r_vector_indexing

Question 1

For the following vector “heights”, multiple each value by 2.54:

• heights <- c(69, 62, 66, 70, 70, 73, 67, 73, 67, 70)

Answer 1

• heights * 2.54

Question 2

For the dataset “murders” with variables “total” and “population”, do the following:

1. compute the “murder_rate” per 100,000
2. order the “states” in decreasing order by “murder_rate”

Answer 2

1. murder_rate <- murders$total/murders$population * 100000
2. murders$state[order(murder_rate,decreasing=TRUE)]

Question 3

Do the following:

1. Create a vector “city” with the following city names:
   
   • “Beijing”, “Lagos”, “Paris”, “Rio de Janeiro”, “San Juan”, “Toronto”
2. Create a vector “temp” with the following temperatures in Fahrenheit
   
   • 35, 88, 42, 84, 81, 30
3. Use vector arithmetics to convert “temp” to Celsius
   • C=5/9×(F−32)
4. Create a data frame called “city_temps” with the city names and temperatures in Celsius.

Answer 3

1. # Assign city names to city
   • city <- c(“Beijing”, “Lagos”, “Paris”, “Rio de Janeiro”, “San Juan”, “Toronto”)
2. # Store temperature values in temp
   • temp <- c(35, 88, 42, 84, 81, 30)
3. # Convert temperature into Celsius and overwrite the original values of ‘temp’ with these Celsius values
   • temp <- (5/9) * (temp-32)
4. # Create a data frame city_temps
   • city_temps <- data.frame(name=city, temperature=temp)

Question 4

Display the values from the ‘murder’ dataset where:

• region == “West”
• murder rate <= 1

Answer 4

• west <- murder$region == “West”
• safe <- murder_rate <= 1
• index <- safe & west

Question 5

Display the index of True values in the following vector:

• x <- c(FALSE, TRUE, FALSE, TRUE, TRUE, FALSE)

Answer 5

• which(x)

Question 6

1. Using the “match” function, find the index of the following states using the “state” variable within the “murders” dataset:
   
   • New York
   • Florida
   • Texas
2. List the murder rate of the three states

Answer 6

1. index <- match(c(“New York”, “Florida”, “Texas”),murders$state)
   
   • index
2. murder_rate[index]

Question 7

Find out if the following states are represented within the “state” variable of the “murders” dataset:

• Boston
• Dakota
• Washington

Answer 7

• c(“Boston”, “Dakota”, “Washington”) %in% murders$state

Question 8

1. Compute the per 100,000 murder rate for each state and store it in an object called murder_rate.
2. Then use the logical operators to create a logical vector, name it low, that tells us which entries of murder_rate are lower than 1.
3. Get the indices of the entries that are below 1
4. Get the names of the states with murder rates lower than 1

Answer 8

1. # Store the murder rate per 100,000 for each state, in murder_rate
   • murder_rate <- murders$total/murders$population*100000
2. # Store the murder_rate < 1 in low
   • low <- murder_rate < 1
3. # Get the indices of entries that are below 1
   • which(low)
4. # Names of states with murder rates lower than 1
   • murders$state[low]

Question 9

1. Compute the per 100,000 murder rate for each state and store it in an object called murder_rate.
2. Then use the logical operators to create a logical vector, name it low, that tells us which entries of murder_rate are lower than 1.
3. Use the & operator to create a new object ind that is true when low is true and the state is in the Northeast
4. Use the brackets [ and ind to show the state names that satisfy this condition

Answer 9

1. # Store the murder rate per 100,000 for each state, in murder_rate
   • murder_rate <- murders$total/murders$population*100000
2. # Store the murder_rate \< 1 in low
   • low <- murder_rate < 1
3. # Create a vector ind for states in the Northeast and with murder rates lower than 1.
   • ind <- (murders$region == “Northeast”) & (low)
4. # Names of states in ind
   • murders$state[ind]

Question 10

1. Compute average murder rate and store in avg using mean
2. How many states have murder rates below avg ? Check using sum

Answer 10

1. # Compute average murder rate and store in avg using mean
   • avg <- mean(murder_rate)
2. # How many states have murder rates below avg ? Check using sum
   • sum(murder_rate < avg)

Question 11

1. Store the 3 abbreviations (AK, MI, and IA) in the variable “abbs” in a vector
2. Match the abbs to the murders$abb and store in ind
3. Print state names from ind

Answer 11

1. # Store the 3 abbreviations in abbs in a vector (remember that they are character vectors and need quotes)
   • abbs <- c(“AK”, “MI”, “IA”)
2. # Match the abbs to the murders$abb and store in ind
   • ind <- match(abbs, murders$abb)
3. # Print state names from ind
   • murders$state[ind]

Question 12

1. Define a character vector with the abbreviations MA, ME, MI, MO, MU.
2. Use the % operator to create a logical vectors that is TRUE when the abbreviation is in murders$abb.

Answer 12

1. # Store the 5 abbreviations in abbs. (remember that they are character vectors)
   • abbs <- c(“MA”,”ME”,”MI”,”MO”,”MU”)
2. # Use the %in% command to check if the entries of abbs are abbreviations in the the murders data frame
   • abbs %in% murders$abb

Question 13

We are again working with the characters abbs <- c(“MA”, “ME”, “MI”, “MO”, “MU”):

1. In a previous exercise we computed the index abbs%in%murders$abb.
   
   • Use which and the ! operator to get the index of the entries of abbs that are not abbreviations.
2. Show the entries of abbs that are not actual abbreviations.

Answer 13

1. # Use the which command and ! operator to find out which abbreviation are not actually part of the dataset and store in ind
   • ind <- which(!abbs%in%murders$abb)
2. # What are the entries of abbs that are not actual abbreviations
   • abbs[ind]