| title | author | date | output |
|---|---|---|---|
Lineup Optimizer |
Jason Lee |
6/2/2018 |
html_document |
If you are like me and love fantasy football but aren't always positive which lineup is the opitmal lineup to use in a given week - Then this optimizer is for you!
Updated Python Version included with Notebook here
A better version will be included in my upcoming Applied Sports Analytics book. Chapter 8 - Lineup Optimization Using Linear Programming
(add versions to R packages)
We are going to solve: maximize f'x subject to A*x with constraints equal to b where:
- x: is the variable to solve for: a vector of 0 or 1:
- 1 when the player is selected, 0 otherwise
- f: is your objective vector
- A: is a matrix
- dir: a vector of "<=", "==", or ">="
- b: a vector defining your linear constraints.
# load library
library(Rglpk)We need to get the weekly fantasy football prices and projected scores.
Here we have data from Yahoo's Daily fantasy tournments.
QB <- read.csv("Data/QBs-Table 1.csv",header=TRUE,stringsAsFactors=FALSE)
RB <- read.csv("Data/RBs-Table 1.csv",header=TRUE,stringsAsFactors=FALSE)
WR <- read.csv("Data/WRs-Table 1.csv",header=TRUE,stringsAsFactors=FALSE)
TE <- read.csv("Data/TEs-Table 1.csv",header=TRUE,stringsAsFactors=FALSE)
DEF <- read.csv("Data/DEFS-Table 1.csv",header=TRUE,stringsAsFactors=FALSE)We need to tag each of the players with their position.
We also need to combine all the players into a single dataset.
QB$pos <- rep("QB")
RB$pos <- rep("RB")
WR$pos <- rep("WR")
TE$pos <- rep("TE")
DEF$pos <- rep("DEF")
ALL <- rbind(QB, RB, WR, TE, DEF) # combind into one data.frameTo give you an idea of what data set you would need to input when you are customizing and updating the equation for your current week.
Here's what my dataset looks like to start
head(ALL)## name cost projPts pos
## 1 Aaron Rodgers 42 20.5550 QB
## 2 Andrew Luck 42 23.9883 QB
## 3 Andy Dalton 31 17.9616 QB
## 4 Ben Roethlisberger 43 19.5577 QB
## 5 Blake Bortles 23 16.5012 QB
## 6 Cam Newton 34 19.7308 QB
As you can see it is very simple. We only need 4 key elements to run the algorithm:
- Player Name - this could be a player name or playerID
- Position - Needs to be either QB, RB, WR, TE, or DEF
- Cost - how much the player costs for the given week
- Projected Points - you can choose whatever site or projections you want
- I will have another project posted showing you how to create your own projections
# count of all the players
num.players <- length(ALL)
# objective:
f <- ALL$projPts
# the variables are all booleans
var.types <- rep("B", num.players)
# the constraints
A <- rbind(as.numeric(ALL$pos == "QB"), # num QB
as.numeric(ALL$pos == "RB"), # num RB
as.numeric(ALL$pos == "WR"), # num WR
as.numeric(ALL$pos == "TE"), # num TE
as.numeric(ALL$pos == "DEF"), # num DEF
ALL$cost) # total cost
dir <- c("==",
"==",
"==",
"==",
"==",
"<=") # this is for the total team salary, which is why it is less than or equalHere is the part that needs to be adjusted depending on how you want your lineup to be and what the total salary limit is for the team.
- QB: It's normal to have only one QB but hey if there's a daily league with more you can change this.
- RB: There is usually 2 required plus a possible flex position.
- If you want that flex position to be a RB then this needs to be at 3.
- If not then set it to 2.
- WR: There is usually 3 required plus a possible flex position.
- If you want that flex position to be a RB then this needs to be at 4.
- If not then set it to 3.
- TE: There is usually 1 required plus a possible flex position.
- If you want that flex position to be a RB then this needs to be at 2.
- If not then set it to 1.
- DEF: There is usually one 1 defense required.
- Salary: The salary limit could be very different depending on the league.
- Yahoo is $200
- Draft Kings is $60,000
- Fan Duel is $50,000
I will run the algorithm 3 different ways.
1. The flex position be an extra RB
2. The flex position be an extra WR
3. The flex position be an extra TE
b <- c(1, # QB
3, # RB
3, # WR
1, # TE
1, # DEF
200) # costThe first thing to check is that the status = 0. If it does not then it did not find an optimal solution.
# Solve the math problem
sol.3rb <- Rglpk_solve_LP(obj = f, mat = A, dir = dir, rhs = b,
types = var.types, max = TRUE)
# Check that Status = 0 and that there is an optimum value
sol.3rb## $optimum
## [1] 130.6174
##
## $solution
## [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
## [36] 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## [71] 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
## [106] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0
## [141] 0 0 0 0 0 0 0 0 0 0 0
##
## $status
## [1] 0
##
## $solution_dual
## [1] NA
##
## $auxiliary
## $auxiliary$primal
## [1] 1 3 3 1 1 200
##
## $auxiliary$dual
## [1] NA
# View the selected players
ALL$name[sol.3rb$solution == 1]## [1] "Andrew Luck" "Ameer Abdullah" "Carlos Hyde"
## [4] "Doug Martin" "Brandin Cooks" "Larry Fitzgerald"
## [7] "Terrance Williams" "Tyler Eifert" "Rams"
b1 <- c(1, # QB
2, # RB
4, # WR
1, # TE
1, # DEF
200) # cost# Solve the math problem
sol.4wr <- Rglpk_solve_LP(obj = f, mat = A, dir = dir, rhs = b1,
types = var.types, max = TRUE)
# Check that Status = 0 and that there is an optimum value
sol.4wr## $optimum
## [1] 130.1095
##
## $solution
## [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## [36] 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## [71] 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
## [106] 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0
## [141] 0 0 0 0 0 0 0 0 0 0 0
##
## $status
## [1] 0
##
## $solution_dual
## [1] NA
##
## $auxiliary
## $auxiliary$primal
## [1] 1 2 4 1 1 200
##
## $auxiliary$dual
## [1] NA
# View the selected players
ALL$name[sol.4wr$solution == 1]## [1] "Andrew Luck" "Chris Johnson" "DeMarco Murray"
## [4] "Brandin Cooks" "Larry Fitzgerald" "Mike Wallace"
## [7] "Terrance Williams" "Tyler Eifert" "Rams"
b2 <- c(1, # QB
2, # RB
3, # WR
2, # TE
1, # DEF
200) # cost# Solve the math problem
sol.2te <- Rglpk_solve_LP(obj = f, mat = A, dir = dir, rhs = b2,
types = var.types, max = TRUE)
# Check that Status = 0 and that there is an optimum value
sol.2te## $optimum
## [1] 130.3089
##
## $solution
## [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## [36] 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## [71] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
## [106] 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0
## [141] 0 0 0 0 0 0 0 0 0 0 0
##
## $status
## [1] 0
##
## $solution_dual
## [1] NA
##
## $auxiliary
## $auxiliary$primal
## [1] 1 2 3 2 1 200
##
## $auxiliary$dual
## [1] NA
# View the selected players
ALL$name[sol.2te$solution == 1]## [1] "Andrew Luck" "DeMarco Murray" "Doug Martin"
## [4] "Larry Fitzgerald" "Mike Wallace" "Terrance Williams"
## [7] "Jordan Cameron" "Tyler Eifert" "Rams"
# 3 RB lineup
sol.3rb$optimum## [1] 130.6174
ALL$name[sol.3rb$solution == 1]## [1] "Andrew Luck" "Ameer Abdullah" "Carlos Hyde"
## [4] "Doug Martin" "Brandin Cooks" "Larry Fitzgerald"
## [7] "Terrance Williams" "Tyler Eifert" "Rams"
# 4 WR lineup
sol.4wr$optimum## [1] 130.1095
ALL$name[sol.4wr$solution == 1]## [1] "Andrew Luck" "Chris Johnson" "DeMarco Murray"
## [4] "Brandin Cooks" "Larry Fitzgerald" "Mike Wallace"
## [7] "Terrance Williams" "Tyler Eifert" "Rams"
# 2 TE lineup
sol.2te$optimum## [1] 130.3089
ALL$name[sol.2te$solution == 1]## [1] "Andrew Luck" "DeMarco Murray" "Doug Martin"
## [4] "Larry Fitzgerald" "Mike Wallace" "Terrance Williams"
## [7] "Jordan Cameron" "Tyler Eifert" "Rams"
Obviously these lineups are from a few years ago but you can change the inputs and try to compete against me next season!