Homework 1 Solutions

Ch. 3, Q1

x <- c("economics", "econometrics", "ECON 4750")

library(stringr)

str_length(x)

[1]  9 12  9

Ch. 3, Q2

Approach 1

sum_one_to_n_1 <- function(n) {
  one_to_n <- seq(1,n)
  sum(one_to_n)
}

Approach 2

sum_one_to_n_2 <- function(n) {
  out <- n*(n+1)/2
  out
}

Approach 3

sum_one_to_n_3 <- function(n) {
  out <- 0
  for (i in 1:n) {
    out <- out + i
  }
  out
}

Let’s check that they all give the same result.

sum_one_to_n_1(100)

[1] 5050

sum_one_to_n_2(100)

[1] 5050

sum_one_to_n_3(100)

[1] 5050

Ch. 3, Q3

Part a

fibonacci <- function(n) {

  # handle cases where n=1 or 2
  if (n == 1) {
    return(0)
  }

  if (n==2) {
    return(1)
  }

  # main code
  fib_seq <- c(0,1)

  for (i in 3:n) {
    fib_seq[i] <- fib_seq[i-1] + fib_seq[i-2]
  }

  fib_seq[n]
}

fibonacci(5)

[1] 3

fibonacci(8)

[1] 13

Part b

alt_seq <- function(a,b,n) {
  # handle cases where n=1 or 2
  if (n == 1) {
    return(a)
  }

  if (n==2) {
    return(b)
  }

  # main code
  this_seq <- c(a,b)

  for (i in 3:n) {
    this_seq[i] <- this_seq[i-1] + this_seq[i-2]
  }

  this_seq[n]
}

alt_seq(3,7,4)

[1] 17

Ch. 3, Q4

Part a

is_prime <- function(x) {
  # handle 1 (which is not prime)
  if (x == 1) {
    return(FALSE)
  }

  # given code below, 2 and 3 are also edge cases that we should handle
  # explicitly (both are prime)
  if (x == 2 | x == 3) {
    return(TRUE)
  }

  # check if x is divisible by any number from 2 to the square root of x
  # (above the square root of x, we would have already caught that it factors)
  for (i in 2:floor(sqrt(x))) {
    if (x %% i == 0) {
      # if we find a factor, it's not prime and we can immediately
      # return FALSE
      return(FALSE)
    }
  }
  # if we didn't find any factors, then x is prime
  return(TRUE)
}

As some examples, consider

is_prime(7)

[1] TRUE

is_prime(10)

[1] FALSE

Part b

prime <- function(n) {
  primes <- c()
  for (i in 1:n) {
    if (is_prime(i)) {
      primes <- c(primes, i)
    }
  }
  primes
}

Check that this works

prime(100)

 [1]  2  3  5  7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Ch. 3, Q5

Part a

nrow(iris)

[1] 150

Part b

mean(iris$Sepal.Length)

[1] 5.843333

Part c

mean(subset(iris, Species=="setosa")$Sepal.Width)

[1] 3.428

Part d

sorted_iris <- iris[order(iris$Petal.Length),]
sorted_iris[1:10,]

   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
23          4.6         3.6          1.0         0.2  setosa
14          4.3         3.0          1.1         0.1  setosa
15          5.8         4.0          1.2         0.2  setosa
36          5.0         3.2          1.2         0.2  setosa
3           4.7         3.2          1.3         0.2  setosa
17          5.4         3.9          1.3         0.4  setosa
37          5.5         3.5          1.3         0.2  setosa
39          4.4         3.0          1.3         0.2  setosa
41          5.0         3.5          1.3         0.3  setosa
42          4.5         2.3          1.3         0.3  setosa

Ch. 3, Q6

quadratic_solver <- function(a, b, c) {
  neg_root <- (-b - sqrt(b^2 - 4*a*c)) / (2*a)
  pos_root <- (-b + sqrt(b^2 - 4*a*c)) / (2*a)
  out <- list(neg_root = neg_root, pos_root = pos_root)
  out
}

quadratic_solver(1,4,3)

$neg_root
[1] -3

$pos_root
[1] -1