In jr-packages/jrTidyverse2: Jumping Rivers: Next Steps in the Tidyverse

First we must load the tidyverse

library("tidyverse")

Question 1 - lists

For this question, there is no purrr. It's a few questions to get you used to lists.

a) Create a list that contains a numeric, a logical, a character, a vector.

b) Name the elements of your list

c) The following code will load a list called toy into your global environment r data(toy, package = "jrTidyverse2") Make sure to take a look at toy before you dive in. How many elements are in toy? r length(toy)

d) Add an extra element called z onto toy. z can be whatever object you like. r toy$z = "fun with lists" # or toy[["z"]] = "fun with lists" # or toy["z"] = "fun with lists"

e) What is the average of the element d within toy? r mean(toy$d)

f) What is the average of the column f within element e of toy? r mean(toy$e$f)

g) What is the average of the column f in the element e, where the values of the column g are "a"? r toy$e %>% filter(g == "a") %>% summarise(mean(f))

Question 2 - purrr beginnings

a) If we have a vector x, we can square root it using the sqrt() function r x = c(1, 4, 9, 25) sqrt(x) Can you do the same but using the map functions? Make sure your output is a vector. r map_dbl(x, sqrt)

Question 3 - Happiness

Now we're going to look at a list containing information such as happiness and economy rankings for countries around the globe in the years 2015, 2016 and 2017.

data(happiness, package = "jrTidyverse2")

a) Using str() to investigate the list and determine: + How long is the list? r str(happiness, max.level = 0) # 146 element in the list + Is the list a recursive list? r str(happiness, max.level = 1, list.len = 3) # Yes, recursive list + How many countries does the list contain information on? + For each country, how many piece of information do we have? r str(happiness, max.level = 2, list.len = 3) # Each element of the list is another list representing a country. # Each list contains elements representative of happiness information # on that country for three successive years. Therefore there is # 146 countries and 12 pieces of information on each.

b) Extract the name of each country using the map functions. To make it a bit easier to read, return the output as a character vector. r country_names = map_chr(happiness, "Country")

c) Try names(happiness), what happens? Use the answer to b) to rename each element of the list after it's representative country. r names(happiness) = country_names

d) What has the UKs happiness rank been over the last 3 years? (You don't have to use purrr for this one.) r happiness[["United Kingdom"]]$`Happiness Rank`

e) Which country has had the highest average happiness score? r happiness %>% map_dbl(~ mean(.x[["Happiness Score"]])) %>% sort(decreasing = TRUE) %>% head()

f) Which country has had the largest increase in happiness score from 2015 - 2017? r happiness %>% map_dbl(~ .x$`Happiness Score`[3] - .x$`Happiness Score`[1]) %>% sort(decreasing = TRUE) %>% head()

g) Which country has had the largest decrease in life expectency? r map_dbl(happiness, ~.x$`Health (Life Expectancy)`[3] - .x$`Health (Life Expectancy)`[1]) %>% sort() %>% head()

Question 4 - Happiness - Advanced

The following two questions are intended to be a bit trickier. Don't worry if you get stuck on them. Just ask!

a) How many countries economies have shrunk from 2015 - 2017? r map_lgl(happiness, ~ (.x$`Economy (GDP per Capita)`[3] - .x$`Economy (GDP per Capita)`[1]) < 0) %>% sum()

b) On average, which region of the world is the most "generous"?
Hint: store the region for each country in a vector, combine it into a data frame with the average generosity score for each country then use dplyr.
```r region = happiness %>% map_chr("Region")

av_gen = happiness %>% 
  map_dbl(~mean(.x$Generosity))

region_gen = tibble(region = region, av_gen = av_gen)

region_gen %>% 
  group_by(region) %>% 
  summarise(av_region_gen = mean(av_gen)) %>% 
  arrange(av_region_gen)
```

c) Using ggplot2 and geom_col(), plot the answer to b) as a bar chart r region_gen %>% group_by(region) %>% summarise(av_region_gen = mean(av_gen)) %>% arrange(av_region_gen) %>% ggplot() + geom_col(aes(x = region, y = av_region_gen)) + coord_flip()

jr-packages/jrTidyverse2 documentation built on Sept. 24, 2020, 4:36 p.m.