LPsolve混合约束

我使用以下R代码为我的幻想体育联盟优化足球阵容。到目前为止，它一直在努力工作，但我想解决的约束列表中添加了新的皱纹。LPsolve混合约束

阵容由8名选手组成。 1GK，2D，2M，2F，& 1 Util。

在创建模型矩阵，我现在必须考虑混合玩家位置如M/F或d/M

在该R是什么在列添加1 M和正确的方式如果球员位置是M/F，则在F列中是1？这是解决这个问题的正确方法吗？或者我应该看看其他想法。

与GK d M F位置工作求解代码占到但不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

这里是大约100名选手的样本池的几个混合球员包括在内。

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))

来源

2016-09-14 sanga collins

您是否收到一些错误？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。如果您对如何建模混合播放器的数据建模有疑问，这听起来像是应该在[stats.se]中提出的统计问题，而不是Stack Overflow。 – MrFlick

我没有收到错误。这不是一个建模问题。问题在于程序运行时和Hybrid播放器在数据中。即使它们是最佳选择，它们也不包含在结果阵容中。我想修改代码来解释混合玩家。我尝试添加两次M和F作为例子的球员，但是在某些情况下，我得到了包含两次球员的结果 –

您所描述的是一个建模问题。您的数据不符合简单的“lp”模型的假设。你有一个更复杂的场景。您可能需要找到一种方法来在'lp'中建模您的附加约束，或者将您的数据转换为与'lp'模型兼容的表单。或者可能找到不同于'lp'的函数来适应这样的优化模型。但正如我已经提到的，我认为有更好的地方可以得到这样的帮助。 – MrFlick

通过使用一个空的矩阵并填充每个位置的正确值的行我能够得到这个工作。

#### 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 ----") 
}

来源

2016-09-20 15:49:50

LPsolve混合约束

回答

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