LPsolve 混合约束答案

【问题标题】：LPsolve Hybrid constraintsLPsolve 混合约束
【发布时间】：2016-09-14 15:19:11
【问题描述】：

我使用以下 R 代码为我的梦幻体育联盟优化足球阵容。到目前为止，它一直运行良好，但在我想要解决的约束列表中添加了一个新问题。

一个阵容由 8 名球员组成。 1GK、2D、2M、2F 和 1 Util。

在创建模型矩阵时，我现在必须考虑混合球员的位置，例如 M/F 或 D/M

在 R 中，如果玩家位置是 M/F，在 M 列中添加 1 并在 F 列中添加 1 的正确方法是什么？这是解决此问题的正确方法还是我应该考虑其他想法。

包含 GK D M F 位置但不包含 D/M 或 M/F 的工作求解器代码

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

这是一个包含大约 100 名玩家的样本池，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))

【问题讨论】：

您遇到错误了吗？或者这是一个关于建模的问题？我在这里没有看到具体的编程问题。如果您对如何为混合玩家建模数据有疑问，这听起来像是一个统计问题，应该在 Cross Validated 而非 Stack Overflow 上提问。
我没有收到错误消息。这不是一个建模问题。问题在于程序何时运行并且混合玩家在数据中。即使它们是最佳选择，它们也不包含在最终的阵容中。我想修改代码以考虑混合玩家。我尝试将玩家添加两次，一个是 M，一个是 F
您所描述的是一个建模问题。您的数据不符合简单 lp 模型的假设。你有一个更复杂的场景。您要么需要找到一种方法来在lp 中对附加约束进行建模，要么将数据转换为与lp 模型兼容的形式。或者可能找到一个除lp 之外的函数来适应这种优化模型。但正如我已经提到的，我认为有更好的地方可以获得此类帮助。
啊，我想我现在明白你的意思了。当您说建模时，我正在考虑回归模型或线性模型，因为您提到了交叉验证的统计问题。（您可能已经注意到，我是统计新手）。我将重新提出这个问题，以便更准确地反映我想要完成的目标

标签： r lpsolve

【解决方案1】：

通过使用一个空矩阵并用每个位置的正确值填充行，我能够让它工作。

#### 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 ----")
}

【讨论】：