Task 2: gapminder country comparison

You have seen the gapminder dataset that has data on life expectancy, population, and GDP per capita for 142 countries from 1952 to 2007. To get a glimpse of the dataframe, namely to see the variable names, variable types, etc., we use the glimpse function. We also want to have a look at the first 20 rows of data.

glimpse(gapminder)

## Rows: 1,704
## Columns: 6
## $ country   <fct> "Afghanistan", "Afghanistan", "Afghanistan", "Afghanistan", …
## $ continent <fct> Asia, Asia, Asia, Asia, Asia, Asia, Asia, Asia, Asia, Asia, …
## $ year      <int> 1952, 1957, 1962, 1967, 1972, 1977, 1982, 1987, 1992, 1997, …
## $ lifeExp   <dbl> 28.801, 30.332, 31.997, 34.020, 36.088, 38.438, 39.854, 40.8…
## $ pop       <int> 8425333, 9240934, 10267083, 11537966, 13079460, 14880372, 12…
## $ gdpPercap <dbl> 779.4453, 820.8530, 853.1007, 836.1971, 739.9811, 786.1134, …

head(gapminder, 20) # look at the first 20 rows of the dataframe

## # A tibble: 20 × 6
##    country     continent  year lifeExp      pop gdpPercap
##    <fct>       <fct>     <int>   <dbl>    <int>     <dbl>
##  1 Afghanistan Asia       1952    28.8  8425333      779.
##  2 Afghanistan Asia       1957    30.3  9240934      821.
##  3 Afghanistan Asia       1962    32.0 10267083      853.
##  4 Afghanistan Asia       1967    34.0 11537966      836.
##  5 Afghanistan Asia       1972    36.1 13079460      740.
##  6 Afghanistan Asia       1977    38.4 14880372      786.
##  7 Afghanistan Asia       1982    39.9 12881816      978.
##  8 Afghanistan Asia       1987    40.8 13867957      852.
##  9 Afghanistan Asia       1992    41.7 16317921      649.
## 10 Afghanistan Asia       1997    41.8 22227415      635.
## 11 Afghanistan Asia       2002    42.1 25268405      727.
## 12 Afghanistan Asia       2007    43.8 31889923      975.
## 13 Albania     Europe     1952    55.2  1282697     1601.
## 14 Albania     Europe     1957    59.3  1476505     1942.
## 15 Albania     Europe     1962    64.8  1728137     2313.
## 16 Albania     Europe     1967    66.2  1984060     2760.
## 17 Albania     Europe     1972    67.7  2263554     3313.
## 18 Albania     Europe     1977    68.9  2509048     3533.
## 19 Albania     Europe     1982    70.4  2780097     3631.
## 20 Albania     Europe     1987    72    3075321     3739.

Your task is to produce two graphs of how life expectancy has changed over the years for the country and the continent you come from.

I have created the country_data and continent_data with the code below.

country_data <- gapminder %>% 
            filter(country == "Greece") # just choosing Greece, as this is where I come from

continent_data <- gapminder %>% 
            filter(continent == "Europe")

First, create a plot of life expectancy over time for the single country you chose. Map year on the x-axis, and lifeExp on the y-axis. You should also use geom_point() to see the actual data points and geom_smooth(se = FALSE) to plot the underlying trendlines. You need to remove the comments # from the lines below for your code to run.

plot1 <- ggplot(data = gapminder, mapping = aes(x = year, y = lifeExp))+
geom_point() +
geom_smooth(se = FALSE)+
NULL 

plot1

## `geom_smooth()` using method = 'gam' and formula 'y ~ s(x, bs = "cs")'

Next we need to add a title. Create a new plot, or extend plot1, using the labs() function to add an informative title to the plot.

plot1 <- plot1 +
labs(title = "Life Expectancy Trends over Time",
     x="Year",y="Life Expectancy") + NULL

plot1

## `geom_smooth()` using method = 'gam' and formula 'y ~ s(x, bs = "cs")'

Secondly, produce a plot for all countries in the continent you come from. (Hint: map the country variable to the colour aesthetic. You also want to map country to the group aesthetic, so all points for each country are grouped together).

ggplot(gapminder, mapping = aes(x = year , y = lifeExp, colour= country, group =country))+
geom_point() + 
geom_smooth(se = FALSE) +
NULL

## `geom_smooth()` using method = 'loess' and formula 'y ~ x'

Finally, using the original gapminder data, produce a life expectancy over time graph, grouped (or faceted) by continent. We will remove all legends, adding the theme(legend.position="none") in the end of our ggplot.

ggplot(data = gapminder , mapping = aes(x = year, y =lifeExp, colour=country ))+
geom_point() + 
geom_smooth(se = FALSE) +
facet_wrap(~continent) +
theme(legend.position="none") +
NULL

## `geom_smooth()` using method = 'loess' and formula 'y ~ x'

Given these trends, what can you say about life expectancy since 1952? Again, don’t just say what’s happening in the graph. Tell some sort of story and speculate about the differences in the patterns.

Type your answer after this blockquote. From the graphs it seems that there has been more of a convergence in life expectancy in the more developed continents, whereas a divergence in the less developed. This may be due to the fact that more developed economies have been able to invest in better healthcare and technology that has been shared across the continent. This factor of information and technology spread is also likely to be correlated with economic development.

Task 3: Brexit vote analysis

We will have a look at the results of the 2016 Brexit vote in the UK. First we read the data using read_csv() and have a quick glimpse at the data

brexit_results <- read_csv(here::here("data","brexit_results.csv"))


glimpse(brexit_results)

## Rows: 632
## Columns: 11
## $ Seat        <chr> "Aldershot", "Aldridge-Brownhills", "Altrincham and Sale W…
## $ con_2015    <dbl> 50.592, 52.050, 52.994, 43.979, 60.788, 22.418, 52.454, 22…
## $ lab_2015    <dbl> 18.333, 22.369, 26.686, 34.781, 11.197, 41.022, 18.441, 49…
## $ ld_2015     <dbl> 8.824, 3.367, 8.383, 2.975, 7.192, 14.828, 5.984, 2.423, 1…
## $ ukip_2015   <dbl> 17.867, 19.624, 8.011, 15.887, 14.438, 21.409, 18.821, 21.…
## $ leave_share <dbl> 57.89777, 67.79635, 38.58780, 65.29912, 49.70111, 70.47289…
## $ born_in_uk  <dbl> 83.10464, 96.12207, 90.48566, 97.30437, 93.33793, 96.96214…
## $ male        <dbl> 49.89896, 48.92951, 48.90621, 49.21657, 48.00189, 49.17185…
## $ unemployed  <dbl> 3.637000, 4.553607, 3.039963, 4.261173, 2.468100, 4.742731…
## $ degree      <dbl> 13.870661, 9.974114, 28.600135, 9.336294, 18.775591, 6.085…
## $ age_18to24  <dbl> 9.406093, 7.325850, 6.437453, 7.747801, 5.734730, 8.209863…

The data comes from Elliott Morris, who cleaned it and made it available through his DataCamp class on analysing election and polling data in R.

Our main outcome variable (or y) is leave_share, which is the percent of votes cast in favour of Brexit, or leaving the EU. Each row is a UK parliament constituency.

To get a sense of the spread, or distribution, of the data, we can plot a histogram, a density plot, and the empirical cumulative distribution function of the leave % in all constituencies.

# histogram
ggplot(brexit_results, aes(x = leave_share)) +
  geom_histogram(binwidth = 2.5)

# density plot-- think smoothed histogram
ggplot(brexit_results, aes(x = leave_share)) +
  geom_density()

# The empirical cumulative distribution function (ECDF) 
ggplot(brexit_results, aes(x = leave_share)) +
  stat_ecdf(geom = "step", pad = FALSE) +
  scale_y_continuous(labels = scales::percent)

One common explanation for the Brexit outcome was fear of immigration and opposition to the EU’s more open border policy. We can check the relationship (or correlation) between the proportion of native born residents (born_in_uk) in a constituency and its leave_share. To do this, let us get the correlation between the two variables

brexit_results %>% 
  select(leave_share, born_in_uk) %>% 
  cor()

##             leave_share born_in_uk
## leave_share   1.0000000  0.4934295
## born_in_uk    0.4934295  1.0000000

The correlation is almost 0.5, which shows that the two variables are positively correlated.

We can also create a scatterplot between these two variables using geom_point. We also add the best fit line, using geom_smooth(method = "lm").

ggplot(brexit_results, aes(x = born_in_uk, y = leave_share)) +
  geom_point(alpha=0.3) +
  
  # add a smoothing line, and use method="lm" to get the best straight-line
  geom_smooth(method = "lm") + 
  
  # use a white background and frame the plot with a black box
  theme_bw() +
  NULL

## `geom_smooth()` using formula 'y ~ x'

You have the code for the plots, I would like you to revisit all of them and use the labs() function to add an informative title, subtitle, and axes titles to all plots.

What can you say about the relationship shown above? Again, don’t just say what’s happening in the graph. Tell some sort of story and speculate about the differences in the patterns.

Type your answer after, and outside, this blockquote. It is evident that areas with a higher proportion of people born in the UK have a higher share of leave voters. An explanation could be that people in these areas are less likely to interact with people foreign to the UK and therefore less appreciative of the benefits of globalisation and free movement of people. Areas with higher diversity of nationalities are more likely to have prople who do not want the UK to be insular and would not want it to be more difficult for foreigners to come and go.

Task 4: Animal rescue incidents attended by the London Fire Brigade

The London Fire Brigade attends a range of non-fire incidents (which we call ‘special services’). These ‘special services’ include assistance to animals that may be trapped or in distress. The data is provided from January 2009 and is updated monthly. A range of information is supplied for each incident including some location information (postcode, borough, ward), as well as the data/time of the incidents. We do not routinely record data about animal deaths or injuries.

Please note that any cost included is a notional cost calculated based on the length of time rounded up to the nearest hour spent by Pump, Aerial and FRU appliances at the incident and charged at the current Brigade hourly rate.

url <- "https://data.london.gov.uk/download/animal-rescue-incidents-attended-by-lfb/8a7d91c2-9aec-4bde-937a-3998f4717cd8/Animal%20Rescue%20incidents%20attended%20by%20LFB%20from%20Jan%202009.csv"

animal_rescue <- read_csv(url,
                          locale = locale(encoding = "CP1252")) %>% 
  janitor::clean_names()


glimpse(animal_rescue)

## Rows: 7,772
## Columns: 31
## $ incident_number               <chr> "139091", "275091", "2075091", "2872091"…
## $ date_time_of_call             <chr> "01/01/2009 03:01", "01/01/2009 08:51", …
## $ cal_year                      <dbl> 2009, 2009, 2009, 2009, 2009, 2009, 2009…
## $ fin_year                      <chr> "2008/09", "2008/09", "2008/09", "2008/0…
## $ type_of_incident              <chr> "Special Service", "Special Service", "S…
## $ pump_count                    <chr> "1", "1", "1", "1", "1", "1", "1", "1", …
## $ pump_hours_total              <chr> "2", "1", "1", "1", "1", "1", "1", "1", …
## $ hourly_notional_cost          <dbl> 255, 255, 255, 255, 255, 255, 255, 255, …
## $ incident_notional_cost        <chr> "510", "255", "255", "255", "255", "255"…
## $ final_description             <chr> "Redacted", "Redacted", "Redacted", "Red…
## $ animal_group_parent           <chr> "Dog", "Fox", "Dog", "Horse", "Rabbit", …
## $ originof_call                 <chr> "Person (land line)", "Person (land line…
## $ property_type                 <chr> "House - single occupancy", "Railings", …
## $ property_category             <chr> "Dwelling", "Outdoor Structure", "Outdoo…
## $ special_service_type_category <chr> "Other animal assistance", "Other animal…
## $ special_service_type          <chr> "Animal assistance involving livestock -…
## $ ward_code                     <chr> "E05011467", "E05000169", "E05000558", "…
## $ ward                          <chr> "Crystal Palace & Upper Norwood", "Woods…
## $ borough_code                  <chr> "E09000008", "E09000008", "E09000029", "…
## $ borough                       <chr> "Croydon", "Croydon", "Sutton", "Hilling…
## $ stn_ground_name               <chr> "Norbury", "Woodside", "Wallington", "Ru…
## $ uprn                          <chr> "NULL", "NULL", "NULL", "100021491149", …
## $ street                        <chr> "Waddington Way", "Grasmere Road", "Mill…
## $ usrn                          <chr> "20500146", "NULL", "NULL", "21401484", …
## $ postcode_district             <chr> "SE19", "SE25", "SM5", "UB9", "RM3", "RM…
## $ easting_m                     <chr> "NULL", "534785", "528041", "504689", "N…
## $ northing_m                    <chr> "NULL", "167546", "164923", "190685", "N…
## $ easting_rounded               <dbl> 532350, 534750, 528050, 504650, 554650, …
## $ northing_rounded              <dbl> 170050, 167550, 164950, 190650, 192350, …
## $ latitude                      <chr> "NULL", "51.39095371", "51.36894086", "5…
## $ longitude                     <chr> "NULL", "-0.064166887", "-0.161985191", …

One of the more useful things one can do with any data set is quick counts, namely to see how many observations fall within one category. For instance, if we wanted to count the number of incidents by year, we would either use group_by()... summarise() or, simply count()

animal_rescue %>% 
  dplyr::group_by(cal_year) %>% 
  summarise(count=n())

## # A tibble: 13 × 2
##    cal_year count
##       <dbl> <int>
##  1     2009   568
##  2     2010   611
##  3     2011   620
##  4     2012   603
##  5     2013   585
##  6     2014   583
##  7     2015   540
##  8     2016   604
##  9     2017   539
## 10     2018   610
## 11     2019   604
## 12     2020   758
## 13     2021   547

animal_rescue %>% 
  count(cal_year, name="count")

## # A tibble: 13 × 2
##    cal_year count
##       <dbl> <int>
##  1     2009   568
##  2     2010   611
##  3     2011   620
##  4     2012   603
##  5     2013   585
##  6     2014   583
##  7     2015   540
##  8     2016   604
##  9     2017   539
## 10     2018   610
## 11     2019   604
## 12     2020   758
## 13     2021   547

Let us try to see how many incidents we have by animal group. Again, we can do this either using group_by() and summarise(), or by using count()

animal_rescue %>% 
  group_by(animal_group_parent) %>% 
  
  #group_by and summarise will produce a new column with the count in each animal group
  summarise(count = n()) %>% 
  
  # mutate adds a new column; here we calculate the percentage
  mutate(percent = round(100*count/sum(count),2)) %>% 
  
  # arrange() sorts the data by percent. Since the default sorting is min to max and we would like to see it sorted
  # in descending order (max to min), we use arrange(desc()) 
  arrange(desc(percent))

## # A tibble: 28 × 3
##    animal_group_parent              count percent
##    <chr>                            <int>   <dbl>
##  1 Cat                               3736   48.1 
##  2 Bird                              1611   20.7 
##  3 Dog                               1213   15.6 
##  4 Fox                                366    4.71
##  5 Unknown - Domestic Animal Or Pet   199    2.56
##  6 Horse                              195    2.51
##  7 Deer                               132    1.7 
##  8 Unknown - Wild Animal               93    1.2 
##  9 Squirrel                            66    0.85
## 10 Unknown - Heavy Livestock Animal    50    0.64
## # … with 18 more rows

animal_rescue %>% 
  
  #count does the same thing as group_by and summarise
  # name = "count" will call the column with the counts "count" ( exciting, I know)
  # and 'sort=TRUE' will sort them from max to min
  count(animal_group_parent, name="count", sort=TRUE) %>% 
  mutate(percent = round(100*count/sum(count),2))

## # A tibble: 28 × 3
##    animal_group_parent              count percent
##    <chr>                            <int>   <dbl>
##  1 Cat                               3736   48.1 
##  2 Bird                              1611   20.7 
##  3 Dog                               1213   15.6 
##  4 Fox                                366    4.71
##  5 Unknown - Domestic Animal Or Pet   199    2.56
##  6 Horse                              195    2.51
##  7 Deer                               132    1.7 
##  8 Unknown - Wild Animal               93    1.2 
##  9 Squirrel                            66    0.85
## 10 Unknown - Heavy Livestock Animal    50    0.64
## # … with 18 more rows

Do you see anything strange in these tables?

Finally, let us have a loot at the notional cost for rescuing each of these animals. As the LFB says,

Please note that any cost included is a notional cost calculated based on the length of time rounded up to the nearest hour spent by Pump, Aerial and FRU appliances at the incident and charged at the current Brigade hourly rate.

There is two things we will do:

1. Calculate the mean and median incident_notional_cost for each animal_group_parent
2. Plot a boxplot to get a feel for the distribution of incident_notional_cost by animal_group_parent.

Before we go on, however, we need to fix incident_notional_cost as it is stored as a chr, or character, rather than a number.

# what type is variable incident_notional_cost from dataframe `animal_rescue`
typeof(animal_rescue$incident_notional_cost)

## [1] "character"

# readr::parse_number() will convert any numerical values stored as characters into numbers
animal_rescue <- animal_rescue %>% 

  # we use mutate() to use the parse_number() function and overwrite the same variable
  mutate(incident_notional_cost = parse_number(incident_notional_cost))

# incident_notional_cost from dataframe `animal_rescue` is now 'double' or numeric
typeof(animal_rescue$incident_notional_cost)

## [1] "double"

Now that incident_notional_cost is numeric, let us quickly calculate summary statistics for each animal group.

animal_rescue %>% 
  
  # group by animal_group_parent
  group_by(animal_group_parent) %>% 
  
  # filter resulting data, so each group has at least 6 observations
  filter(n()>6) %>% 
  
  # summarise() will collapse all values into 3 values: the mean, median, and count  
  # we use na.rm=TRUE to make sure we remove any NAs, or cases where we do not have the incident cos
  summarise(mean_incident_cost = mean (incident_notional_cost, na.rm=TRUE),
            median_incident_cost = median (incident_notional_cost, na.rm=TRUE),
            sd_incident_cost = sd (incident_notional_cost, na.rm=TRUE),
            min_incident_cost = min (incident_notional_cost, na.rm=TRUE),
            max_incident_cost = max (incident_notional_cost, na.rm=TRUE),
            count = n()) %>% 
  
  # sort the resulting data in descending order. You choose whether to sort by count or mean cost.
arrange(desc(mean_incident_cost))

## # A tibble: 16 × 7
##    animal_group_parent      mean_incident_co… median_incident_… sd_incident_cost
##    <chr>                                <dbl>             <dbl>            <dbl>
##  1 Horse                                 740.               596            541. 
##  2 Cow                                   634.               520            475. 
##  3 Deer                                  417.               333            286. 
##  4 Unknown - Wild Animal                 416.               333            324. 
##  5 Unknown - Heavy Livesto…              374.               260            263. 
##  6 Fox                                   373.               328            206. 
##  7 Snake                                 356.               339            105. 
##  8 Dog                                   347.               298            169. 
##  9 Bird                                  344.               328            135. 
## 10 Cat                                   343.               298            160. 
## 11 Unknown - Domestic Anim…              326.               295            117. 
## 12 cat                                   324.               290             94.1
## 13 Hamster                               315.               290             95.0
## 14 Squirrel                              313.               326             57.1
## 15 Ferret                                309.               333             39.4
## 16 Rabbit                                309.               326             32.2
## # … with 3 more variables: min_incident_cost <dbl>, max_incident_cost <dbl>,
## #   count <int>

Compare the mean and the median for each animal group. waht do you think this is telling us? Anything else that stands out? Any outliers?

Finally, let us plot a few plots that show the distribution of incident_cost for each animal group.

# base_plot
base_plot <- animal_rescue %>% 
  group_by(animal_group_parent) %>% 
  filter(n()>6) %>% 
  ggplot(aes(x=incident_notional_cost))+
  facet_wrap(~animal_group_parent, scales = "free")+
  theme_bw()

base_plot + geom_histogram()

base_plot + geom_density()

base_plot + geom_boxplot()

base_plot + stat_ecdf(geom = "step", pad = FALSE) +
  scale_y_continuous(labels = scales::percent)

Which of these four graphs do you think best communicates the variability of the incident_notional_cost values? Also, can you please tell some sort of story (which animals are more expensive to rescue than others, the spread of values) and speculate about the differences in the patterns.

Submit the assignment

Knit the completed R Markdown file as an HTML document (use the “Knit” button at the top of the script editor window) and upload it to Canvas.