Instructions
- Problem sets for the individual exercise is
here
- Upload your individual exercise to your
GitHub repo by Wed 11:59pm.
- The answer key is here in the following code.
# 1. Functions ------------------------------------------------------------
# What is a function that displays all the objects currently stored in
# the memory? Write it below after deleting the line that reads
# "WRITE YOUR ANSWER (code) HERE", and execute it.
ls()
## character(0)
# Create a new object named x5 that is a number 100.
x5 <- 100
# Calculate the square root of x5 using the sqrt() function
sqrt(x5)
## [1] 10
# Calculate the square root of x5 by raising it to the power of 0.5.
# Your numeric answer should be exactly the same as when you used the
# sqrt() function. This is because taking the sqaure root of something
# is equivalent to raising it to the power of 0.5.
x5^0.5
## [1] 10
# Create an object called x6 that is equal to 31.8734.
x6 <- 31.8734
# Use the round() function to get the value of x6 rounded off to
# three decimal places
round(x6, digits = 3)
## [1] 31.873
# Functions floor() and ceiling() can also be used to trim a number
# down to an integer: apply both of these functions to x6 and compare
# the outputs. Can you guess what these functions do?
floor(x6)
## [1] 31
ceiling(x6)
## [1] 32
# The floor function returns the largest integer not greater than the
# number provided.
# The ceiling function returns the smallest integer not less than the
# number provided.
# To find out if your hunch was right, refer to the help file of these
# functions. Write a code to open up the help file for the floor function.
?floor
# 2. Vectors --------------------------------------------------------------
# Create an object called "vec.a" which is a vector consisting of
# the numbers, 1, 3, 5, 7. You need to use the c function.
vec.a <- c(1,3,5,7)
# Create a vector called "vec.b" consisting of the numbers, 2, 4, 6, 8.
vec.b <- c(2,4,6,8)
# Subtract vec.b from vec.a
vec.a - vec.b
## [1] -1 -1 -1 -1
# Create a new vector called vec.c by multiplying vec.a by vector vec.b
vec.c <- vec.a * vec.b
# Create a new vector called vec.d by taking the square root of each
# member of vec.c
vec.d <- sqrt(vec.c)
# What is the third element of the vec.d vector? Find out using
# square bracket. Note that since this is a vector, you only need to
# provide a single number inside the brackets.
vec.d[3]
## [1] 5.477226
# Create a new vector called vec.e consisting of all the integers
# from 1 through 100. You should use the seq function, rather than writing
# down all the 100 integers individually.
vec.e <- seq(from = 1, to = 100)
# The mean function calculates the arithmetic mean of the numbers stored
# in an object. Using the mean function, calculate the mean of the vec.e vector.
mean(vec.e)
## [1] 50.5
# As we saw in the joint exercise, the sum function calculates the sum of all
# the elements in an object. Calculate the sum of the vec.e vector.
sum(vec.e)
## [1] 5050
# The length function returns the number of elements stored in an object.
# Using the length function, find the number of elements stored in the vec.e
# vector.
length(vec.e)
## [1] 100
# The mean of an object can be obtained by sum(X)/length(X) because
# the defininition of the mean is the sum of elements divided by the number of
# elements. Now, using the sum and length functions, calculate the mean of
# the vec.e vector. Compare the answer with that obtained with the mean function
sum(vec.e)/length(vec.e)
## [1] 50.5
# We have used the seq function in the joint and individual exercises.
# In doing so, we have used the from and to arguments so far. There are
# other arguments we can provide. First, the by argument specifies
# an increment. For example,
seq(from = 0, to = 10, by = 2)
## [1] 0 2 4 6 8 10
# This creates a sequence that starts from 0 and ends with 10, and with
# an increment of 2.
# Now, create a new object called olympic which is a sequence that
# starts from 1896 and ends with 2012, with an increment of 4.
olympic <- seq(from = 1896, to = 2012, by = 4)
# How many elements does the olympic vector contain? That is, what is
# the length of this vector? Find out by applying a function (not by
# manually counting the number of elements).
length(olympic)
## [1] 30
# So there are 30 elements in the olympic vector. Display all the
# elements contained in the olympic vector. These are the years
# where olympic games were (supposed to be) held. Display the
# contents of the olympic vector.
olympic
## [1] 1896 1900 1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952
## [16] 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012
# Find out how many olympic games will have been held by the year
# 2400. Use the length and seq functions.
length(seq(from = 1896, to = 2400, by = 4))
## [1] 127
# 127 olympic games.
# 3. Matrices -------------------------------------------------------------
# Create a new vector called "v1" consisting of the following numbers:
# 1, 3, 5, 7, 9, 11
v1 <- c(1,3,5,7,9,11)
# Find out the length of this vector (Don't count the numbers by hand;
# use an appropriate function).
length(v1)
## [1] 6
# We will convert this vector into a matrix. That is, we will rearrange this
# vector so that it will have two dimensions (rows and columns).
# Since this vector has 6 numbers, if we want the matrix to have two
# rows, how many columns will there be?
# 3 columns ( = 6/2)
# Create a matrix called mat.v using the following command:
# matrix(data = v1, nrow = 2)
mat.v <- matrix(data = v1, nrow = 2)
# Take a look at the content of this matrix.
# How many columns are there?
mat.v
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 3 7 11
# Notice how the numbers in vec.v are used to fill up the cells of mat.v.
# We can see that R did it "by column". That is, R first filled up the
# first column of mat.v with the first two elements of vec.v, then moved
# on to the second and third columns.
# You can use the byrow argument to change this. This argument takes
# one of two values, TRUE or FALSE (or T or F). That is, we write
# matrix(data = v1, nrow = 2, byrow = TRUE)
# Now, create an object called mat.w using the command above.
mat.w <- matrix(data = v1, nrow = 2, byrow = TRUE)
# Compare mat.v and mat.w. Do you see that R filled up the cells
# "by row" to create the mat.w matrix ?
mat.v
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 3 7 11
mat.w
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 7 9 11
# Many functions in R have arguments that take TRUE or FALSE like
# the byrow argument we just used. In most cases, functions have a
# default value. In the case of the matrix function, the default
# value for the byrow argument is FALSE, meaning that, if you don't
# specify anything, R will automatically sets byrow = FALSE.
# Find the number in the second row, second column of mat.w
mat.w[2,2]
## [1] 9
# Find the number in the second row, second column of mat.v
mat.v[2,2]
## [1] 7
# Finally, execute the entire contents of this R file by pressing
# Ctrl + A and then pressing Ctrl + Enter.
# Make sure that you don't get any error message. If you get an
# error message, it's probably because you forgot to comment out
# something.
# End of file