Solved: Please create the following data frame: library(tidyverse)student <- tibble( sid = c(66666,66666,66666,66666,66666,66666,22222,22222,22222,22222,22222, 11111,11111,11111,11111,11111,44444,44444,44444,44444,44444, 44444, 55555,55555,55555,55555,55555,55555, 33333,33333,33333,33333,33333,33333, 77777), course = c("COMM101", "COMM102", "COMM103", "COMM201", "COMM204","COMM205", "COMM101", "COMM102", "COMM103", "COMM201","COMM204", "COMM101", "COMM102", "COMM103", "COMM201", "COMM205", "COMM101", "COMM102", "COMM103", "COMM201","COMM204", "COMM205", "COMM101", "COMM102", "COMM103","COMM201", "COMM204", "COMM205", "COMM101", "COMM102", "COMM103", "COMM201", "COMM204","COMM205", "COMM205"), year = c(2016, 2016, 2016, 2017, 2017, 2017, 2016, 2016, 2017, 2017,2017,2016, 2016, 2016, 2017, 2017, 2016, 2016, 2016, 2016, 2017, 2017,2016, 2016, 2016, 2016, 2017, 2017,2016, 2016, 2016, 2017, 2017, 2017, 2017), grade = c(85, 88, 72, 67, 80, 90, 77, 77, 77, 77, 77, 78, 82, 82, 80, 90, 80, 80, 80, 85, 85, 85, 79, 80, 81, 82, 83, 84, 90, 88, 95, 80, 80, 80, 99)) You wanted to find out how many unique students have taken courses in 2016. student %>% filter(year == 2016) %>% XXXX(sid) %>% summarise(XXXX(sid)) The result of your code should look something like this: # A tibble: 1 × 1 XXXXXXXXXX <int> 1 6 Please complete the code below. student %>% filter(year == 2016) %>% [Fill in the blank] (sid) %>% summarise([Fill in the blank] (sid)) | Data manipulation | Computer Science | Black Tom

Question at position 20 ________ is a set of commands used to update and query a database.DDL DPL DCL DML

Assume that you have created the variable bacon correctly (from the previous question). You want to create a column, called bacon_number, which is equal to 1 for each actor who has ever appeared in a movie with Kevin Bacon in Actors data frame. You want bacon_number to be equal to 0 for Kevin Bacon, and you want it to be equal to NA for any other observations. That means that this question does not ask to generate all possible Bacon numbers. It only focuses on Bacon number 0 (for Kevin Bacon) and Bacon number 1 (for those who have ever appeared in a movie with Kevin Bacon). For all others actors, Bacon number should be NA. For example, Tom Cruise appeared with Kevin Bacon in the 1992 movie A Few Good Men, making Tom Cruise’s Bacon Number equal to 1. We want bacon_number to be equal to 1 for all observations for Tom Cruise, even in movies where he did not appear together with Kevin Bacon. For example, we would want to make sure that Tom Cruise’s bacon_number is still equal to 1 when he appears in Rain Man in 1988 or Edge of Tomorrow in 2014. Recall the assumption that you have created the variable bacon successfully before. The following sequence of commands comes right after the previous question. That is, in your Actors data frame you have the 7th column called bacon and it is equal to 1 for every observations of Kevin Bacon and 0 otherwise. Also, recall that Kevin Bacon’s actor_ID is 10000001. Which of the following series of commands will achieve the above task? There is AT LEAST ONE correct option, but you MUST SELECT ALL correct option(s). There is no partial credit. Note that if you put a negative sign (i.e., -) in front of a column name you specified in select(), it means that you want to select all the columns except the one indicated with a negative sign. For example, Actors %>% select(-best_actor) is identical to Actors %>% select( actor_ID, actor_l, actor_f, title, year, best_actress).

Start a fresh of R session by launching RStudio. If RStudio is already open, you can start a fresh R session by going to menu item: Session > Restart R. Note in a fresh R session, there is no library loaded. So, if you execute the following, you will be given an error. df0 <- tibble(X = c("a","b","c"), Y = c(101,102,103), Z = c(10,20,30) Remember tibble function are from tibble library which comes with tidyverse. We often load tidyverse for convenience. If we do load tidyverse by calling library(tidyverse) Links to an external site. , it will load the core tidyverse packages. However, we do not have to load a "mega" library like tidyverse if we are going to use only one or two functions from libraries included in tidyverse. Instead, you can load the specific library, whose function(s) you want to use. In the above case, you only need to load tibble library. library(tibble) df0 <- tibble(X = c("a","b","c"), Y = c(101,102,103), Z = c(10,20,30)) Since tibble is loaded, df0 was successfully created. Now, you wanted to convert the data type of the Y column to character. Here is the code, part of which is redacted: library(XXX) df1 <- df0 %>% mutate( XXX = XXX(Y)) Complete the code below so that df1 is successfully created. library( [ Select ] readr dplyr babynames sauder ) df1 <- df0 %>% mutate( [ Select ] X Y Z = [ Select ] is.character as.character is.numeric as.numeric (Y))

Question textWhich package is used to manipulate your data, such as adding columns or computing summaries? Answer 1 Question 3[select: , fpp3, tidyr, dplyr, lubridate, ggplot2, fable] Which function would you use to: compute a new column for your dataset? Answer 2 Question 3[select: , select, mutate, filter, arrange, summarise] reduce multiple values down to a single summary? Answer 3 Question 3[select: , select, mutate, filter, arrange, summarise] pick variables based on their names? Answer 4 Question 3[select: , select, mutate, filter, arrange, summarise] pick observations based on their values? Answer 5 Question 3[select: , select, mutate, filter, arrange, summarise] changes the ordering of the rows? Answer 6 Question 3[select: , select, mutate, filter, arrange, summarise] The ChickWeight dataset provides the weight chicks on 4 different diets as they age. The weight column is the chick’s body weight in grams, Time is their age in days, Chick is an identifier of each chick, and Diet is the experimental diet given to that chick. as_tibble(ChickWeight) ## # A tibble: 578 x 4 ## weight Time Chick Diet ## <dbl> <dbl> <ord> <fct> ## 1 42 0 1 1 ## 2 51 2 1 1 ## 3 59 4 1 1 ## 4 64 6 1 1 ## 5 76 8 1 1 ## 6 93 10 1 1 ## 7 106 12 1 1 ## 8 125 14 1 1 ## 9 149 16 1 1 ## 10 171 18 1 1 ## # i 568 more rows Complete the code to compute the weight of each chick at 21 days old: FinalChickWeight <- as_tibble(ChickWeight) |> Answer 7 Question 3[select: , select, mutate, filter, arrange, summarise, group_by](Answer 8 Question 3[input] == Answer 9 Question 3[input]) ## # A tibble: 45 x 4 ## weight Time Chick Diet ## <dbl> <dbl> <ord> <fct> ## 1 205 21 1 1 ## 2 215 21 2 1 ## 3 202 21 3 1 ## 4 157 21 4 1 ## 5 223 21 5 1 ## 6 157 21 6 1 ## 7 305 21 7 1 ## 8 98 21 9 1 ## 9 124 21 10 1 ## 10 175 21 11 1 ## # i 35 more rows Complete the code to compute the average weight of 21 day old chicks for each diet type: FinalChickWeight |> Answer 10 Question 3[select: , select, mutate, filter, arrange, summarise, group_by](Answer 11 Question 3[input]) |> Answer 12 Question 3[select: , select, mutate, filter, arrange, summarise, group_by](Answer 13 Question 3[input](Answer 14 Question 3[input])) ## # A tibble: 4 x 2 ## Diet `mean(weight)` ## <fct> <dbl> ## 1 1 178. ## 2 2 215. ## 3 3 270. ## 4 4 239.Check Question 3

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