幻想足球线性规划在R与RGLPK

本文介绍了幻想足球线性规划在R与RGLPK的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 长时间监听器第一次呼叫者SO .. 我问一个问题，以前已经问过非常相似，但我不相信我足够聪明解密如何实施的解决方案，对于这个我道歉。这里是我发现的问题的链接： R中的约束多整数线性规划 我超过了我的投资幻想点（FPTS_PREDICT_RF），受到50,000工资帽，同时最小化一个风险计算，我已经想出了。 现在，问题出在flex位置。该团队需要由9个职位组成， 1 QB 2 RB 3 WR 1 TE 1 DEF 1 FLEX flex可以是RB，WR或TE。 那么，我们可以得到： 1 QB 2- 3 RB 3-4 WR 1-2 TE 1 DEF 我试图实现约束#RB + #WR + #TE == 7。 以下是相关代码： library（Rglpk） ＃变量数量 num.players ＃objective ： obj< - final $ FPTS_PREDICT_RF ＃vars表示为布尔值 var.types< - rep（B，num.players）＃ matrix< - rbind（as.numeric（final $ position ==QB），＃num QB as.numeric（final $ position ==RB），＃num RB as.numeric（final $ position ==WR），＃num WR as.numeric（final $ position ==TE），＃num TE as.numeric（final $位置==DEF），＃num DEF diag（final $ riskNormalized），＃player's risk final $ Salary）＃total cost direction< ==，< =，< =，< =，==， rep < =，num.players），< =） rhs< - c（1，＃Quartbacks 3，＃Running Backs 2，＃Wide Receivers 1，＃Tight Ends 1，＃Defense rep（10，num.players），#HERE，您需要输入一个数字， b $ b #risk你愿意，1是低风险，＃10是高风险。 10是最大。 50000）＃默认情况下，您会花50K，所以请不要离开这个号码。 sol< - Rglpk_solve_LP（obj = obj，mat = matrix，dir = direction，rhs = rhs， types = var.types，max = TRUE） sol＃投射的幻想点 有人可以帮助我实现这个约束吗？ 编辑：链接到数据集final是csv格式： https://www.dropbox.com/s/qp35wc4d380hep1/final.csv?dl=0 另一个问题：对于任何你梦幻足球运动员在那里，我计算我的风险因素直接从SD的玩家的历史幻想点，并在[0,10]的支持下规范化这个数字。 解决方案您可以通过添加以下约束来做到这一点： / p> RB数量= 2 RB数量 WR数> = 3 WR数 TE数> = 1 TE数量 RB数量+ WR + TEs == 7 以下是更新后的代码： b $ b＃变量数 num.players ＃objective： obj ＃ vars表示为布尔值 var.types< - rep（B，num.players）＃约束矩阵< - rbind（as.numeric ==QB），＃num QB as.numeric（final $ position ==RB），＃num RB as.numeric（final $ position ==RB），＃ num RB as.numeric（final $ position ==WR），＃num WR as.numeric（final $ position ==WR），＃num WR as。数字（final $ position ==TE），＃num TE as.numeric（final $ position ==TE），＃num TE as.numeric c（RB，WR，TE）），＃Num RB / WR / TE as.numeric（final $ position ==DEF），＃num DEF diag final $ riskNormalized），＃player's risk final $ Salary）＃total cost direction< - c（==，> =， ; =，> =，< =，> =，< =， =，==， rep（< =，num.players），< =） rhs 2，＃RB Min 3，＃RB Max 3，＃WR Min 4，＃WR Max 1，＃TE Min 2，＃TE Max 7，＃RB / WR / TE 1，＃Defense rep（10，num.players），#HERE，您需要输入一个数字表示您愿意如何 #risk，1是低风险，＃10是高风险。 10是最大。 50000）＃默认情况下，您会花50K，所以请不要离开这个号码。 sol< - Rglpk_solve_LP（obj = obj，mat = matrix，dir = direction，rhs = rhs， types = var.types，max = TRUE） 最后，您可以通过子集 final 来评估您的解决方案： final [sol $ solution == 1，] ＃X PLAYER FPTS_PREDICT_LIN FPTS_PREDICT_RF工资位置＃1 1 AJ绿色20.30647 20.885558 5900 WR ＃17 18 Andre Holmes 13.26369 15.460503 4100 WR ＃145 156 Giovani Bernard 17.05857 19.521157 6100 RB ＃148 160 Greg Olsen 17.08808 17.831687 5500 TE ＃199 222 Jordy Nelson 22.12326 24.077787 7800 WR ＃215 239 Kelvin Benjamin 16.12116 17.132573 5000 WR ＃233 262 Le'Veon Bell 20.51564 18.565763 6300 RB ＃303 340 Ryan Tannehill 17.92518 19.134305 6700 QB ＃362 3641 SD 5.00000 6.388666 2600 DEF ＃风险风险正规化＃1 5.131601 3.447990 ＃17 9.859006 6.624396 ＃145 9.338094 6.274388 ＃148 6.517376 4.379111 ＃199 9.651055 6.484670 ＃215 7.081162 4.757926 ＃233 6.900656 4.636641 ＃303 4.857983 3.264143 ＃362 2.309401 0.000000 对于这个问题数据，您选择了一个宽接收器到flex位置。 long time listener first time caller to S.O...I am asking a question that has been asked very similarly before, however I don't believe I am smart enough to decipher how to implement the solution, for this I apologize. Here is the link to the question I found:Constraints in R Multiple Integer Linear ProgrammingI am maxing over my projected fantasy points(FPTS_PREDICT_RF), subject to a 50,000 salary cap, while minimizing a 'risk' calculation that I have came up with. Now, the problem lies in the "flex" position. The team needs to be made up of 9 positions, 1 QB2 RB3 WR1 TE1 DEF1 FLEXThe flex can be a RB, WR, or TE.So, we can then have:1 QB2-3 RB3-4 WR1-2 TE1 DEFI am trying to implement the constraint that #RB + #WR + #TE ==7.Here is the relevant code:library(Rglpk)# number of variablesnum.players <- length(final$PLAYER)# objective:obj <- final$FPTS_PREDICT_RF# the vars are represented as booleansvar.types <- rep("B", num.players)# the constraintsmatrix <- rbind(as.numeric(final$position == "QB"), # num QB as.numeric(final$position == "RB"), # num RB as.numeric(final$position == "WR"), # num WR as.numeric(final$position == "TE"), # num TE as.numeric(final$position == "DEF"),# num DEF diag(final$riskNormalized), # player's risk final$Salary) # total costdirection <- c("==", "<=", "<=", "<=", "==", rep("<=", num.players), "<=")rhs <- c(1, # Quartbacks 3, # Running Backs 2, # Wide Receivers 1, # Tight Ends 1, # Defense rep(10, num.players), #HERE, you need to enter a number that indicates how #risk you are willing to be, 1 being low risk, # 10 being high risk. 10 is max. 50000) # By default, you get 50K to spend, so leave this number alone. sol <- Rglpk_solve_LP(obj = obj, mat = matrix, dir = direction, rhs = rhs, types = var.types, max = TRUE)sol #Projected Fantasy PointsCan someone help me implement this constraint?Any help is much, much appreciated!EDIT: Link to dataset 'final' is csv format: https://www.dropbox.com/s/qp35wc4d380hep1/final.csv?dl=0SIDE QUESTION: For any of you fantasy footballers out there, I am calculating my 'risk' factor directly from the S.D. of the player's historical fantasy points, and normalizing this number over the support of [0,10]. Can you think of a better way to calculate a given players risk? 解决方案 You can do this by adding the following constraints:The number of RBs >= 2The number of RBs <= 3The number of WRs >= 3The number of WRs <= 4The number of TEs >= 1The number of TEs <= 2The number of RBs + WRs + TEs == 7Here's the updated code:library(Rglpk)# number of variablesnum.players <- length(final$PLAYER)# objective:obj <- final$FPTS_PREDICT_RF# the vars are represented as booleansvar.types <- rep("B", num.players)# the constraintsmatrix <- rbind(as.numeric(final$position == "QB"), # num QB as.numeric(final$position == "RB"), # num RB as.numeric(final$position == "RB"), # num RB as.numeric(final$position == "WR"), # num WR as.numeric(final$position == "WR"), # num WR as.numeric(final$position == "TE"), # num TE as.numeric(final$position == "TE"), # num TE as.numeric(final$position %in% c("RB", "WR", "TE")), # Num RB/WR/TE as.numeric(final$position == "DEF"),# num DEF diag(final$riskNormalized), # player's risk final$Salary) # total costdirection <- c("==", ">=", "<=", ">=", "<=", ">=", "<=", "==", "==", rep("<=", num.players), "<=")rhs <- c(1, # Quartbacks 2, # RB Min 3, # RB Max 3, # WR Min 4, # WR Max 1, # TE Min 2, # TE Max 7, # RB/WR/TE 1, # Defense rep(10, num.players), #HERE, you need to enter a number that indicates how #risk you are willing to be, 1 being low risk, # 10 being high risk. 10 is max. 50000) # By default, you get 50K to spend, so leave this number alone. sol <- Rglpk_solve_LP(obj = obj, mat = matrix, dir = direction, rhs = rhs, types = var.types, max = TRUE)Finally, you can evaluate your solution by subsetting final:final[sol$solution==1,]# X PLAYER FPTS_PREDICT_LIN FPTS_PREDICT_RF Salary position# 1 1 A.J. Green 20.30647 20.885558 5900 WR# 17 18 Andre Holmes 13.26369 15.460503 4100 WR# 145 156 Giovani Bernard 17.05857 19.521157 6100 RB# 148 160 Greg Olsen 17.08808 17.831687 5500 TE# 199 222 Jordy Nelson 22.12326 24.077787 7800 WR# 215 239 Kelvin Benjamin 16.12116 17.132573 5000 WR# 233 262 Le'Veon Bell 20.51564 18.565763 6300 RB# 303 340 Ryan Tannehill 17.92518 19.134305 6700 QB# 362 3641 SD 5.00000 6.388666 2600 DEF# risk riskNormalized# 1 5.131601 3.447990# 17 9.859006 6.624396# 145 9.338094 6.274388# 148 6.517376 4.379111# 199 9.651055 6.484670# 215 7.081162 4.757926# 233 6.900656 4.636641# 303 4.857983 3.264143# 362 2.309401 0.000000For this problem data you have selected a wide receiver to the flex position. 这篇关于幻想足球线性规划在R与RGLPK的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！