Weekly R Exercise

Author

Affiliation

Dr Wei Miao

UCL School of Management

Published

September 27, 2024

1 Induction Week

Question 1
Answer

Create a sequence of {1,1,2,2,3,3,3}.

Code

# solution 1

c(1, 1, 2, 2, 3, 3, 3)

[1] 1 1 2 2 3 3 3

Code

# solution 2

c(rep(1, 2), rep(2, 2), rep(3, 3))

[1] 1 1 2 2 3 3 3

Question 2
Answer

Create a geometric sequence {2,4,8,16,32} using seq().

We can see that the sequence is a geometric sequence with a common ratio of 2.

The seq() function generates a sequence from 1 to 5 with a step of 1.

Code

# solution

seq(1, 5, 1) # this generates a sequence from 1 to 5 with a step of 1

[1] 1 2 3 4 5

The ^ operator calculates the power of 2 raised to the power of the sequence generated by seq(). Remember that R is vectorized, so the ^ operator will apply to each element in the sequence.

Code

# solution

2^seq(1, 5, 1) # this generates a geometric sequence {2^1, 2^2, 2^3, 2^4, 2^5}

[1]  2  4  8 16 32

Question 3
Answer

Create a vector of 10 numbers from 1 to 10, and extract the 2nd, 4th, and 6th elements.

Code

# solution

x <- 1:10 # create a vector of 10 numbers from 1 to 10

x[c(2, 4, 6)] # use [] to extract the 2nd, 4th, and 6th elements

[1] 2 4 6

Question 4
Answer

Create a vector of 5 numbers from 1 to 5, and check if 3 is in the vector.

Code

# solution

x <- 1:5 # create a vector of 5 numbers from 1 to 5

3 %in% x # check if 3 is in the vector

[1] TRUE

Question 5
Answer

Now the interest rate is 0.1, and you have 1000 pounds in your bank account. Calculate the amount in your bank account after 1 year, 2 years, and 3 years, respectively.

First, set the interest rate to 0.1 and the initial amount to 1000.

Then, calculate the amount in your bank account after 1 year, 2 years, and 3 years, respectively.

Since the interest is compounded annually, the formula is:

[ A = P(1 + r)^n ]

where:

(A) is the amount in your bank account after (n) years
(P) is the initial amount
(r) is the interest rate
(n) is the number of years

Code

# solution

interest_rate <- 0.1 # set the interest rate to 0.1

initial_amount <- 1000 # set the initial amount to 1000

# calculate the amount in your bank account after 1 year, 2 years, and 3 years, respectively

# generate the geometric sequence from 1 to 3 years

initial_amount * (1 + interest_rate)^(1:3) # use the formula A = P(1 + r)^n

[1] 1100 1210 1331

2 Week 1

The after-class exercise solutions are in the Week 1 case study

--- author: Dr Wei Miao date: "`r (lubridate::ymd('20240927'))`" date-format: long institute: UCL School of Management title: "Weekly R Exercise" format: html: default knitr: opts_chunk: echo: true warning: true message: true error: true execute: freeze: auto cache: true editor_options: chunk_output_type: inline --- # Induction Week ```{r} #| echo: false question_index <- 1 ``` ::: {.content-visible when-format='html'} ::: {.panel-tabset} ### **Question `r question_index<-question_index+1; question_index-1`** Create a sequence of {1,1,2,2,3,3,3}. ### Answer ```{r} # solution 1 c(1, 1, 2, 2, 3, 3, 3) # solution 2 c(rep(1, 2), rep(2, 2), rep(3, 3)) ``` ::: ::: ::: {.content-visible when-format='html'} ::: {.panel-tabset} ### **Question `r question_index<-question_index+1; question_index-1`** Create a geometric sequence {2,4,8,16,32} using seq(). ### Answer We can see that the sequence is a geometric sequence with a common ratio of 2. - The `seq()` function generates a sequence from 1 to 5 with a step of 1. ```{r} # solution seq(1, 5, 1) # this generates a sequence from 1 to 5 with a step of 1 ``` - The `^` operator calculates the power of 2 raised to the power of the sequence generated by `seq()`. Remember that R is vectorized, so the `^` operator will apply to each element in the sequence. ```{r} # solution 2^seq(1, 5, 1) # this generates a geometric sequence {2^1, 2^2, 2^3, 2^4, 2^5} ``` ::: ::: ::: {.content-visible when-format='html'} ::: {.panel-tabset} ### **Question `r question_index<-question_index+1; question_index-1`** Create a vector of 10 numbers from 1 to 10, and extract the 2nd, 4th, and 6th elements. ### Answer ```{r} # solution x <- 1:10 # create a vector of 10 numbers from 1 to 10 x[c(2, 4, 6)] # use [] to extract the 2nd, 4th, and 6th elements ``` ::: ::: ::: {.content-visible when-format='html'} ::: {.panel-tabset} ### **Question `r question_index<-question_index+1; question_index-1`** Create a vector of 5 numbers from 1 to 5, and check if 3 is in the vector. ### Answer ```{r} # solution x <- 1:5 # create a vector of 5 numbers from 1 to 5 3 %in% x # check if 3 is in the vector ``` ::: ::: ::: {.content-visible when-format='html'} ::: {.panel-tabset} ### **Question `r question_index<-question_index+1; question_index-1`** Now the interest rate is 0.1, and you have 1000 pounds in your bank account. Calculate the amount in your bank account after 1 year, 2 years, and 3 years, respectively. ### Answer First, set the interest rate to 0.1 and the initial amount to 1000. Then, calculate the amount in your bank account after 1 year, 2 years, and 3 years, respectively. Since the interest is compounded annually, the formula is: \[ A = P(1 + r)^n \] where: - \(A\) is the amount in your bank account after \(n\) years - \(P\) is the initial amount - \(r\) is the interest rate - \(n\) is the number of years ```{r} # solution interest_rate <- 0.1 # set the interest rate to 0.1 initial_amount <- 1000 # set the initial amount to 1000 # calculate the amount in your bank account after 1 year, 2 years, and 3 years, respectively # generate the geometric sequence from 1 to 3 years initial_amount * (1 + interest_rate)^(1:3) # use the formula A = P(1 + r)^n ``` ::: ::: # Week 1 - The after-class exercise solutions are in the [Week 1 case study](Case-ProfitabilityAnalysis.qmd)