assignment4_Perreault.rmd

---
title: 'Bios 6301: Assignment 4'
author: 'Andrea Perreault'
output: pdf_document
---

*Due Tuesday, 01 November, 1:00 PM*

$5^{n=day}$ points taken off for each day late.

50 points total.

Submit a single knitr file (named `homework4.rmd`), along with a valid PDF output file. Inside the file, clearly indicate which parts of your responses go with which problems (you may use the original homework document as a template). Add your name as `author` to the file's metadata section. Raw R code/output or word processor files are not acceptable.

Failure to name file `homework4.rmd` or include author name may result in 5 points taken off.

### Question 1 ###

**15 points**

A problem with the Newton-Raphson algorithm is that it needs the derivative $f'$. If the derivative is hard to compute or does not exist, then we can use the *secant method*, which only requires that the function $f$ is continuous.

Like the Newton-Raphson method, the **secant method** is based on a linear approximation to the function $f$. Suppose that $f$ has a root at $a$. For this method we assume that we have *two* current guesses, $x_0$ and $x_1$, for the value of $a$. We will think of $x_0$ as an older guess and we want to replace the pair $x_0$, $x_1$ by the pair $x_1$, $x_2$, where $x_2$ is a new guess.

To find a good new guess $x_2$ we first draw the straight line from $(x_0,f(x_0))$ to $(x_1,f(x_1))$, which is called a secant of the curve $y = f(x)$. Like the tangent, the secant is a linear approximation of the behavior of $y = f(x)$, in the region of the points $x_0$ and $x_1$. As the new guess we will use the x-coordinate $x_2$ of the point at which the secant crosses the x-axis.

The general form of the recurrence equation for the secant method is:

$$x_{i+1} = x_i - f(x_i)\frac{x_i - x_{i-1}}{f(x_i) - f(x_{i-1})}$$

Notice that we no longer need to know $f'$ but in return we have to provide *two* initial points, $x_0$ and $x_1$.

**Write a function that implements the secant algorithm.** Validate your program by finding the root of the function $f(x) = \cos(x) - x$. Compare its performance with the Newton-Raphson method -- which is faster, and by how much?  For this example $f'(x) = -\sin(x) - 1$.

```{r}
# secant method
secant <- function(g1, g2, f, tol=10e-7, maxiter=1000) {
  i <- 0 
  while(abs(g1-g2) > tol && i < maxiter) {
    x <- g2 - f(g2)*(g2-g1)/(f(g2)-f(g1))
    g2 <- g1
    x <- g2
    i <- i + 1
  }
  if(i==maxiter) {
    print('failed to converge')
    return(NULL)
  } else {
    print(sprintf('converges at %s after %s iterations', x, i))
  }
}

f <- function(x) cos(x) - x

# compare to Newton-Raphson method
newton <- function(guess, f, fp, tol=10e-7, maxiter=1000) {
  i <- 1 # counter
  while(abs(f(guess)) > tol && i < maxiter) {
    guess <- guess - f(guess)/fp(guess)
    i <- i + 1
  }
  if(i==maxiter) {
    print('failed to converge')
    return(NULL)
  } else {
    print(sprintf('converges at %s after %s iterations', guess, i))
  }
}

f <- function(x) cos(x) - x
fp <- function(x) (-sin(x))-1
```

```{r eval=FALSE}
# test 1
(system.time(replicate(1000, secant(0.5, 0.75, f))))
(system.time(replicate(1000, newton(0.5, f, fp))))
# test 2
(system.time(replicate(1000, secant(0, 0.2, f))))
(system.time(replicate(1000, newton(0.25, f, fp))))
```

```
The Newton-Raphson method is faster, taking 0.155 seconds to complete 1000 replicates. 
This is compared to 0.189 seconds for the secant method. However, the amount of time it 
takes to find the root for these methods varies depending on the guesses input by the 
user. For example, it takes 0.175 seconds for the secant method to find the root if the
user gives 0 and 0.2 as initial guesses, compared to 0.210 with the Newton-Raphson method. 
```

### Question 2 ###

**18 points**

The game of craps is played as follows. First, you roll two six-sided dice; let x be the sum of the dice on the first roll. If x = 7 or 11 you win, otherwise you keep rolling until either you get x again, in which case you also win, or until you get a 7 or 11, in which case you lose.

Write a program to simulate a game of craps. You can use the following snippet of code to simulate the roll of two (fair) dice:

```{r}
x <- sum(ceiling(6*runif(2)))
```

1. The instructor should be able to easily import and run your program (function), and obtain output that clearly shows how the game progressed.  Set the RNG seed with `set.seed(100)` and show the output of three games. (lucky 13 points)

```{r}
set.seed(100)

craps <- function(x, numGames) {
  results <- matrix(0, nrow = numGames, ncol = 2, dimnames = list(NULL, c("Win", "Lose")))
  for(i in 1:numGames) {
    win <- FALSE 
    lose <- FALSE
    if(x==7||x==11){
    win <- TRUE
    print('Yay! You win :)')
  } else {
    x1 <- 0
    while(x1!=x && x1!=7 && x1!=11){
      print('Cross your fingers and roll again.')
      x1 <- sum(ceiling(6*runif(2)))
      print(x1)
    }
    if(x1==7||x1==11){
      lose <- TRUE
      print("Sorry, you lose :(")
    } else {
      win <- TRUE
      print(sprintf('You rolled %s again, you win!', x))
    }
  }
    if(win) { 
    results[i, 1] <- 1
  } else {
    results[i, 2] <- 1
  }
  }
  return(results)
}

x=sum(ceiling(6*runif(2)))
results <- craps(x, 3)
results
```

2. Find a seed that will win ten straight games.  Consider adding an argument to your function that disables output. Show the output of the ten games. (5 points)

```{r}
set.seed(500)

craps <- function(x, numGames) {
  results <- matrix(0, nrow = numGames, ncol = 2, dimnames = list(NULL, c("Win", "Lose")))
  for(i in 1:numGames) {
    win <- FALSE 
    lose <- FALSE
    if(x==7||x==11){
    win <- TRUE
  } else {
    x1 <- 0
    while(x1!=x && x1!=7 && x1!=11){
      x1 <- sum(ceiling(6*runif(2)))
    }
    if(x1==7||x1==11){
      lose <- TRUE
    } else {
      win <- TRUE
    }
  }
    if(win) { 
    results[i, 1] <- 1
  } else {
    results[i, 2] <- 1
  }
  }
  return(results)
}

x=sum(ceiling(6*runif(2)))
results <- craps(x, 10)
results
```

### Question 3 ###

**12 points**

Obtain a copy of the [football-values lecture](https://github.com/couthcommander/football-values). Save the five 2016 CSV files in your working directory.

Modify the code to create a function.  This function will create dollar values given information (as arguments) about a league setup.  It will return a data.frame and write this data.frame to a CSV file.  The final data.frame should contain the columns 'PlayerName', 'pos', 'points', 'value' and be orderd by value descendingly.  Do not round dollar values.

Note that the returned data.frame should have `sum(posReq)*nTeams` rows.

Define the function as such (6 points):

```{r}
# path: directory path to input files
# file: name of the output file; it should be written to path
# nTeams: number of teams in league
# cap: money available to each team
# posReq: number of starters for each position
# points: point allocation for each category
ffvalues <- function(path, file='outfile.csv', nTeams=12, cap=200, posReq=c(qb=1, rb=2, wr=3, te=1, k=1),
                     points=c(fg=4, xpt=1, pass_yds=1/25, pass_tds=4, pass_ints=-2, rush_yds=1/10, rush_tds=6, fumbles=-2, rec_yds=1/20, rec_tds=6)) {
  # read in files
  k <- read.csv('proj_k16.csv', header=TRUE, stringsAsFactors=FALSE)
  qb <- read.csv('proj_qb16.csv', header=TRUE, stringsAsFactors=FALSE)
  rb <- read.csv('proj_rb16.csv', header=TRUE, stringsAsFactors=FALSE)
  te <- read.csv('proj_te16.csv', header=TRUE, stringsAsFactors=FALSE)
  wr <- read.csv('proj_wr16.csv', header=TRUE, stringsAsFactors=FALSE)
  
  # generate unique list of column names
  cols <- unique(c(names(k), names(qb), names(rb), names(te), names(wr)))
  k[,'pos'] <- 'k'
  qb[,'pos'] <- 'qb'
  rb[,'pos'] <- 'rb'
  te[,'pos'] <- 'te'
  wr[,'pos'] <- 'wr'

  # append 'pos' to unique column list
  cols <- c(cols, 'pos')

  # create common columns in each data.frame, initialize values to zero
  k[,setdiff(cols, names(k))] <- 0
  qb[,setdiff(cols, names(qb))] <- 0
  rb[,setdiff(cols, names(rb))] <- 0
  te[,setdiff(cols, names(te))] <- 0
  wr[,setdiff(cols, names(wr))] <- 0

  # combine data.frames by row, using consistent column order
  x <- rbind(k[,cols], qb[,cols], rb[,cols], te[,cols], wr[,cols])
  
  # convert NFL stat to fantasy points
  x[,'p_fg'] <- x[,'fg']*points[[1]]
  x[,'p_xpt'] <- x[,'xpt']*points[[2]]
  x[,'p_pass_yds'] <- x[,'pass_yds']*points[[3]]
  x[,'p_pass_tds'] <- x[,'pass_tds']*points[[4]]
  x[,'p_pass_ints'] <- x[,'pass_ints']*points[[5]]
  x[,'p_rush_yds'] <- x[,'rush_yds']*points[[6]]
  x[,'p_rush_tds'] <- x[,'rush_tds']*points[[7]]
  x[,'p_fumbles'] <- x[,'fumbles']*points[[8]]
  x[,'p_rec_yds'] <- x[,'rec_yds']*points[[9]]
  x[,'p_rec_tds'] <- x[,'rec_tds']*points[[10]]

  # sum selected column values for every row
  x[,'points'] <- rowSums(x[,grep("^p_", names(x))])
  
  # create new data.frame ordered by points descendingly
  x2 <- x[order(x[,'points'], decreasing=TRUE),]

  # determine the row indeces for each position
  k.idx <- which(x2[,'pos']=='k')
  qb.idx <- which(x2[,'pos']=='qb')
  rb.idx <- which(x2[,'pos']=='rb')
  te.idx <- which(x2[,'pos']=='te')
  wr.idx <- which(x2[,'pos']=='wr')
  
  # calculate marginal points by subtracting "baseline" player's points
  x2[k.idx, 'marg'] <- x2[k.idx,'points'] - x2[k.idx[12],'points']
  x2[qb.idx, 'marg'] <- x2[qb.idx,'points'] - x2[qb.idx[12],'points']
  x2[rb.idx, 'marg'] <- x2[rb.idx,'points'] - x2[rb.idx[24],'points']
  x2[te.idx, 'marg'] <- x2[te.idx,'points'] - x2[te.idx[12],'points']
  x2[wr.idx, 'marg'] <- x2[wr.idx,'points'] - x2[wr.idx[36],'points']

  # create a new data.frame subset by non-negative marginal points
  x3 <- x2[x2[,'marg'] >= 0,]

  # re-order by marginal points
  x3 <- x3[order(x3[,'marg'], decreasing=TRUE),]

  # reset the row names
  rownames(x3) <- NULL

  # calculation for player value
  x3[,'value'] <- x3[,'marg']*(nTeams*cap-nrow(x3))/sum(x3[,'marg']) + 1
  
  # save dollar values to a csv file
  x4 <- x3[,c('PlayerName','pos','points','value')]
  write.csv(x4, file)
  # return data.frame with dollar values 
  as.data.frame(x4, row.names=NULL)
}
```

1. Call `x1 <- ffvalues('.')`

    a.  How many players are worth more than $20? (1 point)
    ```{r}
    x1 <- ffvalues('.')
    which(x1[,'value']>20)
    ```
    ```
    There are 46 players that are worth more than $20.
    ```
    
    b.  Who is 15th most valuable running back (rb)? (1 point)
    ```{r}
    which(x1[,'pos']=='rb')[15]
    ```
    ```
    The 15th most valuable running back is at position 47, which is Carlos Hyde.
    ```
    

2. Call `x2 <- ffvalues(getwd(), '16team.csv', nTeams=16, cap=150)`

    a.  How many players are worth more than $20? (1 point)
    ```{r}
    x2 <- ffvalues(getwd(), '16team.csv', nTeams=16, cap=150)
    which(x2[,'value']>20)
    ```
    ```
    There are 46 players that are worth more than $20. 
    ```
    
    b.  How many wide receivers (wr) are in the top 40? (1 point)
    ```{r}
    which(x2[,'pos']=='wr')[1:40]
    ```
    ```
    There are 36 wide receivers in the top 40.
    ```

3. Call:

    ```{r eval=FALSE}
    x3 <- ffvalues('.', 'qbheavy.csv', posReq=c(qb=2, rb=2, wr=3, te=1, k=0),
            points=c(fg=0, xpt=0, pass_yds=1/25, pass_tds=6, pass_ints=-2,
                    rush_yds=1/10, rush_tds=6, fumbles=-2, rec_yds=1/20, rec_tds=6))
    ```

    a.  How many players are worth more than $20? (1 point)
    ```{r}
    x3 <- ffvalues('.', 'qbheavy.csv', posReq=c(qb=2, rb=2, wr=3, te=1, k=0), points=c(fg=0, xpt=0,  
    pass_yds=1/25, pass_tds=6, pass_ints=-2, rush_yds=1/10, rush_tds=6, fumbles=-2, rec_yds=1/20, rec_tds=6))
    which(x3[,'value']>20)
    ```
    ```
    There are 46 players that are worth more than $20.
    ```
    
    b.  How many quarterbacks (qb) are in the top 30? (1 point)
    ```{r}
    which(x3[,'pos']=='qb')[1:30]
    ```
    ```
    There are 12 quarterbacks in the top 30.
    ```
    

### Question 4 ###

**5 points**

This code makes a list of all functions in the base package:

```{r}
objs <- mget(ls("package:base"), inherits = TRUE)
funs <- Filter(is.function, objs)
```

Using this list, write code to answer these questions.

1. Which function has the most arguments? (3 points)
```{r}
numArgs <- matrix(0, nrow = length(funs), ncol = 2, dimnames = list(NULL, c("Function", "Arguments")))
numArgs <- as.data.frame(numArgs)
x1 <- lapply(funs, function(x) length(formals(x)))
numArgs[,1] <- names(x1)
numArgs[,2] <- as.integer(x1)
numArgs <- numArgs[order(numArgs[,'Arguments'], decreasing=TRUE),]
head(numArgs)
```
```
The scan function has 22 arguments.
```

2. How many functions have no arguments? (2 points)
```{r}
which(numArgs[,2]==0)
```
```
There are 225 functions with no arguments.
```