# 9 Programs

In this chapter, you will build a real, working slot machine that you can play by running an R function. When you’re finished, you’ll be able to play it like this:

play()
## 0 0 DD
## $0 play() ## 7 7 7 ##$80

The play function will need to do two things. First, it will need to randomly generate three symbols; and, second, it will need to calculate a prize based on those symbols.

The first step is easy to simulate. You can randomly generate three symbols with the sample function—just like you randomly “rolled” two dice in Project 1: Weighted Dice. The following function generates three symbols from a group of common slot machine symbols: diamonds (DD), sevens (7), triple bars (BBB), double bars (BB), single bars (B), cherries (C), and zeroes (0). The symbols are selected randomly, and each symbol appears with a different probability:

get_symbols <- function() {
wheel <- c("DD", "7", "BBB", "BB", "B", "C", "0")
sample(wheel, size = 3, replace = TRUE,
prob = c(0.03, 0.03, 0.06, 0.1, 0.25, 0.01, 0.52))
}

You can use get_symbols to generate the symbols used in your slot machine:

get_symbols()
## "BBB" "0"   "C"

get_symbols()
## "0" "0" "0"

get_symbols()
## "7" "0" "B"

get_symbols uses the probabilities observed in a group of video lottery terminals from Manitoba, Canada. These slot machines became briefly controversial in the 1990s, when a reporter decided to test their payout rate. The machines appeared to pay out only 40 cents on the dollar, even though the manufacturer claimed they would pay out 92 cents on the dollar. The original data collected on the machines and a description of the controversy is available online in a journal article by W. John Braun. The controversy died down when additional testing showed that the manufacturer was correct.

The Manitoba slot machines use the complicated payout scheme shown in Table 9.1. A player will win a prize if he gets:

• Three of the same type of symbol (except for three zeroes)
• Three bars (of mixed variety)
• One or more cherries

Otherwise, the player receives no prize.

The monetary value of the prize is determined by the exact combination of symbols and is further modified by the presence of diamonds. Diamonds are treated like “wild cards,” which means they can be considered any other symbol if it would increase a player’s prize. For example, a player who rolls 7 7 DD would earn a prize for getting three sevens. There is one exception to this rule, however: a diamond cannot be considered a cherry unless the player also gets one real cherry. This prevents a dud roll like, 0 DD 0 from being scored as 0 C 0.

Diamonds are also special in another way. Every diamond that appears in a combination doubles the amount of the final prize. So 7 7 DD would actually be scored higher than 7 7 7. Three sevens would earn you $80, but two sevens and a diamond would earn you$160. One seven and two diamonds would be even better, resulting in a prize that has been doubled twice, or $320. A jackpot occurs when a player rolls DD DD DD. Then a player earns$100 doubled three times, which is $800. Table 9.1: Each play of the slot machine costs$1. A player’s symbols determine how much they win. Diamonds (DD) are wild, and each diamond doubles the final prize. * = any symbol.
Combination Prize($) DD DD DD 100 7 7 7 80 BBB BBB BBB 40 BB BB BB 25 B B B 10 C C C 10 Any combination of bars 5 C C * 5 C * C 5 * C C 5 C * * 2 * C * 2 * * C 2 To create your play function, you will need to write a program that can take the output of get_symbols and calculate the correct prize based on Table 9.1. In R, programs are saved either as R scripts or as functions. We’ll save your program as a function named score. When you are finished, you will be able to use score to calculate a prize like this: score(c("DD", "DD", "DD")) ## 800 After that it will be easy to create the full slot machine, like this: play <- function() { symbols <- get_symbols() print(symbols) score(symbols) } The print command prints its output to the console window, which makes print a useful way to display messages from within the body of a function. You may notice that play calls a new function, print. This will help play display the three slot machine symbols, since they do not get returned by the last line of the function. The print command prints its output to the console window – even if R calls it from within a function. In Project 1: Weighted Dice, I encouraged you to write all of your R code in an R script, a text file where you can compose and save code. That advice will become very important as you work through this chapter. Remember that you can open an R script in RStudio by going to the menu bar and clicking on File > New File > R Script. ## 9.1 Strategy Scoring slot-machine results is a complex task that will require a complex algorithm. You can make this, and other coding tasks, easier by using a simple strategy: • Break complex tasks into simple subtasks. • Use concrete examples. • Describe your solutions in English, then convert them to R. Let’s start by looking at how you can divide a program into subtasks that are simple to work with. A program is a set of step-by-step instructions for your computer to follow. Taken together, these instructions may accomplish something very sophisticated. Taken apart, each individual step will likely be simple and straightforward. You can make coding easier by identifying the individual steps or subtasks within your program. You can then work on each subtask separately. If a subtask seems complicated, try to divide it again into even subtasks that are even more simple. You can often reduce an R program into substasks so simple that each can be performed with a preexisting function. R programs contain two types of subtasks: sequential steps and parallel cases. ### 9.1.1 Sequential Steps One way to subdivide a program is into a series of sequential steps. The play function takes the approach, shown in Figure 9.1. First, it generates three symbols (step 1), then it displays them in the console window (step 2), and then it scores them (step 3): play <- function() { # step 1: generate symbols symbols <- get_symbols() # step 2: display the symbols print(symbols) # step 3: score the symbols score(symbols) } To have R execute steps in sequence, place the steps one after another in an R script or function body. Figure 9.1: The play function uses a series of steps. ### 9.1.2 Parallel Cases Another way to divide a task is to spot groups of similar cases within the task. Some tasks require different algorithms for different groups of input. If you can identify those groups, you can work out their algorithms one at a time. For example, score will need to calculate the prize one way if symbols contains three of a kind (In that case, score will need to match the common symbol to a prize). score will need to calculate the prize a second way if the symbols are all bars (In that case, score can just assign a prize of$5). And, finally, score will need to calculate the prize in a third way if the symbols do not contain three of a kind or all bars (In that case, score must count the number of cherries present). score will never use all three of these algorithms at once; it will always choose just one algorithm to run based on the combination of symbols.

Diamonds complicate all of this because diamonds can be treated as wild cards. Let’s ignore that for now and focus on the simpler case where diamonds double the prize but are not wilds. score can double the prize as necessary after it runs one of the following algorithms, as shown in Figure 9.2.

Adding the score cases to the play steps reveals a strategy for the complete slot machine program, as shown in Figure 9.3.

We’ve already solved the first few steps in this strategy. Our program can get three slot machine symbols with the get_symbols function. Then it can display the symbols with the print function. Now let’s examine how the program can handle the parallel score cases. Figure 9.2: The score function must distinguish between parallel cases. Figure 9.3: The complete slot machine simulation will involve subtasks that are arranged both in series and in parallel.

## 9.2 if Statements

Linking cases together in parallel requires a bit of structure; your program faces a fork in the road whenever it must choose between cases. You can help the program navigate this fork with an if statement.

An if statement tells R to do a certain task for a certain case. In English you would say something like, “If this is true, do that.” In R, you would say:

if (this) {
that
}

The this object should be a logical test or an R expression that evaluates to a single TRUE or FALSE. If this evaluates to TRUE, R will run all of the code that appears between the braces that follow the if statement (i.e., between the { and } symbols). If this evaluates to FALSE, R will skip the code between the braces without running it.

For example, you could write an if statement that ensures some object, num, is positive:

if (num < 0) {
num <- num * -1
}

If num < 0 is TRUE, R will multiply num by negative one, which will make num positive:

num <- -2

if (num < 0) {
num <- num * -1
}

num
## 2

If num < 0 is FALSE, R will do nothing and num will remain as it is—positive (or zero):

num <- 4

if (num < 0) {
num <- num * -1
}

num
## 4

The condition of an if statement must evaluate to a single TRUE or FALSE. If the condition creates a vector of TRUEs and FALSEs (which is easier to make than you may think), your if statement will print a warning message and use only the first element of the vector. Remember that you can condense vectors of logical values to a single TRUE or FALSE with the functions any and all.

You don’t have to limit your if statements to a single line of code; you can include as many lines as you like between the braces. For example, the following code uses many lines to ensure that num is positive. The additional lines print some informative statements if num begins as a negative number. R will skip the entire code block—print statements and all—if num begins as a positive number:

num <- -1

if (num < 0) {
print("num is negative.")
print("Don't worry, I'll fix it.")
num <- num * -1
print("Now num is positive.")
}
## "num is negative."
## "Don't worry, I'll fix it."
## "Now num is positive."

num
## 1

Try the following quizzes to develop your understanding of if statements.

Exercise 9.1 (Quiz A) What will this return?
x <- 1
if (3 == 3) {
x <- 2
}
x
Solution. The code will return the number 2. x begins as 1, and then R encounters the if statement. Since the condition evaluates to TRUE, R will run x <- 2, changing the value of x.
Exercise 9.2 (Quiz B) What will this return?
x <- 1
if (TRUE) {
x <- 2
}
x
Solution. This code will also return the number 2. It works the same as the code in Quiz A, except the condition in this statement is already TRUE. R doesn’t even need to evaluate it. As a result, the code inside the if statement will be run, and x will be set to 2.
Exercise 9.3 (Quiz C) What will this return?
x <- 1
if (x == 1) {
x <- 2
if (x == 1) {
x <- 3
}
}
x
Solution. Once again, the code will return the number 2. x starts out as 1, and the condition of the first if statement will evaluate to TRUE, which causes R to run the code in the body of the if statement. First, R sets x equal to 2, then R evaluates the second if statement, which is in the body of the first. This time x == 1 will evaluate to FALSE because x now equals 2. As a result, R ignores x <- 3 and exits both if statements.

## 9.3 else Statements

if statements tell R what to do when your condition is true, but you can also tell R what to do when the condition is false. else is a counterpart to if that extends an if statement to include a second case. In English, you would say, “If this is true, do plan A; else do plan B.” In R, you would say:

if (this) {
Plan A
} else {
Plan B
}

When this evaluates to TRUE, R will run the code in the first set of braces, but not the code in the second. When this evaluates to FALSE, R will run the code in the second set of braces, but not the first. You can use this arrangement to cover all of the possible cases. For example, you could write some code that rounds a decimal to the nearest integer.

a <- 3.14

Then isolate the decimal component with trunc:

dec <- a - trunc(a)
dec
## 0.14
trunc takes a number and returns only the portion of the number that appears to the left of the decimal place (i.e., the integer part of the number).
a - trunc(a) is a convenient way to return the decimal part of a.

Then use an if else tree to round the number (either up or down):

if (dec >= 0.5) {
a <- trunc(a) + 1
} else {
a <- trunc(a)
}

a
## 3

If your situation has more than two mutually exclusive cases, you can string multiple if and else statements together by adding a new if statement immediately after else. For example:

a <- 1
b <- 1

if (a > b) {
print("A wins!")
} else if (a < b) {
print("B wins!")
} else {
print("Tie.")
}
## "Tie."

R will work through the if conditions until one evaluates to TRUE, then R will ignore any remaining if and else clauses in the tree. If no conditions evaluate to TRUE, R will run the final else statement.

If two if statements describe mutually exclusive events, it is better to join the if statements with an else if than to list them separately. This lets R ignore the second if statement whenever the first returns a TRUE, which saves work.

You can use if and else to link the subtasks in your slot-machine function. Open a fresh R script, and copy this code into it. The code will be the skeleton of our final score function. Compare it to the flow chart for score in Figure 9.2:

if ( # Case 1: all the same <1>) {
prize <- # look up the prize <3>
} else if ( # Case 2: all bars <2> ) {
prize <- # assign $5 <4> } else { # count cherries <5> prize <- # calculate a prize <7> } # count diamonds <6> # double the prize if necessary <8> Our skeleton is rather incomplete; there are many sections that are just code comments instead of real code. However, we’ve reduced the program to eight simple subtasks: <1> - Test whether the symbols are three of a kind. <2> - Test whether the symbols are all bars. <3> - Look up the prize for three of a kind based on the common symbol. <4> - Assign a prize of$5.
<5> - Count the number of cherries.
<6> - Count the number of diamonds.
<7> - Calculate a prize based on the number of cherries.
<8> - Adjust the prize for diamonds.

If you like, you can reorganize your flow chart around these tasks, as in Figure 9.4. The chart will describe the same strategy, but in a more precise way. I’ll use a diamond shape to symbolize an if else decision. Figure 9.4: score can navigate three cases with two if else decisions. We can also break some of our tasks into two steps.

Now we can work through the subtasks one at a time, adding R code to the if tree as we go. Each subtask will be easy to solve if you set up a concrete example to work with and try to describe a solution in English before coding in R.

The first subtask asks you to test whether the symbols are three of a kind. How should you begin writing the code for this subtask?

You know that the final score function will look something like this:

score <- function(symbols) {

# calculate a prize

prize
}

Its argument, symbols, will be the output of get_symbols, a vector that contains three character strings. You could start writing score as I have written it, by defining an object named score and then slowly filling in the body of the function. However, this would be a bad idea. The eventual function will have eight separate parts, and it will not work correctly until all of those parts are written (and themselves work correctly). This means you would have to write the entire score function before you could test any of the subtasks. If score doesn’t work—which is very likely—you will not know which subtask needs fixed.

You can save yourself time and headaches if you focus on one subtask at a time. For each subtask, create a concrete example that you can test your code on. For example, you know that score will need to work on a vector named symbols that contains three character strings. If you make a real vector named symbols, you can run the code for many of your subtasks on the vector as you go:

symbols <- c("7", "7", "7")

If a piece of code does not work on symbols, you will know that you need to fix it before you move on. You can change the value of symbols from subtask to subtask to ensure that your code works in every situation:

symbols <- c("B", "BB", "BBB")
symbols <- c("C", "DD", "0")

Only combine your subtasks into a score function once each subtask works on a concrete example. If you follow this plan, you will spend more time using your functions and less time trying to figure out why they do not work.

After you set up a concrete example, try to describe how you will do the subtask in English. The more precisely you can describe your solution, the easier it will be to write your R code.

Our first subtask asks us to “test whether the symbols are three of a kind.” This phrase does not suggest any useful R code to me. However, I could describe a more precise test for three of a kind: three symbols will be the same if the first symbol is equal to the second and the second symbol is equal to the third. Or, even more precisely:

A vector named symbols will contain three of the same symbol if the first element of symbols is equal to the second element of symbols and the second element of symbols is equal to the third element of symbols.

Exercise 9.4 (Write a Test) Turn the preceding statement into a logical test written in R. Use your knowledge of logical tests, Boolean operators, and subsetting from R Notation. The test should work with the vector symbols and return a TRUE if and only if each element in symbols is the same. Be sure to test your code on symbols.
Solution. Here are a couple of ways to test that symbols contains three of the same symbol. The first method parallels the English suggestion above, but there are other ways to do the same test. There is no right or wrong answer, so long as your solution works, which is easy to check because you’ve created a vector named symbols:
symbols
##  "7" "7" "7"

symbols == symbols & symbols == symbols
## TRUE

symbols == symbols & symbols == symbols
## TRUE

all(symbols == symbols)
## TRUE

As your vocabulary of R functions broadens, you’ll think of more ways to do basic tasks. One method that I like for checking three of a kind is:

length(unique(symbols) == 1)

The unique function returns every unique term that appears in a vector. If your symbols vector contains three of a kind (i.e., one unique term that appears three times), then unique(symbols) will return a vector of length 1.

Now that you have a working test, you can add it to your slot-machine script:

same <- symbols == symbols && symbols == symbols

if (same) {
prize <- # look up the prize
} else if ( # Case 2: all bars ) {
prize <- # assign $5 } else { # count cherries prize <- # calculate a prize } # count diamonds # double the prize if necessary && and || behave like & and | but can sometimes be more efficient. The double operators will not evaluate the second test in a pair of tests if the first test makes the result clear. For example, if symbols does not equal symbols in the next expression, && will not evaluate symbols == symbols; it can immediately return a FALSE for the whole expression (because FALSE & TRUE and FALSE & FALSE both evaluate to FALSE). This efficiency can speed up your programs; however, double operators are not appropriate everywhere. && and || are not vectorized, which means they can only handle a single logical test on each side of the operator. The second prize case occurs when all the symbols are a type of bar, for example, B, BB, and BBB. Let’s begin by creating a concrete example to work with: symbols <- c("B", "BBB", "BB") Exercise 9.5 (Test for All Bars) Use R’s logical and Boolean operators to write a test that will determine whether a vector named symbols contains only symbols that are a type of bar. Check whether your test works with our example symbols vector. Remember to describe how the test should work in English, and then convert the solution to R. Solution. As with many things in R, there are multiple ways to test whether symbols contains all bars. For example, you could write a very long test that uses multiple Boolean operators, like this: symbols == "B" | symbols == "BB" | symbols == "BBB" & symbols == "B" | symbols == "BB" | symbols == "BBB" & symbols == "B" | symbols == "BB" | symbols == "BBB" ## TRUE However, this is not a very efficient solution, because R has to run nine logical tests (and you have to type them). You can often replace multiple | operators with a single %in%. Also, you can check that a test is true for each element in a vector with all. These two changes shorten the preceding code to: all(symbols %in% c("B", "BB", "BBB")) ## TRUE Let’s add this code to our script: same <- symbols == symbols && symbols == symbols bars <- symbols %in% c("B", "BB", "BBB") if (same) { prize <- # look up the prize } else if (all(bars)) { prize <- # assign$5
} else {
# count cherries
prize <- # calculate a prize
}

# count diamonds
# double the prize if necessary

You may have noticed that I split this test up into two steps, bars and all(bars). That’s just a matter of personal preference. Wherever possible, I like to write my code so it can be read with function and object names conveying what they do.

You also may have noticed that our test for Case 2 will capture some symbols that should be in Case 1 because they contain three of a kind:

symbols <- c("B", "B", "B")
all(symbols %in% c("B", "BB", "BBB"))
## TRUE

That won’t be a problem, however, because we’ve connected our cases with else if in the if tree. As soon as R comes to a case that evaluates to TRUE, it will skip over the rest of the tree. Think of it this way: each else tells R to only run the code that follows it if none of the previous conditions have been met. So when we have three of the same type of bar, R will evaluate the code for Case 1 and then skip the code for Case 2 (and Case 3).

Our next subtask is to assign a prize for symbols. When the symbols vector contains three of the same symbol, the prize will depend on which symbol there are three of. If there are three DDs, the prize will be $100; if there are three 7s, the prize will be$80; and so on.

This suggests another if tree. You could assign a prize with some code like this:

if (same) {
symbol <- symbols
if (symbol == "DD") {
prize <- 800
} else if (symbol == "7") {
prize <- 80
} else if (symbol == "BBB") {
prize <- 40
} else if (symbol == "BB") {
prize <- 5
} else if (symbol == "B") {
prize <- 10
} else if (symbol == "C") {
prize <- 10
} else if (symbol == "0") {
prize <- 0
}
}

While this code will work, it is a bit long to write and read, and it may require R to perform multiple logical tests before delivering the correct prize. We can do better with a different method.

## 9.4 Lookup Tables

Very often in R, the simplest way to do something will involve subsetting. How could you use subsetting here? Since you know the exact relationship between the symbols and their prizes, you can create a vector that captures this information. This vector can store symbols as names and prize values as elements:

payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
payouts
##  DD   7 BBB  BB   B   C   0
## 100  80  40  25  10  10   0 

Now you can extract the correct prize for any symbol by subsetting the vector with the symbol’s name:

payouts["DD"]
##  DD
## 100

payouts["B"]
##  B
## 10

If you want to leave behind the symbol’s name when subsetting, you can run the unname function on the output:

unname(payouts["DD"])
## 100 

unname returns a copy of an object with the names attribute removed.

payouts is a type of lookup table, an R object that you can use to look up values. Subsetting payouts provides a simple way to find the prize for a symbol. It doesn’t take many lines of code, and it does the same amount of work whether your symbol is DD or 0. You can create lookup tables in R by creating named objects that can be subsetted in clever ways.

Sadly, our method is not quite automatic; we need to tell R which symbol to look up in payouts. Or do we? What would happen if you subsetted payouts by symbols? Give it a try:

symbols <- c("7", "7", "7")
symbols
## "7"

payouts[symbols]
##  7
## 80

symbols <- c("C", "C", "C")
payouts[symbols]
##  C
## 10 

You don’t need to know the exact symbol to look up because you can tell R to look up whichever symbol happens to be in symbols. You can find this symbol with symbols, symbols, or symbols, because each contains the same symbol in this case. You now have a simple automated way to calculate the prize when symbols contains three of a kind. Let’s add it to our code and then look at Case 2:

same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- # assign $5 } else { # count cherries prize <- # calculate a prize } # count diamonds # double the prize if necessary Case 2 occurs whenever the symbols are all bars. In that case, the prize will be$5, which is easy to assign:

same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- 5
} else {
# count cherries
prize <- # calculate a prize
}

# count diamonds
# double the prize if necessary

Now we can work on the last case. Here, you’ll need to know how many cherries are in symbols before you can calculate a prize.

Exercise 9.6 (Find C’s) How can you tell which elements of a vector named symbols are a C? Devise a test and try it out.

Challenge

How might you count the number of Cs in a vector named symbols? Remember R’s coercion rules.
Solution. As always, let’s work with a real example:
symbols <- c("C", "DD", "C")

One way to test for cherries would be to check which, if any, of the symbols are a C:

symbols == "C"
## TRUE FALSE  TRUE

It’d be even more useful to count how many of the symbols are cherries. You can do this with sum, which expects numeric input, not logical. Knowing this, R will coerce the TRUEs and FALSEs to 1s and 0s before doing the summation. As a result, sum will return the number of TRUEs, which is also the number of cherries:

sum(symbols == "C")
## 2

You can use the same method to count the number of diamonds in symbols:

sum(symbols == "DD")
## 1

Let’s add both of these subtasks to the program skeleton:

same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- 5
} else {
cherries <- sum(symbols == "C")
prize <- # calculate a prize
}

diamonds <- sum(symbols == "DD")
# double the prize if necessary

Since Case 3 appears further down the if tree than Cases 1 and 2, the code in Case 3 will only be applied to players that do not have three of a kind or all bars. According to the slot machine’s payout scheme, these players will win $5 if they have two cherries and$2 if they have one cherry. If the player has no cherries, she gets a prize of $0. We don’t need to worry about three cherries because that outcome is already covered in Case 1. As in Case 1, you could write an if tree that handles each combination of cherries, but just like in Case 1, this would be an inefficient solution: if (cherries == 2) { prize <- 5 } else if (cherries == 1) { prize <- 2 } else {} prize <- 0 } Again, I think the best solution will involve subsetting. If you are feeling ambitious, you can try to work this solution out on your own, but you will learn just as quickly by mentally working through the following proposed solution. We know that our prize should be$0 if we have no cherries, $2 if we have one cherry, and$5 if we have two cherries. You can create a vector that contains this information. This will be a very simple lookup table:

c(0, 2, 5)

Now, like in Case 1, you can subset the vector to retrieve the correct prize. In this case, the prize’s aren’t identified by a symbol name, but by the number of cherries present. Do we have that information? Yes, it is stored in cherries. We can use basic integer subsetting to get the correct prize from the prior lookup table, for example, c(0, 2, 5).

cherries isn’t exactly suited for integer subsetting because it could contain a zero, but that’s easy to fix. We can subset with cherries + 1. Now when cherries equals zero, we have:

cherries + 1
## 1

c(0, 2, 5)[cherries + 1]
## 0

When cherries equals one, we have:

cherries + 1
## 2

c(0, 2, 5)[cherries + 1]
## 2

And when cherries equals two, we have:

cherries + 1
## 3

c(0, 2, 5)[cherries + 1]
## 5

Examine these solutions until you are satisfied that they return the correct prize for each number of cherries. Then add the code to your script, as follows:

same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- 5
} else {
cherries <- sum(symbols == "C")
prize <- c(0, 2, 5)[cherries + 1]
}

diamonds <- sum(symbols == "DD")
# double the prize if necessary

Lookup Tables Versus if Trees

This is the second time we’ve created a lookup table to avoid writing an if tree. Why is this technique helpful and why does it keep appearing? Many if trees in R are essential. They provide a useful way to tell R to use different algorithms in different cases. However, if trees are not appropriate everywhere.

if trees have a couple of drawbacks. First, they require R to run multiple tests as it works its way down the if tree, which can create unnecessary work. Second, as you’ll see in Speed, it can be difficult to use if trees in vectorized code, a style of code that takes advantage of R’s programming strengths to create fast programs. Lookup tables do not suffer from either of these drawbacks.

You won’t be able to replace every if tree with a lookup table, nor should you. However, you can usually use lookup tables to avoid assigning variables with if trees. As a general rule, use an if tree if each branch of the tree runs different code. Use a lookup table if each branch of the tree only assigns a different value.

To convert an if tree to a lookup table, identify the values to be assigned and store them in a vector. Next, identify the selection criteria used in the conditions of the if tree. If the conditions use character strings, give your vector names and use name-based subsetting. If the conditions use integers, use integer-based subsetting.

The final subtask is to double the prize once for every diamond present. This means that the final prize will be some multiple of the current prize. For example, if no diamonds are present, the prize will be:

prize * 1      # 1 = 2 ^ 0

If one diamond is present, it will be:

prize * 2      # 2 = 2 ^ 1

If two diamonds are present, it will be:

prize * 4      # 4 = 2 ^ 2

And if three diamonds are present, it will be:

prize * 8      # 8 = 2 ^ 3

Can you think of an easy way to handle this? How about something similar to these examples?

Exercise 9.7 (Adjust for Diamonds) Write a method for adjusting prize based on diamonds. Describe a solution in English first, and then write your code.
Solution. Here is a concise solution inspired by the previous pattern. The adjusted prize will equal:
prize * 2 ^ diamonds

which gives us our final score script:

same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- 5
} else {
cherries <- sum(symbols == "C")
prize <- c(0, 2, 5)[cherries + 1]
}

diamonds <- sum(symbols == "DD")
prize * 2 ^ diamonds

## 9.5 Code Comments

You now have a working score script that you can save to a function. Before you save your script, though, consider adding comments to your code with a #. Comments can make your code easier to understand by explaining why the code does what it does. You can also use comments to break long programs into scannable chunks. For example, I would include three comments in the score code:

# identify case
same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

# get prize
if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- 5
} else {
cherries <- sum(symbols == "C")
prize <- c(0, 2, 5)[cherries + 1]
}

# adjust for diamonds
diamonds <- sum(symbols == "DD")
prize * 2 ^ diamonds

Now that each part of your code works, you can wrap it into a function with the methods you learned in Writing Your Own Functions. Either use RStudio’s Extract Function option in the menu bar under Code, or use the function function. Ensure that the last line of the function returns a result (it does), and identify any arguments used by your function. Often the concrete examples that you used to test your code, like symbols, will become the arguments of your function. Run the following code to start using the score function:

score <- function (symbols) {
# identify case
same <- symbols == symbols && symbols == symbols
bars <- symbols %in% c("B", "BB", "BBB")

# get prize
if (same) {
payouts <- c("DD" = 100, "7" = 80, "BBB" = 40, "BB" = 25,
"B" = 10, "C" = 10, "0" = 0)
prize <- unname(payouts[symbols])
} else if (all(bars)) {
prize <- 5
} else {
cherries <- sum(symbols == "C")
prize <- c(0, 2, 5)[cherries + 1]
}

# adjust for diamonds
diamonds <- sum(symbols == "DD")
prize * 2 ^ diamonds
}

Once you have defined the score function, the play function will work as well:

play <- function() {
symbols <- get_symbols()
print(symbols)
score(symbols)
}

Now it is easy to play the slot machine:

play()
## "0"  "BB" "B"
## 0

play()
## "DD"  "0" "B"
## 0

play()
## "BB" "BB" "B"
## 25

## 9.6 Summary

An R program is a set of instructions for your computer to follow that has been organized into a sequence of steps and cases. This may make programs seem simple, but don’t be fooled: you can create complicated results with the right combination of simple steps (and cases).

As a programmer, you are more likely to be fooled in the opposite way. A program may seem impossible to write when you know that it must do something impressive. Do not panic in these situations. Divide the job before you into simple tasks, and then divide the tasks again. You can visualize the relationship between tasks with a flow chart if it helps. Then work on the subtasks one at a time. Describe solutions in English, then convert them to R code. Test each solution against concrete examples as you go. Once each of your subtasks works, combine your code into a function that you can share and reuse.

R provides tools that can help you do this. You can manage cases with if and else statements. You can create a lookup table with objects and subsetting. You can add code comments with #. And you can save your programs as a function with function.

Things often go wrong when people write programs. It will be up to you to find the source of any errors that occur and to fix them. It should be easy to find the source of your errors if you use a stepwise approach to writing functions, writing—and then testing—one bit at a time. However, if the source of an error eludes you, or you find yourself working with large chunks of untested code, consider using R’s built in debugging tools, described in Debugging R Code.

The next two chapters will teach you more tools that you can use in your programs. As you master these tools, you will find it easier to write R programs that let you do whatever you wish to your data. In S3, you will learn how to use R’s S3 system, an invisible hand that shapes many parts of R. You will use the system to build a custom class for your slot machine output, and you will tell R how to display objects that have your class.