Nathan Redmond", "Nordin Amrabat", "Pape Souare", 
"Papy Djilobodji", "Patrick van Aanholt", "Paul Pogba", "Phil Bardsley", 
"Pierre-Emile Højbjerg", "Ramiro Funes Mori", "Riyad Mahrez", 
"Robert Snodgrass", "Ross Barkley", "Ryan Fraser", "Sam Clucas", 
"Sam Vokes", "Santiago Cazorla", "Serge Gnabry", "Shane Long", 
"Shaun Maloney", "Simon Francis", "Stephen Kingsley", "Stephen Ward", 
"Steven Davis", "Steven Defour", "Theo Walcott", "Thibaut Courtois", 
"Tom Heaton", "Wayne Rooney", "Wayne Routledge", "Wilfried Zaha", 
"Xherdan Shaqiri", "Zlatan Ibrahimovic"), class = "factor"), 
    Salary = c(7000L, 9600L, 5700L, 7100L, 6500L, 3200L, 7800L, 
    4200L, 3300L, 8600L, 4200L, 7900L, 9900L, 8700L, 7700L, 4300L, 
    6700L, 5600L, 3700L, 6600L, 4700L, 5700L, 6600L, 7200L, 3500L, 
    7300L, 5900L, 4300L, 7700L, 7100L, 4000L, 9100L, 7400L, 4000L, 
    5800L, 5700L, 5600L, 6300L, 6800L, 4500L, 5100L, 3400L, 5700L, 
    5100L, 8000L, 7800L, 7000L, 5100L, 4900L, 4500L, 3300L, 8300L, 
    3200L, 6600L, 4900L, 6300L, 4400L, 4200L, 4800L, 5200L, 5200L, 
    4500L, 4300L, 7100L, 6500L, 4100L, 3000L, 3800L, 4700L, 4600L, 
    5800L, 4600L, 4200L, 6100L, 3500L, 6800L, 5800L, 4800L, 7300L, 
    5000L, 5000L, 3300L, 4200L, 3900L, 6100L, 5500L, 5400L, 4700L, 
    4700L, 4600L, 4400L, 3400L, 4300L, 4900L, 4600L, 4000L, 3500L, 
    3600L, 3300L, 4800L, 9300L, 7900L, 3700L, 3400L, 2800L), 
    Position = structure(c(5L, 3L, 2L, 5L, 5L, 5L, 5L, 5L, 1L, 
    6L, 4L, 3L, 6L, 3L, 3L, 4L, 6L, 6L, 1L, 3L, 1L, 1L, 5L, 5L, 
    1L, 6L, 6L, 5L, 5L, 3L, 6L, 5L, 5L, 5L, 6L, 6L, 4L, 3L, 5L, 
    1L, 2L, 5L, 5L, 6L, 5L, 3L, 3L, 2L, 5L, 4L, 1L, 5L, 1L, 5L, 
    1L, 6L, 1L, 6L, 6L, 4L, 1L, 5L, 5L, 3L, 5L, 1L, 1L, 1L, 5L, 
    5L, 4L, 1L, 1L, 3L, 1L, 3L, 2L, 1L, 6L, 3L, 6L, 1L, 1L, 5L, 
    1L, 2L, 4L, 5L, 1L, 1L, 5L, 5L, 1L, 5L, 5L, 1L, 5L, 1L, 5L, 
    6L, 6L, 5L, 1L, 1L, 1L), .Label = c("D", "D/M", "F", "GK", 
    "M", "M/F"), class = "factor"), FP = c(23.5, 21.75, 21, 19.75, 
    17.5, 17.333, 16.625, 16.5, 16.5, 16.25, 16, 15.25, 14.875, 
    14.25, 13.75, 13.5, 13.375, 13.25, 12.875, 12.75, 12.75, 
    12.5, 12.375, 12, 11.75, 11.625, 11.375, 11, 10.875, 10.625, 
    10.5, 10.375, 10.125, 10, 9.625, 9.625, 9.5, 9.25, 9.125, 
    9.125, 9, 9, 8.875, 8.875, 8.75, 8.75, 8.5, 8.5, 8.5, 8.5, 
    8.5, 8.25, 8.25, 8, 8, 7.875, 7.875, 7.875, 7.75, 7.5, 7.5, 
    7.5, 7.5, 7.25, 7.25, 7.125, 7, 6.875, 6.625, 6.625, 6.5, 
    6.5, 6.5, 6.25, 6.25, 6.125, 6.125, 6.125, 6, 6, 6, 6, 5.875, 
    5.875, 5.75, 5.75, 5.75, 5.75, 5.75, 5.75, 5.75, 5.75, 5.625, 
    5.5, 5.5, 5.5, 5.5, 5.375, 5.375, 5.25, 5.125, 5, 5, 5, 5
    ), teamAbbrev = structure(c(11L, 9L, 7L, 5L, 7L, 4L, 6L, 
    14L, 1L, 4L, 3L, 9L, 8L, 4L, 3L, 7L, 1L, 6L, 13L, 7L, 2L, 
    12L, 9L, 1L, 7L, 9L, 10L, 13L, 10L, 4L, 14L, 13L, 12L, 6L, 
    1L, 11L, 9L, 7L, 4L, 10L, 1L, 1L, 5L, 13L, 9L, 12L, 3L, 4L, 
    3L, 13L, 6L, 4L, 13L, 1L, 10L, 5L, 5L, 13L, 8L, 8L, 7L, 13L, 
    8L, 13L, 4L, 7L, 10L, 3L, 10L, 6L, 4L, 2L, 12L, 2L, 6L, 10L, 
    6L, 11L, 2L, 12L, 1L, 12L, 9L, 11L, 8L, 2L, 6L, 10L, 1L, 
    9L, 13L, 2L, 5L, 7L, 12L, 1L, 11L, 3L, 14L, 2L, 1L, 8L, 11L, 
    3L, 13L), .Label = c("ARS", "BOU", "BUR", "CHE", "CRY", "EVE", 
    "HUL", "LEI", "MU", "SOU", "STK", "SUN", "SWA", "WAT"), class = "factor")), .Names = c("Name", 
"Salary", "Position", "FP", "teamAbbrev"), class = "data.frame", row.names = c(NA, 
-105L))

1 个答案:

答案 0 :(得分:1)

通过使用空矩阵并使用每个位置的正确值填充行,我能够使其工作。

#### SOLVER ##### ----
mm <- matrix(0, nrow = 8, ncol = nrow(df))
# Goal Keeper
j<-1
i<-1
for (i in 1:nrow(df)){
  if (df$Pos[i]=="GK")
    mm[j,i]<-1
}
# Defender
j<-2
i<-1
for (i in 1:nrow(df)){
  if (df$Pos[i]=="D")
    mm[j,i]<-1
}
# Midfielder
j<-3
i<-1
for (i in 1:nrow(df)){
  if (df$Pos[i]=="M"   ||
      df$Pos[i]=="M/F")
    mm[j,i]<-1
}
# Forward
j<-4
i<-1
for (i in 1:nrow(df)){
  if (df$Pos[i]=="F"   ||
      df$Pos[i]=="M/F")
    mm[j,i]<-1
}
# Utility
j<-5
i<-1
for (i in 1:nrow(df)){
  if (!df$Pos[i]=="GK")
    mm[j,i]<-1
}
# Salary
mm[6, ] <- df$Salary
mm[7, ] <- df$FP
mm[8, ] <- 1
# rbind existing matrix to itself to set minimum constraints
mm <- rbind(mm, mm[1:5,])
i<-1

objective.in <- df$FP
const.mat <- mm
const.dir <- c("<=", "<=", "<=", "<=", "<=", "<=", "<=", "==",
               ">=", ">=", ">=", ">=", ">=")

x <- 20000
vals <- c()

for(i in 1:5){
  const.rhs <- c(1, 4, 4, 4, 7, 50000, x, 8, # max for each contraint
                 1, 2, 2, 2, 7)              # min for each constraint
  sol <- lp(direction = "max", objective.in, # maximize objective function
            const.mat, const.dir, const.rhs, # constraints
            all.bin = TRUE)
  vals <- c(vals, sol$objval)
  x <- sol$objval - 0.00001
  inds <- which(sol$solution == 1)
  sum(df$salary[inds])
  solution<-df[inds, ]
  solution <- solution[,-c(8)]
  solution <- solution %>%
    arrange(Pos)
  print("---- Start ----")
  print(i)
  print(solution)
  print(sum(solution$FP))
  print(sum(solution$Salary))
  print(sum(solution$S))
  print("---- END ----")
}