LPsolve混合約束

我使用以下R代碼爲我的幻想體育聯盟優化足球陣容。到目前爲止，它一直在努力工作，但我想解決的約束列表中添加了新的皺紋。LPsolve混合約束

陣容由8名選手組成。 1GK，2D，2M，2F，& 1 Util。

在創建模型矩陣，我現在必須考慮混合玩家位置如M/F或d/M

在該R是什麼在列添加1 M和正確的方式如果球員位置是M/F，則在F列中是1？這是解決這個問題的正確方法嗎？或者我應該看看其他想法。

與GK d M F位置工作求解代碼佔到但不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

這裏是大約100名選手的樣本池的幾個混合球員包括在內。

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))

來源

2016-09-14 sanga collins

您是否收到一些錯誤？或者這是一個關於建模的問題？我在這裏沒有看到具體的編程問題。如果您對如何建模混合播放器的數據建模有疑問，這聽起來像是應該在[stats.se]中提出的統計問題，而不是Stack Overflow。 – MrFlick

我沒有收到錯誤。這不是一個建模問題。問題在於程序運行時和Hybrid播放器在數據中。即使它們是最佳選擇，它們也不包含在結果陣容中。我想修改代碼來解釋混合玩家。我嘗試添加兩次M和F作爲例子的球員，但是在某些情況下，我得到了包含兩次球員的結果 –

您所描述的是一個建模問題。您的數據不符合簡單的「lp」模型的假設。你有一個更復雜的場景。您可能需要找到一種方法來在'lp'中建模您的附加約束，或者將您的數據轉換爲與'lp'模型兼容的表單。或者可能找到不同於'lp'的函數來適應這樣的優化模型。但正如我已經提到的，我認爲有更好的地方可以得到這樣的幫助。 – MrFlick

通過使用一個空的矩陣並填充每個位置的正確值的行我能夠得到這個工作。

#### 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 ----") 
}

來源

2016-09-20 15:49:50

LPsolve混合約束

回答

相關問題