3-11_functions_practicals_answers.qmd

# Functions: Answers {.unnumbered}

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Make sure that you try the exercises yourself first before looking at the answers
:::

## Function Arguments

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### Question 1

Look at the help file for the function `mean()`.

How many arguments does the function have?

What types of vectors are accepted?

What is the default setting for dealing with NA values?

### Answer

Three arguments (plus further arguments).

Numerical and logical vectors are accepted.

The default setting is to NOT remove NA (missing) values.
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### Question 2

Use the function `mean()` to calculate the mean of the following values:

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note the `NA` and use named argument matching
:::

```{r, eval = F}
c(1, 2, NA, 6)
```

### Answer

```{r}
mean(x = c(1, 2, NA, 6), na.rm = TRUE)
```
:::

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### Question 3

Do Q2 again but rearrange the arguments.

### Answer

```{r}
mean(na.rm = TRUE, x = c(1, 2, NA, 6))
```
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### Question 4

Do Q2 again using positional matching.

### Answer

```{r}
mean(c(1, 2, NA, 6), 0, TRUE)
```
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### Question 5

Determine the class of `mean()` using `class()`.

### Answer

```{r}
class(mean)
```

The class is `function`.
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### Question 6

Determine the class of `mean()` using `str()`.

### Answer

```{r}
str(mean)
```

The class is `function`.
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### Question 7

Determine the class of the value output in Q4 using `class()`.

### Answer

```{r}
class(mean(c(1, 2, NA, 6), 0, TRUE))
```

The class is `numeric`
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### Question 8

Determine the class of the value output in Q4 using str().

### Answer

```{r}
str(mean(c(1, 2, NA, 6), 0, TRUE))
```

The `num` means the class is `numeric`.
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## Function environment and scoping

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### Question 9

For each of the following sets of commands, give the value that will be returned by the last command. Try to answer without using R.

a)  

```{r, eval=F}
w <- 5
f <- function(y) {
  return(w + y)
}
f(y = 2)
```

b)  

```{r, eval=F}
w <- 5
f <- function(y) {
  w <- 4
  return(w + y)
}
f(y = 2)
```

### Answer

a)  This will return 7 because `w` is 5 and we are evaluating the function at `y = 2`

b)  This will return 6 because `w` is reassigned as 4 inside the function and we are evaluating the function at `y = 2`.
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### Question 10

Among the variables `w`, `d`, and `y`, which are global to `f()` and which are local? What is the value of z when executing `f(w)`

```{r}
w <- 2
f <- function(y) {
  d <- 3
  h <- function(z) {
    return(z + d)
  }
  return(y * h(y))
}
```

### Answer

The object `w` is global to `f()` while `d` and `y` are local to `f()`.

`z` is 2, because it takes the value of `y` when executing `h(y)` in function `f()`, which takes the value of global variable `w` when executing `f(w)`
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### Question 11

Do the following in R:

a)  Try:

```{r, eval=F}
myFun1 <- function(a) {
  b <- 3
  myFun2(a)
}

myFun2 <- function(y) {
  return(y + a + b)
}

myFun1(10)
```

What happens?

b)  Now try:

```{r, eval=F}
a <- 1
b <- 2
myFun1(10)
```

What happens?

### Answer

a)  We get an error message because `a` and `b` are local to `myFun1` so the function `myFun2` can't find them in the global environment.

b)  We get get the value 13 because the values `a` and `b` are global so `myFun2` can find them and use them in its commands.
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## if(), else, and ifelse() and Vectorization

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### Question 12

Write a function called 'evenOrOdd' involving `if` and `else` that takes an argument `x` and returns "Even" or "Odd" depending on whether or not `x` is divisible by 2. (Do not use the ifelse() function).

### Anwser

```{r}
evenOrOdd <- function(x) {
  if(x %% 2 == 0) {
    return("Even")
  } else {
    return("Odd")
  }
}
```
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### Question 13

Is your function 'evenOrOdd' vectorized? Check by passing it the vector:

```{r}
w <- c(3, 6, 6, 4, 7, 9, 11, 6)
```

### Answer

```{r,error=T}
evenOrOdd(w)
```

An error is given, because x in `if(x %% 2 == 0)` is longer than 1.
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### Question 14

Another way to determine if each element of a vector is even or odd is to use the `ifelse()` function, which serves as a vectorized version `if` and `else`. Use `ifelse()` to obtain "Even" or "Odd" for each element of `w`.

### Answer

```{r}
ifelse(w %% 2 == 0, "Even", "Odd")
```
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## Terminating a function with returns, errors, and warnings

The functions `warning()` and `stop()` are used to print a warning message and to stop the execution of the function call and print an error message. For example:

```{r}
noNegMean <- function(x) {
  if(all(x < 0)) {
    stop("All values in x are negative")
  }
  
  if(any(x < 0)) {
    x[x < 0] <- 0
    warning("Negative values in x replaced by zero")
  }
  
  return(mean(x))
}
```

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### Question 15

Copy the above code and then pass `noNegMean()` a vector containing some negative and some positive values. What happens?

### Answer

```{r}
noNegMean(c(-1,0,1))
```

We get the warning message and it returned 0.3333, which is the average of `c(0, 0, 1)`.
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### Question 16

What happens when you pass `noNegMean()` a vector containing all negative values?

### Answer

```{r,error=T}
noNegMean(c(-1,-1,-1))
```

We get the error message and nothing is returned.
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### Question 17

Write a function `ratio()` that takes two arguments, `x` and `y`, and attempts to compute the ratio `x/y`.

If both `x == 0 & y == 0`, the function should stop and print an error message about dividing 0 by 0.

If `y == 0` (but not x), the function should print a warning message about dividing by 0, and then return `x/y` (which will be `Inf`).

In all other cases, it should return `x/y`.

Test your `ratio()` function first using two nonzero values for `x` and `y`, then using a nonzero `x` but `y = 0`, and finally using `x = 0` and `y = 0`.

### Answer

```{r}
ratio <- function(x,y) {
  if(x == 0 & y == 0) {
    stop("Cannot divide zero by zero.")
  }

  if(y == 0) {
    warning("Cannot divide by zero.")
  }

  ratio <- x/y
  return(ratio)
}
```

```{r, error=T}
ratio(2,3)

ratio(0,0)

ratio(1,0)
```
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## looping using for() loops and the apply functions

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### Question 18

Copy this is a function to determine if a number is a prime number:

```{r}
isPrime <- function(num){
  if (num == 2) {
    return(TRUE)
  }
  if(num > 1) {
    for(i in 2:(num-1)) {
      if ((num %% i) == 0) {
        return(FALSE)
      }
    }
  } else {
    return(FALSE)
  }
  
  return(TRUE)
}
```

Copy this matrix for which we would like to check if a number is a prime number:

```{r}
mat <- matrix(1:100, nrow=10)
```

Use the `apply()` function to calculate the prime number for each number in the matrix.

What numbers from 1 until 100 are prime numbers?

### Answer

```{r}
apply(mat, c(1,2), isPrime)
```

```{r}
mat[apply(mat, c(1,2), isPrime)]
```
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### Question 19

Copy the following command to create a list containing two generations of the famous Kennedy family:

```{r}
Kennedys <- list(
    JosephJr = character(0),
    John = c("Caroline", "JohnJr", "Patrick"),
    Rosemary = character(0),
    Kathleen = character(0),
    Eunice = c("RobertIII", "Maria", "Timothy", "Mark", "Anthony"),
    Patricia = c("Christopher", "Sydney", "Victoria", "Robin"),
    Robert = c("Kathleen", "JosephII", "RobertJr", "David", 
               "MaryC", "Michael", "MaryK", "Christopher", 
               "Matthew", "Douglas", "Rory"),
    Jean = c("Stephen", "William", "Amanda", "Kym"),
    Edward = c("Kara", "EdwardJr", "Patrick")
)
```

Use a `for()` loop to loop over the list of the first generation of Kennedys, keeping track of how many children each one has in a vector.

### Answer

```{r}
children <- NULL
for(i in Kennedys){
  children <- c(children, length(i))
}
children
```
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### Question 20

Now, using the `lapply()` function, loop over the list of the first generation of Kennedys and keep track of how many children each Kennedy has. What is the class of the output?

### Answer

```{r}
result <- lapply(Kennedys, length)
result

class(result)
```
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### Question 21

Answer Question 20 again using the `sapply()` function. What is the class of the output?

### Answer

```{r}
result <- sapply(Kennedys, length)
result

class(result)
```
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### Question 22

Load the "diamonds" dataset from the ggplot2 package by running `library(gglot2)` and calculate the average price of diamonds by color and clarity using the `tapply()` function.

### Answer

```{r}
library(ggplot2)
tapply(diamonds$price, list(diamonds$color, diamonds$clarity), mean)
```
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