About Me
Story of my life
I was born in Belgium in may 2001 from italian parents. I then moved to Switewrland and later on to China. I finally completed my bachelor’s at Bocconi univeristy in milan. To summarize, my education went a bit like this:
- kindergarden in french in belgium
- primary school in italian in blgium
- middle school in both english and french in switwerland
- high school in enlgish in china
- bachelor’s in english in italy
Studies aside I enjoy playing tennis and the lesser known padel
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 == "Italy") # 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 = country_data, mapping = aes(x = year, y = lifeExp))+
geom_point() +
geom_smooth(se = FALSE)+
NULL
plot1## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
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 in Italy",
x = "Year",
y = "Life expectancy") +
NULL
plot1## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
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(continent_data, 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=continent ))+
geom_point() +
geom_smooth(se = FALSE) +
facet_wrap(~continent) +
theme(legend.position="none") + #remove all legends
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.
In all continets there has been an increase in life expectancy since 1952. It started off with a rapid increase but has slowed down in the past decades, although still increasing. The post war period saw vast improvements in helthcare and child care, drastically imporving life expectancy, but this is getting harder and harder to improve so the increments in life expectancy are less dramatic.
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)+
labs(title = 'Histogram' , x='leave share', y='count', subtitle = 'count of leave share by constituency')
# density plot-- think smoothed histogram
ggplot(brexit_results, aes(x = leave_share)) +
geom_density()+
labs(title = 'Density Plot' , x='leave share', y='density', subtitle = 'density of leave share by constituency')
# 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)+
labs(title = 'CDF' , x='leave share', y='total share', subtitle = 'cumulative function of the leave percentage in all constituencies')
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 1.0000000The 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() +
labs(title = 'Scatter Plot' , x='proportion of native born', y='leave share', subtitle = 'Correlation between native born and leave share')+
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.
We notice that constituencies with a higher share of native born residents voted more for brexit than those with a lower share. This leads to the conclusion that constutuencies with more migrants and diverse environments skew the vote towards remain, pobbibly because they benefited themselves from EU regulations which allowed them to migrate; and influenced friends to think alike. However there are a lot more dara points as the proportion of native born increases, which could skew the data.
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 547animal_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 547Let 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 rowsanimal_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 rowsDo 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:
- Calculate the mean and median
incident_notional_costfor eachanimal_group_parent - Plot a boxplot to get a feel for the distribution of
incident_notional_costbyanimal_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 cost
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?
Some categories have very high standard deviations, pointing to towards the presence of outliers, and this set is mostly comprised of heavy and bulky animals (horse, cow, deer, heavy livestock, wild animal) as they are the ones that can get into the most intricate and dangerous situations. For these anumals the mean is higher than the median once again indicating presence of outliers. The smaller and more docile the animal the closer the median and mean become, with the mean being below the mediam for the smallest ones.
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.
In my opinion the histogram is the easiest to interpret and gives a rapid and efficient idea of the variability of the data. The density chart is similar but can be a mit misleading for certain animals (like ferret). Horses seem to be very expensive to rescue on average, due to their size, weight and strength. Horses also have the most expensive rescue, together with cats. The different patterns arise from the variability of situations the animal can get stuck in and the ease of rescue (usually related to size and weight). For example a cow is unlikely to end up in a dangerous situation, but it is still relatively expensive to rescue due to its size. The horse, which is of similar size and weight costs considerably more to rescue due to its agility and speed which can push it into dangerous situations.
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.
Details
If you want to, please answer the following
- Who did you collaborate with: no one
- Approximately how much time did you spend on this problem set: 4 hours
- What, if anything, gave you the most trouble: writing the descriptions