Chapter 9 - Tidy Data with tidyr

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options(scipen=10000)
library(tidyverse)
library(flair)
library(emo)
library(lubridate)
library(magrittr)
library(tidyquant)
theme_set(theme_tq())

What’s tidy data anyway?

1. Each variable must have its own column.
2. Each observation must have its own row.
3. Each value must have its own cell.

In all the examples tidyr::table to tidyr::table4b, only tidyr::table1 is tidy.

(
  # practising the read_csv function to create table1
  # just note however that table1 is in tidyr ;)
  # tidyr::table1 etc. 
  # In all honesty, I only figured this out after "practising" :P
  table1 <- read_csv("country, year, cases, population
                     Afghanistan, 1999, 745, 19987071
                     Afghanistan, 2000, 2666, 20595360
                     Brazil, 1999, 37737, 172006362
                     Brazil, 2000, 80488, 174504898
                     China, 1999, 212258, 1272915272
                     China, 2000, 213766, 1280428583")
)

# A tibble: 6 x 4
  country      year  cases population
  <chr>       <dbl>  <dbl>      <dbl>
1 Afghanistan  1999    745   19987071
2 Afghanistan  2000   2666   20595360
3 Brazil       1999  37737  172006362
4 Brazil       2000  80488  174504898
5 China        1999 212258 1272915272
6 China        2000 213766 1280428583

Working with tidy data

table1 %>% 
  mutate(rate = cases/population * 10000)

# A tibble: 6 x 5
  country      year  cases population  rate
  <chr>       <dbl>  <dbl>      <dbl> <dbl>
1 Afghanistan  1999    745   19987071 0.373
2 Afghanistan  2000   2666   20595360 1.29 
3 Brazil       1999  37737  172006362 2.19 
4 Brazil       2000  80488  174504898 4.61 
5 China        1999 212258 1272915272 1.67 
6 China        2000 213766 1280428583 1.67 

table1 %>% 
  count(year, wt=cases) # same as group_by and sum

# A tibble: 2 x 2
   year      n
  <dbl>  <dbl>
1  1999 250740
2  2000 296920

table1 %>% 
  group_by(year) %>% 
  summarise(sum(cases))

# A tibble: 2 x 2
   year `sum(cases)`
  <dbl>        <dbl>
1  1999       250740
2  2000       296920

ggplot(table1, aes(year, cases)) +
  geom_line(aes(group = country), colour = "grey50") +
  geom_point(aes(colour = country)) +
  scale_colour_tq()

Exercises

1. Using prose, describe how the variables and observations are organised in each of the sample tables.

   tidyr::table1

   # A tibble: 6 x 4
     country      year  cases population
     <chr>       <int>  <int>      <int>
   1 Afghanistan  1999    745   19987071
   2 Afghanistan  2000   2666   20595360
   3 Brazil       1999  37737  172006362
   4 Brazil       2000  80488  174504898
   5 China        1999 212258 1272915272
   6 China        2000 213766 1280428583

   • Each column is a variable ✅
   • Each observation is in a row ✅
   • Each value is in a cell ✅
   • The table describes the number of cases, and the population count (each in its own column) for each country and year combination.

   
   

   tidyr::table2

   # A tibble: 12 x 4
      country      year type            count
      <chr>       <int> <chr>           <int>
    1 Afghanistan  1999 cases             745
    2 Afghanistan  1999 population   19987071
    3 Afghanistan  2000 cases            2666
    4 Afghanistan  2000 population   20595360
    5 Brazil       1999 cases           37737
    6 Brazil       1999 population  172006362
    7 Brazil       2000 cases           80488
    8 Brazil       2000 population  174504898
    9 China        1999 cases          212258
   10 China        1999 population 1272915272
   11 China        2000 cases          213766
   12 China        2000 population 1280428583

   • Each column is a variable ❌
   • Each observation is in a row ✅
   • Each value is in a cell ✅
   • Each row contains either the number of cases, or the population count for each country and year combination.

   
   

   tidyr::table3

   # A tibble: 6 x 3
     country      year rate             
   * <chr>       <int> <chr>            
   1 Afghanistan  1999 745/19987071     
   2 Afghanistan  2000 2666/20595360    
   3 Brazil       1999 37737/172006362  
   4 Brazil       2000 80488/174504898  
   5 China        1999 212258/1272915272
   6 China        2000 213766/1280428583

   • Each column is a variable ❌
   • Each observation is in a row ✅
   • Each value is in a cell ❌
   • Each row considers the country and year combination, and shows the number of cases and the population count (separated by a /) in one variable named rate.

   
   

   tidyr::table4a

   # A tibble: 3 x 3
     country     `1999` `2000`
   * <chr>        <int>  <int>
   1 Afghanistan    745   2666
   2 Brazil       37737  80488
   3 China       212258 213766

   • Each column is a variable ✅
   • Each observation is in a row ❌
   • Each value is in a cell ✅

   
   

   tidyr::table4b

   # A tibble: 3 x 3
     country         `1999`     `2000`
   * <chr>            <int>      <int>
   1 Afghanistan   19987071   20595360
   2 Brazil       172006362  174504898
   3 China       1272915272 1280428583

   • Each column is a variable ✅
   • Each observation is in a row ❌
   • Each value is in a cell ✅
     
   • Each table houses either the number of cases, or the population count for each country, in separated columns for each year.

   
   

   tidyr::table5

   # A tibble: 6 x 4
     country     century year  rate             
   * <chr>       <chr>   <chr> <chr>            
   1 Afghanistan 19      99    745/19987071     
   2 Afghanistan 20      00    2666/20595360    
   3 Brazil      19      99    37737/172006362  
   4 Brazil      20      00    80488/174504898  
   5 China       19      99    212258/1272915272
   6 China       20      00    213766/1280428583

   • Each column is a variable ❌
   • Each observation is in a row ✅
   • Each value is in a cell ❌
   • The table considers each year separated into century and year for each country and then similar to table3 it combines the cases and population count in one variable rate (separated within the column by /)

   
   

2. Compute the rate for table2, and table4a + table4b. You will need to perform four operations:

   1. Extract the number of TB cases per country per year.
   2. Extract the matching population per country per year.
   3. Divide cases by population, and multiply by 10000.
   4. Store back in the appropriate place.

   Which representation is easiest to work with? Which is hardest? Why?

   (
     tbl1 <- tidyr::table2 %>% 
     filter(type == "cases") %>% 
     group_by(country, year) %>% 
     mutate(cases = count) %>% 
     ungroup() %>% 
     select(country, year, cases) %>% 
     arrange(country, year)
   )

   # A tibble: 6 x 3
     country      year  cases
     <chr>       <int>  <int>
   1 Afghanistan  1999    745
   2 Afghanistan  2000   2666
   3 Brazil       1999  37737
   4 Brazil       2000  80488
   5 China        1999 212258
   6 China        2000 213766

   (
     tbl2 <- tidyr::table2 %>% 
     filter(type == "population") %>% 
     group_by(country, year) %>% 
     mutate(population = count) %>% 
     ungroup() %>% 
     select(country_temp = country, 
            year_temp = year, 
            population) %>% 
     arrange(country_temp, year_temp)
   )

   # A tibble: 6 x 3
     country_temp year_temp population
     <chr>            <int>      <int>
   1 Afghanistan       1999   19987071
   2 Afghanistan       2000   20595360
   3 Brazil            1999  172006362
   4 Brazil            2000  174504898
   5 China             1999 1272915272
   6 China             2000 1280428583

   (
   tbl3 <- tbl1 %>% 
     bind_cols(tbl2) %>% 
     select(c(1:3,6)) %>% 
     mutate(rate = (cases / population) * 10000) %>% 
     arrange(country, year) %>% 
     select(country, year, rate) %>% 
     mutate(type = "rate",
            count = rate) %>% 
     select(c(1,2,4,5))
   )

   # A tibble: 6 x 4
     country      year type  count
     <chr>       <int> <chr> <dbl>
   1 Afghanistan  1999 rate  0.373
   2 Afghanistan  2000 rate  1.29 
   3 Brazil       1999 rate  2.19 
   4 Brazil       2000 rate  4.61 
   5 China        1999 rate  1.67 
   6 China        2000 rate  1.67 

   tidyr::table2 %>% 
     bind_rows(tbl3) %>% 
     mutate(count = round(count, 2)) %>% 
     arrange(country, year, type) %>% 
     gt::gt()

   country	year	type	count
   Afghanistan	1999	cases	745.00
   Afghanistan	1999	population	19987071.00
   Afghanistan	1999	rate	0.37
   Afghanistan	2000	cases	2666.00
   Afghanistan	2000	population	20595360.00
   Afghanistan	2000	rate	1.29
   Brazil	1999	cases	37737.00
   Brazil	1999	population	172006362.00
   Brazil	1999	rate	2.19
   Brazil	2000	cases	80488.00
   Brazil	2000	population	174504898.00
   Brazil	2000	rate	4.61
   China	1999	cases	212258.00
   China	1999	population	1272915272.00
   China	1999	rate	1.67
   China	2000	cases	213766.00
   China	2000	population	1280428583.00
   China	2000	rate	1.67

   
   
   

   (
     tbl1_cases <- tidyr::table4a %>% 
     select(country, `1999`) %>% 
     mutate(year = 1999,
            cases  = `1999`) %>% 
     select(country, year, cases)
   )

   # A tibble: 3 x 3
     country      year  cases
     <chr>       <dbl>  <int>
   1 Afghanistan  1999    745
   2 Brazil       1999  37737
   3 China        1999 212258

   (
     tbl2_cases <- tidyr::table4a %>% 
     select(country, "2000") %>% 
     mutate(year = 2000, 
            cases = `2000`) %>% 
     select(country, year, cases)
   )

   # A tibble: 3 x 3
     country      year  cases
     <chr>       <dbl>  <int>
   1 Afghanistan  2000   2666
   2 Brazil       2000  80488
   3 China        2000 213766

   (
     tbl_cases <- tbl1_cases %>% 
     bind_rows(tbl2_cases) %>% 
     arrange(country, year)
   )

   # A tibble: 6 x 3
     country      year  cases
     <chr>       <dbl>  <int>
   1 Afghanistan  1999    745
   2 Afghanistan  2000   2666
   3 Brazil       1999  37737
   4 Brazil       2000  80488
   5 China        1999 212258
   6 China        2000 213766

   (
     tbl1_pop <- tidyr::table4b %>% 
     select(country, `1999`) %>% 
     mutate(year = 1999,
            population  = `1999`) %>% 
     select(country, year, population)
   )

   # A tibble: 3 x 3
     country      year population
     <chr>       <dbl>      <int>
   1 Afghanistan  1999   19987071
   2 Brazil       1999  172006362
   3 China        1999 1272915272

   (
     tbl2_pop <- tidyr::table4b %>% 
     select(country, "2000") %>% 
     mutate(year = 2000, 
            population = `2000`) %>% 
     select(country, year, population)
   )

   # A tibble: 3 x 3
     country      year population
     <chr>       <dbl>      <int>
   1 Afghanistan  2000   20595360
   2 Brazil       2000  174504898
   3 China        2000 1280428583

   (
     tbl_pop <- tbl1_pop %>% 
     bind_rows(tbl2_pop) %>% 
     arrange(country, year)
   )

   # A tibble: 6 x 3
     country      year population
     <chr>       <dbl>      <int>
   1 Afghanistan  1999   19987071
   2 Afghanistan  2000   20595360
   3 Brazil       1999  172006362
   4 Brazil       2000  174504898
   5 China        1999 1272915272
   6 China        2000 1280428583

   (
     tbl_rate <- tbl_cases %>% 
       bind_cols(tbl_pop) %>% 
       janitor::clean_names() %>% 
       select(country = country_1,
              year = year_2,
              cases, population) %>% 
       mutate(rate = cases / population * 10000)
   )

   # A tibble: 6 x 5
     country      year  cases population  rate
     <chr>       <dbl>  <int>      <int> <dbl>
   1 Afghanistan  1999    745   19987071 0.373
   2 Afghanistan  2000   2666   20595360 1.29 
   3 Brazil       1999  37737  172006362 2.19 
   4 Brazil       2000  80488  174504898 4.61 
   5 China        1999 212258 1272915272 1.67 
   6 China        2000 213766 1280428583 1.67 

   (
     tbl_1999 <- tbl_rate %>% 
       select(country, year, rate) %>% 
       filter(year == 1999) %>% 
       mutate(`1999` = rate) %>% 
       select(country, `1999`)
   )

   # A tibble: 3 x 2
     country     `1999`
     <chr>        <dbl>
   1 Afghanistan  0.373
   2 Brazil       2.19 
   3 China        1.67 

   (
     tbl_2000 <- tbl_rate %>% 
       select(country, year, rate) %>% 
       filter(year == 2000) %>% 
       mutate(`2000` = rate) %>% 
       select(country_temp = country, `2000`)
   )

   # A tibble: 3 x 2
     country_temp `2000`
     <chr>         <dbl>
   1 Afghanistan    1.29
   2 Brazil         4.61
   3 China          1.67

   (
     tbl_4c <- 
       tbl_1999 %>% 
       bind_cols(tbl_2000) %>% 
       select(country, `1999`, `2000`)
   )

   # A tibble: 3 x 3
     country     `1999` `2000`
     <chr>        <dbl>  <dbl>
   1 Afghanistan  0.373   1.29
   2 Brazil       2.19    4.61
   3 China        1.67    1.67

3. Recreate the plot showing change in cases over time using table2 instead of table1. What do you need to do first?

   tidyr::table1

   # A tibble: 6 x 4
     country      year  cases population
     <chr>       <int>  <int>      <int>
   1 Afghanistan  1999    745   19987071
   2 Afghanistan  2000   2666   20595360
   3 Brazil       1999  37737  172006362
   4 Brazil       2000  80488  174504898
   5 China        1999 212258 1272915272
   6 China        2000 213766 1280428583

   ggplot(table1, aes(year, cases)) +
     geom_line(aes(group = country), colour = "grey50") +
     geom_point(aes(colour = country)) +
     scale_colour_tq()

   

   tidyr::table2

   # A tibble: 12 x 4
      country      year type            count
      <chr>       <int> <chr>           <int>
    1 Afghanistan  1999 cases             745
    2 Afghanistan  1999 population   19987071
    3 Afghanistan  2000 cases            2666
    4 Afghanistan  2000 population   20595360
    5 Brazil       1999 cases           37737
    6 Brazil       1999 population  172006362
    7 Brazil       2000 cases           80488
    8 Brazil       2000 population  174504898
    9 China        1999 cases          212258
   10 China        1999 population 1272915272
   11 China        2000 cases          213766
   12 China        2000 population 1280428583

   table2 %>% 
     filter(type == "cases") %>% 
     ggplot(aes(year, count)) +
     geom_line(aes(group = country), colour = "grey50") +
     geom_point(aes(colour = country)) +
     scale_colour_tq()

   

Sometimes you will have to resolve one of two common problems:

1. One variable might be spread across multiple columns.

2. One observation might be scattered across multiple rows.

Pivot Longer / Gather

pivot_longer() makes datasets longer by increasing the number of rows and decreasing the number of columns.

table4a

# A tibble: 3 x 3
  country     `1999` `2000`
* <chr>        <int>  <int>
1 Afghanistan    745   2666
2 Brazil       37737  80488
3 China       212258 213766

table4a %>% 
  # gather(list out columns you want to gather like dplyr::select() style,
  #        key = what do you want to call the column
  #              these column names go into,
  #        value = the values of the columns will go here)
  gather(`1999`, `2000`, 
         key = "year",
         value = "cases" )

# A tibble: 6 x 3
  country     year   cases
  <chr>       <chr>  <int>
1 Afghanistan 1999     745
2 Brazil      1999   37737
3 China       1999  212258
4 Afghanistan 2000    2666
5 Brazil      2000   80488
6 China       2000  213766

(tidy_4a <- table4a %>% 
  # cols = list the columns you want to pivot
  # names_to = what will you call the new column these
  #            column names go into
  # values_to = the values in the columns will go here
  pivot_longer(cols = c(`1999`, `2000`),
               names_to = "year",
               values_to = "cases"))

# A tibble: 6 x 3
  country     year   cases
  <chr>       <chr>  <int>
1 Afghanistan 1999     745
2 Afghanistan 2000    2666
3 Brazil      1999   37737
4 Brazil      2000   80488
5 China       1999  212258
6 China       2000  213766

table4b

# A tibble: 3 x 3
  country         `1999`     `2000`
* <chr>            <int>      <int>
1 Afghanistan   19987071   20595360
2 Brazil       172006362  174504898
3 China       1272915272 1280428583

table4b %>% 
  gather(`1999`, `2000`, 
         key = "year",
         value = "population")

# A tibble: 6 x 3
  country     year  population
  <chr>       <chr>      <int>
1 Afghanistan 1999    19987071
2 Brazil      1999   172006362
3 China       1999  1272915272
4 Afghanistan 2000    20595360
5 Brazil      2000   174504898
6 China       2000  1280428583

(tidy_4b <- table4b %>% 
  pivot_longer(cols = c(`1999`, `2000`),
               names_to = "year",
               values_to = "population"))

# A tibble: 6 x 3
  country     year  population
  <chr>       <chr>      <int>
1 Afghanistan 1999    19987071
2 Afghanistan 2000    20595360
3 Brazil      1999   172006362
4 Brazil      2000   174504898
5 China       1999  1272915272
6 China       2000  1280428583

left_join(tidy_4a, tidy_4b) %>% 
  arrange(country, year)

# A tibble: 6 x 4
  country     year   cases population
  <chr>       <chr>  <int>      <int>
1 Afghanistan 1999     745   19987071
2 Afghanistan 2000    2666   20595360
3 Brazil      1999   37737  172006362
4 Brazil      2000   80488  174504898
5 China       1999  212258 1272915272
6 China       2000  213766 1280428583

Pivot Wider / Spread

pivot_wider() is the opposite of pivot_longer(). You use it when an observation is scattered across multiple rows.

table2

# A tibble: 12 x 4
   country      year type            count
   <chr>       <int> <chr>           <int>
 1 Afghanistan  1999 cases             745
 2 Afghanistan  1999 population   19987071
 3 Afghanistan  2000 cases            2666
 4 Afghanistan  2000 population   20595360
 5 Brazil       1999 cases           37737
 6 Brazil       1999 population  172006362
 7 Brazil       2000 cases           80488
 8 Brazil       2000 population  174504898
 9 China        1999 cases          212258
10 China        1999 population 1272915272
11 China        2000 cases          213766
12 China        2000 population 1280428583

table2 %>% 
  # key = column with the variable name, here `type`
  spread(key = type, 
  # value = column with the value that will be assigned
  # to new columns
         value = count)

# A tibble: 6 x 4
  country      year  cases population
  <chr>       <int>  <int>      <int>
1 Afghanistan  1999    745   19987071
2 Afghanistan  2000   2666   20595360
3 Brazil       1999  37737  172006362
4 Brazil       2000  80488  174504898
5 China        1999 212258 1272915272
6 China        2000 213766 1280428583

table2 %>% 
  pivot_wider(names_from = type,
              values_from = count)

# A tibble: 6 x 4
  country      year  cases population
  <chr>       <int>  <int>      <int>
1 Afghanistan  1999    745   19987071
2 Afghanistan  2000   2666   20595360
3 Brazil       1999  37737  172006362
4 Brazil       2000  80488  174504898
5 China        1999 212258 1272915272
6 China        2000 213766 1280428583

Exercises

1. Why are pivot_longer() and pivot_wider() not perfectly symmetrical?
   Carefully consider the following example:

   (stocks <- tibble(
     year   = c(2015, 2015, 2016, 2016),
     half  = c(   1,    2,     1,    2),
     return = c(1.88, 0.59, 0.92, 0.17)
   ))

   # A tibble: 4 x 3
      year  half return
     <dbl> <dbl>  <dbl>
   1  2015     1   1.88
   2  2015     2   0.59
   3  2016     1   0.92
   4  2016     2   0.17

   stocks %>% 
     pivot_wider(names_from = year, values_from = return) %>% 
     pivot_longer(`2015`:`2016`, names_to = "year", values_to = "return")

   # A tibble: 4 x 3
      half year  return
     <dbl> <chr>  <dbl>
   1     1 2015    1.88
   2     1 2016    0.92
   3     2 2015    0.59
   4     2 2016    0.17

   (Hint: look at the variable types and think about column names.)

   pivot_longer() has a names_ptypes argument, e.g.  names_ptypes = list(year = double()). What does it do?

   # vignette("pivot")
   stocks %>% 
       pivot_wider(names_from = year, values_from = return)

   # A tibble: 2 x 3
      half `2015` `2016`
     <dbl>  <dbl>  <dbl>
   1     1   1.88   0.92
   2     2   0.59   0.17

   Let’s have a look at the first part - here we take the year and make it a variable. That means that 2015 and 2016 become variables (new columns) in our new tibble, and the return gets pulled into the appropriate column (2015/2016) against the appropriate half. By nature of this move we changed year which was a double into two new column names which are 2015 and 2016 and hence “character”.

   (stocks_ <- stocks %>% 
     pivot_wider(names_from = year, values_from = return) %>% 
     pivot_longer(`2015`:`2016`, names_to = "year", values_to = "return"))

   # A tibble: 4 x 3
      half year  return
     <dbl> <chr>  <dbl>
   1     1 2015    1.88
   2     1 2016    0.92
   3     2 2015    0.59
   4     2 2016    0.17

   colnames(stocks)

   [1] "year"   "half"   "return"

   colnames(stocks_)

   [1] "half"   "year"   "return"

   So following on that we take these new columns and then collapse them into a column year again. But now we have changed the type given we made them columns in the pivot_wider() step. So they keep their “character” nature when they are made longer again. Final result is year started off double (when we created it) but ends up character (after the pivot_wider and pivot_longer steps).

   Th columns also get rearranged since the pivot_wider spreads the year column into 2015 and 2016 which come after half in that initial step. When we subsequently pivot_longer half remains as the first column, followed by the names_to = column (year in this case), and finally the values_to = column (return in this case).

   Q: pivot_longer() has a names_ptypes argument, e.g.  names_ptypes = list(year = double()). What does it do?

   Okay so upon reading the help page and the info I expected that this function would convert my character column year created after the pivot_wider() step into a double, but instead it throws an error. 😕

   stocks %>% 
     pivot_wider(names_from = year, values_from = return) %>% 
     pivot_longer(`2015`:`2016`, 
                  names_to = "year", 
                  names_ptypes = list(year = double()),
                  values_to = "return"
                  )

   Error: Can't convert <character> to <double>.

   We use this to confirm that the columns we create are of the type / class we expect - so here it provides a check it seems 🤷.
   

   To transform the column from character to double you would need to use the names_transform.

   (stocks_ptypes <- stocks %>% 
     pivot_wider(names_from = year, values_from = return) %>% 
     pivot_longer(`2015`:`2016`, 
                  names_to = "year", 
                  names_transform = list(year = as.double),
                  values_to = "return",
                  # is the value column of the type expected
                  values_ptypes = list(return = double())
                  ))

   # A tibble: 4 x 3
      half  year return
     <dbl> <dbl>  <dbl>
   1     1  2015   1.88
   2     1  2016   0.92
   3     2  2015   0.59
   4     2  2016   0.17

   Strangely though I would expect that if I transform a column from x to y (using names_transform), and then use names_ptypes to check if my name column is indeed now of type y that would be fine? It still throws an error, so my thinking is flawed here.

   stocks %>% 
     pivot_wider(names_from = year, values_from = return) %>% 
     pivot_longer(`2015`:`2016`, 
                  names_to = "year", 
                  names_transform = list(year = as.double),
                  names_ptypes = list(year = double()),
                  values_to = "return",
                  # is the value column of the type expected
                  values_ptypes = list(return = double())
                  )

   Error: Can't convert <character> to <double>.

2. Why does this code fail?

   table4a %>% 
     pivot_longer(c(1999, 2000), names_to = "year", values_to = "cases")

   Error: Can't subset columns that don't exist.
   x Locations 1999 and 2000 don't exist.
   i There are only 3 columns.

     # Error: Can't subset columns that don't exist.
     # x Locations 1999 and 2000 don't exist.
     # i There are only 3 columns.

   # fixing it
   table4a %>% 
     pivot_longer(c("1999", `2000`), names_to = "year", values_to = "cases")

   # A tibble: 6 x 3
     country     year   cases
     <chr>       <chr>  <int>
   1 Afghanistan 1999     745
   2 Afghanistan 2000    2666
   3 Brazil      1999   37737
   4 Brazil      2000   80488
   5 China       1999  212258
   6 China       2000  213766

   The 1999 and 2000 are non-syntactically named columns. These have to be surrounded by backticks (``) or quotations "". Here tidyr is trying to read columns numbered 1999, and 2000 which don’t exist.

3. What would happen if you widen this table? Why? How could you add a new column to uniquely identify each value?

   people <- tribble(
     ~name,             ~names,  ~values,
     #-----------------|--------|------
     "Phillip Woods",   "age",       45,
     "Phillip Woods",   "height",   186,
     "Phillip Woods",   "age",       50,
     "Jessica Cordero", "age",       37,
     "Jessica Cordero", "height",   156
   )

   You get a warning and it has a list for each variable age and height since Philips Woods has two ages which are different.

   people %>% 
     pivot_wider(names_from = names,
                 values_from = "values")

   # A tibble: 2 x 3
     name            age       height   
     <chr>           <list>    <list>   
   1 Phillip Woods   <dbl [2]> <dbl [1]>
   2 Jessica Cordero <dbl [1]> <dbl [1]>

   people2 <- tribble(
     ~name,             ~names,  ~values,
     #-----------------|--------|------
     "Phillip Woods",   "age",       45,
     "Phillip Woods",   "height",   186,
     "Phillip Woods",   "age2",      50, # second age gets diff col name
     "Jessica Cordero", "age",       37,
     "Jessica Cordero", "height",   156
   )

   people2 %>% 
     pivot_wider(names_from = names,
                 values_from = "values")

   # A tibble: 2 x 4
     name              age height  age2
     <chr>           <dbl>  <dbl> <dbl>
   1 Phillip Woods      45    186    50
   2 Jessica Cordero    37    156    NA

4. Tidy the simple tibble below. Do you need to make it wider or longer? What are the variables?

   (preg <- tribble(
     ~pregnant, ~male, ~female,
     "yes",     NA,    10,
     "no",      20,    12
   ))

   # A tibble: 2 x 3
     pregnant  male female
     <chr>    <dbl>  <dbl>
   1 yes         NA     10
   2 no          20     12

   We need to make it longer. The variable’s are pregnant (yes or no), and the number of male(s)/female(s) in each outcome of pregnant.

   preg %>% 
     pivot_longer(c('male', 'female'), 
                  names_to = 'sex',
                  values_to = 'count')

   # A tibble: 4 x 3
     pregnant sex    count
     <chr>    <chr>  <dbl>
   1 yes      male      NA
   2 yes      female    10
   3 no       male      20
   4 no       female    12

Separating and Uniting

Separate

separate() pulls apart one column into multiple columns, by splitting wherever a separator character appears, and by default, separate() will split values wherever it sees a non-alphanumeric character (i.e. a character that isn’t a number or letter e.g. ,, ;, etc.).

In tidyr::table3 we have a column rate that has a combination of the cases and population in one column separated by a /.

table3

# A tibble: 6 x 3
  country      year rate             
* <chr>       <int> <chr>            
1 Afghanistan  1999 745/19987071     
2 Afghanistan  2000 2666/20595360    
3 Brazil       1999 37737/172006362  
4 Brazil       2000 80488/174504898  
5 China        1999 212258/1272915272
6 China        2000 213766/1280428583

No sep specified

table3 %>% 
  # separate without specifying the separator
  separate(rate, into = c("cases", "population"))

# A tibble: 6 x 4
  country      year cases  population
  <chr>       <int> <chr>  <chr>     
1 Afghanistan  1999 745    19987071  
2 Afghanistan  2000 2666   20595360  
3 Brazil       1999 37737  172006362 
4 Brazil       2000 80488  174504898 
5 China        1999 212258 1272915272
6 China        2000 213766 1280428583

sep specified as single value

sep is a regular expression, so more complicated sequences than “/” maybe supplied, e.g. sep = '[0-9]' will separate on any digit encountered.

table3 %>% 
  # separate with specifying the separator
  separate(rate, into = c("cases", "population"),
           sep = '/')

# A tibble: 6 x 4
  country      year cases  population
  <chr>       <int> <chr>  <chr>     
1 Afghanistan  1999 745    19987071  
2 Afghanistan  2000 2666   20595360  
3 Brazil       1999 37737  172006362 
4 Brazil       2000 80488  174504898 
5 China        1999 212258 1272915272
6 China        2000 213766 1280428583

as_tibble("This0is1a2test.with3a4deliberate5mistake.") %>%
  separate(value, into = c(glue::glue("col_{c(1:7)}")),
         sep = '[0-9]')

# A tibble: 1 x 7
  col_1 col_2 col_3 col_4     col_5 col_6      col_7   
  <chr> <chr> <chr> <chr>     <chr> <chr>      <chr>   
1 This  is    a     test.with a     deliberate mistake.

sep specified as vector of integers / position

You can also pass a vector of integers to sep. separate() will interpret the integers as positions to split at.

• Positive values start at 1 and start counting from the left of the string
• Negative values start at -1 and start counting from the right of the string

table3 %>% 
  # separate with specifying the separator
  separate(year, into = c("century", "year"),
           # here we use a single value
           # use c(2, 5) to split in multiple places
           sep = 2,
           convert = TRUE)

# A tibble: 6 x 4
  country     century  year rate             
  <chr>         <int> <int> <chr>            
1 Afghanistan      19    99 745/19987071     
2 Afghanistan      20     0 2666/20595360    
3 Brazil           19    99 37737/172006362  
4 Brazil           20     0 80488/174504898  
5 China            19    99 212258/1272915272
6 China            20     0 213766/1280428583

convert = TRUE

separate() retains the character type of the original field. Use convert = TRUE to get separate() to make a guess at a better type for the resulting columns.

table3 %>% 
  # separate with specifying the separator
  separate(rate, into = c("cases", "population"),
           sep = '/',
           convert = TRUE)

# A tibble: 6 x 4
  country      year  cases population
  <chr>       <int>  <int>      <int>
1 Afghanistan  1999    745   19987071
2 Afghanistan  2000   2666   20595360
3 Brazil       1999  37737  172006362
4 Brazil       2000  80488  174504898
5 China        1999 212258 1272915272
6 China        2000 213766 1280428583

Unite

unite() is the inverse of separate(): it combines multiple columns into a single column.

The year column was deliberately made untidy by splitting it into century and year in tidyr::table5. We can concat these using unite.

table5 %>% 
  unite(new, century, year)

# A tibble: 6 x 3
  country     new   rate             
  <chr>       <chr> <chr>            
1 Afghanistan 19_99 745/19987071     
2 Afghanistan 20_00 2666/20595360    
3 Brazil      19_99 37737/172006362  
4 Brazil      20_00 80488/174504898  
5 China       19_99 212258/1272915272
6 China       20_00 213766/1280428583

If you look at the help page of unite (?unite) you will see that the default separator is ___ (sep = "_"). To override this behaviour we need to specify the sep argument.

table5 %>% 
  unite(new, century, year, sep = '')

# A tibble: 6 x 3
  country     new   rate             
  <chr>       <chr> <chr>            
1 Afghanistan 1999  745/19987071     
2 Afghanistan 2000  2666/20595360    
3 Brazil      1999  37737/172006362  
4 Brazil      2000  80488/174504898  
5 China       1999  212258/1272915272
6 China       2000  213766/1280428583

Exercises

1. What do the extra and fill arguments do in separate()? Experiment with the various options for the following two toy datasets.

   tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
     separate(x, c("one", "two", "three"))
   
   tibble(x = c("a,b,c", "d,e", "f,g,i")) %>% 
     separate(x, c("one", "two", "three"))

   • The default for extra is “warn”.
   • extra warns if during the splitting more columns are found than those specified in into. For example I may have into = c("year_of_birth", "month_of_birth") but upon splitting on - three columns are found. If I don’t specify extra then separate function will warn me there are more columns found, but these will be missing from my result.
   • extra can be warn (default), drop (drop any extra values without a warning) or merge (only splits at most length(into) times).

   tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
         separate(x, c("one", "two", "three"))

   
       Warning: Expected 3 pieces. Additional pieces discarded in 1 rows [2].
   

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 d     e     f    
       3 h     i     j    
   

   tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
         separate(x, c("one", "two", "three"),
                  extra = 'drop')

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 d     e     f    
       3 h     i     j    
   

   tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
         separate(x, c("one", "two", "three"),
                  extra = 'merge')

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 d     e     f,g  
       3 h     i     j    
   

   • The default for fill is “warn”.

   • fill warns if during the splitting less columns are found than those specified in into.
     For example I may have into = c("year_of_birth", "month_of_birth", "day_of_birth", "sex") but upon splitting on - only three columns are found. If I don’t specify fill then the separate function will warn me there are less columns than expected and these will be filled with NA.

   • fill can be:

     • warn (default)
     • right (fill with missing values on the right) - i.e. if there are not enough values for the columns in the tibble the right hand side columns must be filled with NA. You’d use this if you expect that more of the end columns are missing.
     • left (fill with missing values on the left) - i.e. if there are not enough values for the columns in the tibble the left hand side columns must be filled with NA. You’d use this if you expect that more of the beginning columns you’re making are likely to be missing.

   tibble(x = c("a,b,c", "d,e", "f,g,i")) %>% 
         separate(x, c("one", "two", "three"))

   
       Warning: Expected 3 pieces. Missing pieces filled with `NA` in 1 rows [2].
   

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 d     e     <NA> 
       3 f     g     i    
   

   tibble(x = c("a,b,c", "d,e", "f,g,i")) %>% 
         separate(x, c("one", "two", "three"),
                  fill = 'right')

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 d     e     <NA> 
       3 f     g     i    
   

   tibble(x = c("a,b,c", "d,e", "f,g,i")) %>% 
         separate(x, c("one", "two", "three"),
                  fill = 'left')

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 <NA>  d     e    
       3 f     g     i    
   

2. Both unite() and separate() have a remove argument. What does it do? Why would you set it to FALSE?

   • remove is by default TRUE.
   • It means remove the column(s) from which you are separating / uniting.
   • You would set it to FALSE if you’d like to perform the action (separate / unite) but also keep the original column(s).

   tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
         separate(x, c("one", "two", "three"))

   
       Warning: Expected 3 pieces. Additional pieces discarded in 1 rows [2].
   

   
       # A tibble: 3 x 3
         one   two   three
         <chr> <chr> <chr>
       1 a     b     c    
       2 d     e     f    
       3 h     i     j    
   

   tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
         separate(x, c("one", "two", "three"),
                  remove = FALSE)

   
       Warning: Expected 3 pieces. Additional pieces discarded in 1 rows [2].
   

   
       # A tibble: 3 x 4
         x       one   two   three
         <chr>   <chr> <chr> <chr>
       1 a,b,c   a     b     c    
       2 d,e,f,g d     e     f    
       3 h,i,j   h     i     j    
   

   Notice that initially x does not exist. In the second call we added the remove = FALSE and x is kept.

3. Compare and contrast separate() and extract(). Why are there three variations of separation (by position, by separator, and with groups), but only one unite?

   • separate has by position, by separator, and with groups because it takes a string and splits it into multiple parts; unite on the other hand combines separate columns and there is only one way to do it - specify the columns you want to join, in the order they should be joined, and with the separator that is appropriate.

   • separate does not throw away any part of the string unless it “falls over the edge” so to say. I.e. if we have extra string left over after the splitting, and it has run out of columns to place it in. You may kinda simulate this by saying “Hey, keep only the second column” as this help example shows. While sep is a regex it tells you how to split the string.

     # If you want to split by any non-alphanumeric value (the default):
     df <- data.frame(x = c(NA, "a.b", "a.d", "b.c"))
     df %>% separate(x, c("A", "B"))

          A    B
     1 <NA> <NA>
     2    a    b
     3    a    d
     4    b    c

     # If you just want the second variable:
     df %>% separate(x, c(NA, "B"))

          B
     1 <NA>
     2    b
     3    d
     4    c

   • extract will give you back only the “groups” you are interested in, and “throws” away the rest, and maybe better to use when you have complex patterns you want to extract. 🤷

   df <- tibble(x = c(NA, "a-b", "a-d", "b-c", "d-e"))
   
   df %>% tidyr::extract(x,
                        # make 2 columns
                        c("A", "B"), 
                        # here's the groups I want - surrounded by ()
                        "([a-d]+)-([a-d]+)")

   # A tibble: 5 x 2
     A     B    
     <chr> <chr>
   1 <NA>  <NA> 
   2 a     b    
   3 a     d    
   4 b     c    
   5 <NA>  <NA> 

   df %>% 
     tidyr::extract(col = x,
                   into = c("A"),
                   regex = "([a-d]+)")

   # A tibble: 5 x 1
     A    
     <chr>
   1 <NA> 
   2 a    
   3 a    
   4 b    
   5 d    

   df <- tibble(x = c(NA, "a-b-c", "a-d-e", "b-c-d", "d-e-f"))
   
   df %>% tidyr::extract(x,
                        # make 2 columns
                        c("A", "C"), 
                        # here's the groups I want - surrounded by ()
                        # only first, and last
                        "([a-z]+)-[a-z]-([a-z]+)")

   # A tibble: 5 x 2
     A     C    
     <chr> <chr>
   1 <NA>  <NA> 
   2 a     c    
   3 a     e    
   4 b     d    
   5 d     f    

   df %>% tidyr::separate(x,
                        # You can also do it using separate
                        # make 2 columns
                        c("A", "C"), 
                        # here's the groups I want - surrounded by ()
                        # only first, and last
                        "-[a-z]-")

   # A tibble: 5 x 2
     A     C    
     <chr> <chr>
   1 <NA>  <NA> 
   2 a     c    
   3 a     e    
   4 b     d    
   5 d     f    

Missing values

A value can be missing in one of two possible ways:

• Explicitly, i.e. flagged with NA.
• Implicitly, i.e. simply not present in the data.

For example here we have explicit NA as well as 2016 is missing quarter 1.

(stocks <- tibble(
  year   = c(2015, 2015, 2015, 2015, 2016, 2016, 2016),
  qtr    = c(   1,    2,    3,    4,    2,    3,    4),
  return = c(1.88, 0.59, 0.35,   NA, 0.92, 0.17, 2.66)
))

# A tibble: 7 x 3
   year   qtr return
  <dbl> <dbl>  <dbl>
1  2015     1   1.88
2  2015     2   0.59
3  2015     3   0.35
4  2015     4  NA   
5  2016     2   0.92
6  2016     3   0.17
7  2016     4   2.66

Representing the dataset can make the missing values more clear.

1. Show implicit missing values.

   stocks %>% 
     pivot_wider(names_from = year, values_from = return)

   # A tibble: 4 x 3
       qtr `2015` `2016`
     <dbl>  <dbl>  <dbl>
   1     1   1.88  NA   
   2     2   0.59   0.92
   3     3   0.35   0.17
   4     4  NA      2.66

2. Turn explicit missing values, into implicit missing values.

   stocks %>% 
     pivot_wider(names_from = year, values_from = return) %>% 
     pivot_longer(
       cols = c(`2015`, `2016`), 
       names_to = "year", 
       values_to = "return", 
       values_drop_na = TRUE
     )

   # A tibble: 6 x 3
       qtr year  return
     <dbl> <chr>  <dbl>
   1     1 2015    1.88
   2     2 2015    0.59
   3     2 2016    0.92
   4     3 2015    0.35
   5     3 2016    0.17
   6     4 2016    2.66

3. Use complete() : make missing values explicit, by looking at a set of columns and finding all combinations. The gaps are filled with NA explicitly.

   stocks %>% 
     complete(year, qtr)

   # A tibble: 8 x 3
      year   qtr return
     <dbl> <dbl>  <dbl>
   1  2015     1   1.88
   2  2015     2   0.59
   3  2015     3   0.35
   4  2015     4  NA   
   5  2016     1  NA   
   6  2016     2   0.92
   7  2016     3   0.17
   8  2016     4   2.66

4. Sometimes when working with data the missing are an indication that the previous rows value must be filled down.

   (treatment <- tribble(
     ~ person,           ~ treatment, ~response,
     "Derrick Whitmore", 1,           7,
     NA,                 2,           10,
     NA,                 3,           9,
     "Katherine Burke",  1,           4
   ))

   # A tibble: 4 x 3
     person           treatment response
     <chr>                <dbl>    <dbl>
   1 Derrick Whitmore         1        7
   2 <NA>                     2       10
   3 <NA>                     3        9
   4 Katherine Burke          1        4

   Using fill() allows you to fill a columns values with the most recent non-missing value.

   treatment %>% 
     fill(person)

   # A tibble: 4 x 3
     person           treatment response
     <chr>                <dbl>    <dbl>
   1 Derrick Whitmore         1        7
   2 Derrick Whitmore         2       10
   3 Derrick Whitmore         3        9
   4 Katherine Burke          1        4

Exercises

1. Compare and contrast the fill arguments to pivot_wider() and complete().

   • Both fill missing values with something other than NA.
   • In pivot_wider() it is called values_fill, and the spec is either a scalar or named list.
   • In complete() it is called fill, and the spec is a named list.

   (df <- tibble(
        group = c(1:2, NA_real_, 1),
        item_id = c(1:2, 1, 2),
        item_name = c("a", "NA", "b", "b"),
        value1 = 1:4,
        value2 = 4:7
    ))

   # A tibble: 4 x 5
     group item_id item_name value1 value2
     <dbl>   <dbl> <chr>      <int>  <int>
   1     1       1 a              1      4
   2     2       2 NA             2      5
   3    NA       1 b              3      6
   4     1       2 b              4      7

   df %>% 
     pivot_wider(names_from = item_name, 
                 values_from = value1,
                 # the values are filled with 0 if missing
                 values_fill = 0)

   # A tibble: 4 x 6
     group item_id value2     a  `NA`     b
     <dbl>   <dbl>  <int> <int> <int> <int>
   1     1       1      4     1     0     0
   2     2       2      5     0     2     0
   3    NA       1      6     0     0     3
   4     1       2      7     0     0     4

   df %>% 
     complete(group, item_id,
              # explicitly list cols you want filled
              # with alternate to NA
              fill = list(group = 0,
              item_name = "0"))

   # A tibble: 6 x 5
     group item_id item_name value1 value2
     <dbl>   <dbl> <chr>      <int>  <int>
   1     1       1 a              1      4
   2     1       2 b              4      7
   3     2       1 0             NA     NA
   4     2       2 NA             2      5
   5     0       1 b              3      6
   6     0       2 0             NA     NA

2. What does the direction argument to fill() do?

   The argument is .direction and tells you in which direction the fill() must occur, and can be one of 4 different values.

   .direction = c(“down”, “up”, “downup”, “updown”)

   # Value (n_squirrels) is missing above and below within a group
   (squirrels <- tibble::tribble(
     ~group,    ~name,     ~role,     ~n_squirrels,
     1,      "Sam",    "Observer",   NA,
     1,     "Mara", "Scorekeeper",    8,
     1,    "Jesse",    "Observer",   NA,
     1,      "Tom",    "Observer",   NA,
     2,     "Mike",    "Observer",   NA,
     2,  "Rachael",    "Observer",   NA,
     2,  "Sydekea", "Scorekeeper",   14,
     2, "Gabriela",    "Observer",   NA,
     3,  "Derrick",    "Observer",   NA,
     3,     "Kara", "Scorekeeper",    9,
     3,    "Emily",    "Observer",   NA,
     3, "Danielle",    "Observer",   NA
   ))

   # A tibble: 12 x 4
      group name     role        n_squirrels
      <dbl> <chr>    <chr>             <dbl>
    1     1 Sam      Observer             NA
    2     1 Mara     Scorekeeper           8
    3     1 Jesse    Observer             NA
    4     1 Tom      Observer             NA
    5     2 Mike     Observer             NA
    6     2 Rachael  Observer             NA
    7     2 Sydekea  Scorekeeper          14
    8     2 Gabriela Observer             NA
    9     3 Derrick  Observer             NA
   10     3 Kara     Scorekeeper           9
   11     3 Emily    Observer             NA
   12     3 Danielle Observer             NA

   # The values are inconsistently missing by position within the group
   # Use .direction = "downup" to fill missing values in both directions
   squirrels %>%
     dplyr::group_by(group) %>%
     fill(n_squirrels, .direction = "downup") %>%
     dplyr::ungroup()

   # A tibble: 12 x 4
      group name     role        n_squirrels
      <dbl> <chr>    <chr>             <dbl>
    1     1 Sam      Observer              8
    2     1 Mara     Scorekeeper           8
    3     1 Jesse    Observer              8
    4     1 Tom      Observer              8
    5     2 Mike     Observer             14
    6     2 Rachael  Observer             14
    7     2 Sydekea  Scorekeeper          14
    8     2 Gabriela Observer             14
    9     3 Derrick  Observer              9
   10     3 Kara     Scorekeeper           9
   11     3 Emily    Observer              9
   12     3 Danielle Observer              9

   # contrast with "updown"
   squirrels %>%
     dplyr::group_by(group) %>%
     # should be the same in this e.g.
     fill(n_squirrels, .direction = "updown") %>%
     dplyr::ungroup()

   # A tibble: 12 x 4
      group name     role        n_squirrels
      <dbl> <chr>    <chr>             <dbl>
    1     1 Sam      Observer              8
    2     1 Mara     Scorekeeper           8
    3     1 Jesse    Observer              8
    4     1 Tom      Observer              8
    5     2 Mike     Observer             14
    6     2 Rachael  Observer             14
    7     2 Sydekea  Scorekeeper          14
    8     2 Gabriela Observer             14
    9     3 Derrick  Observer              9
   10     3 Kara     Scorekeeper           9
   11     3 Emily    Observer              9
   12     3 Danielle Observer              9

   # contrast with "down"
   squirrels %>%
     dplyr::group_by(group) %>%
     fill(n_squirrels, .direction = "down") %>%
     dplyr::ungroup()

   # A tibble: 12 x 4
      group name     role        n_squirrels
      <dbl> <chr>    <chr>             <dbl>
    1     1 Sam      Observer             NA
    2     1 Mara     Scorekeeper           8
    3     1 Jesse    Observer              8
    4     1 Tom      Observer              8
    5     2 Mike     Observer             NA
    6     2 Rachael  Observer             NA
    7     2 Sydekea  Scorekeeper          14
    8     2 Gabriela Observer             14
    9     3 Derrick  Observer             NA
   10     3 Kara     Scorekeeper           9
   11     3 Emily    Observer              9
   12     3 Danielle Observer              9

   # contrast with "up"
   squirrels %>%
     dplyr::group_by(group) %>%
     fill(n_squirrels, .direction = "up") %>%
     dplyr::ungroup()

   # A tibble: 12 x 4
      group name     role        n_squirrels
      <dbl> <chr>    <chr>             <dbl>
    1     1 Sam      Observer              8
    2     1 Mara     Scorekeeper           8
    3     1 Jesse    Observer             NA
    4     1 Tom      Observer             NA
    5     2 Mike     Observer             14
    6     2 Rachael  Observer             14
    7     2 Sydekea  Scorekeeper          14
    8     2 Gabriela Observer             NA
    9     3 Derrick  Observer              9
   10     3 Kara     Scorekeeper           9
   11     3 Emily    Observer             NA
   12     3 Danielle Observer             NA

Case Study

who

# A tibble: 7,240 x 60
   country iso2  iso3   year new_sp_m014 new_sp_m1524 new_sp_m2534 new_sp_m3544
   <chr>   <chr> <chr> <int>       <int>        <int>        <int>        <int>
 1 Afghan~ AF    AFG    1980          NA           NA           NA           NA
 2 Afghan~ AF    AFG    1981          NA           NA           NA           NA
 3 Afghan~ AF    AFG    1982          NA           NA           NA           NA
 4 Afghan~ AF    AFG    1983          NA           NA           NA           NA
 5 Afghan~ AF    AFG    1984          NA           NA           NA           NA
 6 Afghan~ AF    AFG    1985          NA           NA           NA           NA
 7 Afghan~ AF    AFG    1986          NA           NA           NA           NA
 8 Afghan~ AF    AFG    1987          NA           NA           NA           NA
 9 Afghan~ AF    AFG    1988          NA           NA           NA           NA
10 Afghan~ AF    AFG    1989          NA           NA           NA           NA
# ... with 7,230 more rows, and 52 more variables: new_sp_m4554 <int>,
#   new_sp_m5564 <int>, new_sp_m65 <int>, new_sp_f014 <int>,
#   new_sp_f1524 <int>, new_sp_f2534 <int>, new_sp_f3544 <int>,
#   new_sp_f4554 <int>, new_sp_f5564 <int>, new_sp_f65 <int>,
#   new_sn_m014 <int>, new_sn_m1524 <int>, new_sn_m2534 <int>,
#   new_sn_m3544 <int>, new_sn_m4554 <int>, new_sn_m5564 <int>,
#   new_sn_m65 <int>, new_sn_f014 <int>, new_sn_f1524 <int>,
#   new_sn_f2534 <int>, new_sn_f3544 <int>, new_sn_f4554 <int>,
#   new_sn_f5564 <int>, new_sn_f65 <int>, new_ep_m014 <int>,
#   new_ep_m1524 <int>, new_ep_m2534 <int>, new_ep_m3544 <int>,
#   new_ep_m4554 <int>, new_ep_m5564 <int>, new_ep_m65 <int>,
#   new_ep_f014 <int>, new_ep_f1524 <int>, new_ep_f2534 <int>,
#   new_ep_f3544 <int>, new_ep_f4554 <int>, new_ep_f5564 <int>,
#   new_ep_f65 <int>, newrel_m014 <int>, newrel_m1524 <int>,
#   newrel_m2534 <int>, newrel_m3544 <int>, newrel_m4554 <int>,
#   newrel_m5564 <int>, newrel_m65 <int>, newrel_f014 <int>,
#   newrel_f1524 <int>, newrel_f2534 <int>, newrel_f3544 <int>,
#   newrel_f4554 <int>, newrel_f5564 <int>, newrel_f65 <int>

(
  who1 <- who %>% 
          # change the wide data to long data
          pivot_longer(
            cols = new_sp_m014:newrel_f65, 
            names_to = "key", 
            values_to = "cases", 
            values_drop_na = TRUE
          )
)

# A tibble: 76,046 x 6
   country     iso2  iso3   year key          cases
   <chr>       <chr> <chr> <int> <chr>        <int>
 1 Afghanistan AF    AFG    1997 new_sp_m014      0
 2 Afghanistan AF    AFG    1997 new_sp_m1524    10
 3 Afghanistan AF    AFG    1997 new_sp_m2534     6
 4 Afghanistan AF    AFG    1997 new_sp_m3544     3
 5 Afghanistan AF    AFG    1997 new_sp_m4554     5
 6 Afghanistan AF    AFG    1997 new_sp_m5564     2
 7 Afghanistan AF    AFG    1997 new_sp_m65       0
 8 Afghanistan AF    AFG    1997 new_sp_f014      5
 9 Afghanistan AF    AFG    1997 new_sp_f1524    38
10 Afghanistan AF    AFG    1997 new_sp_f2534    36
# ... with 76,036 more rows

who1 %>% 
  # how many keys do we have?
  count(key, sort = TRUE) %>% 
  DT::datatable()

To parse the columns, refer to Chaper 9 (Chapter 12 online).

(who2 <- who1 %>% 
  # Look in key. Do you see any strings that have "newrel"?
  # If you do replace these with new_rel for consistent
  # key values for easier parsing
  mutate(key = stringr::str_replace(key,
                                    "newrel",
                                    "new_rel")))

# A tibble: 76,046 x 6
   country     iso2  iso3   year key          cases
   <chr>       <chr> <chr> <int> <chr>        <int>
 1 Afghanistan AF    AFG    1997 new_sp_m014      0
 2 Afghanistan AF    AFG    1997 new_sp_m1524    10
 3 Afghanistan AF    AFG    1997 new_sp_m2534     6
 4 Afghanistan AF    AFG    1997 new_sp_m3544     3
 5 Afghanistan AF    AFG    1997 new_sp_m4554     5
 6 Afghanistan AF    AFG    1997 new_sp_m5564     2
 7 Afghanistan AF    AFG    1997 new_sp_m65       0
 8 Afghanistan AF    AFG    1997 new_sp_f014      5
 9 Afghanistan AF    AFG    1997 new_sp_f1524    38
10 Afghanistan AF    AFG    1997 new_sp_f2534    36
# ... with 76,036 more rows

(
  who3 <- who2 %>% 
    separate(key,
             c("new", "type", "sexage"),
             sep = "_")
)

# A tibble: 76,046 x 8
   country     iso2  iso3   year new   type  sexage cases
   <chr>       <chr> <chr> <int> <chr> <chr> <chr>  <int>
 1 Afghanistan AF    AFG    1997 new   sp    m014       0
 2 Afghanistan AF    AFG    1997 new   sp    m1524     10
 3 Afghanistan AF    AFG    1997 new   sp    m2534      6
 4 Afghanistan AF    AFG    1997 new   sp    m3544      3
 5 Afghanistan AF    AFG    1997 new   sp    m4554      5
 6 Afghanistan AF    AFG    1997 new   sp    m5564      2
 7 Afghanistan AF    AFG    1997 new   sp    m65        0
 8 Afghanistan AF    AFG    1997 new   sp    f014       5
 9 Afghanistan AF    AFG    1997 new   sp    f1524     38
10 Afghanistan AF    AFG    1997 new   sp    f2534     36
# ... with 76,036 more rows

Ok, from the chapter text it seems we only have new cases, so let us verify that.

who3 %>% 
  count(new)

# A tibble: 1 x 2
  new       n
  <chr> <int>
1 new   76046

This is indeed the case! The count == the dataset count so we can drop this column

(
  who4 <- who3 %>% 
    select(-new, -iso2, iso3)
)

# A tibble: 76,046 x 6
   country     iso3   year type  sexage cases
   <chr>       <chr> <int> <chr> <chr>  <int>
 1 Afghanistan AFG    1997 sp    m014       0
 2 Afghanistan AFG    1997 sp    m1524     10
 3 Afghanistan AFG    1997 sp    m2534      6
 4 Afghanistan AFG    1997 sp    m3544      3
 5 Afghanistan AFG    1997 sp    m4554      5
 6 Afghanistan AFG    1997 sp    m5564      2
 7 Afghanistan AFG    1997 sp    m65        0
 8 Afghanistan AFG    1997 sp    f014       5
 9 Afghanistan AFG    1997 sp    f1524     38
10 Afghanistan AFG    1997 sp    f2534     36
# ... with 76,036 more rows

Let’s separate the sexage column.

(
  who5 <- who4 %>% 
    separate(sexage, c("sex", "age"),
             sep = 1)
)

# A tibble: 76,046 x 7
   country     iso3   year type  sex   age   cases
   <chr>       <chr> <int> <chr> <chr> <chr> <int>
 1 Afghanistan AFG    1997 sp    m     014       0
 2 Afghanistan AFG    1997 sp    m     1524     10
 3 Afghanistan AFG    1997 sp    m     2534      6
 4 Afghanistan AFG    1997 sp    m     3544      3
 5 Afghanistan AFG    1997 sp    m     4554      5
 6 Afghanistan AFG    1997 sp    m     5564      2
 7 Afghanistan AFG    1997 sp    m     65        0
 8 Afghanistan AFG    1997 sp    f     014       5
 9 Afghanistan AFG    1997 sp    f     1524     38
10 Afghanistan AFG    1997 sp    f     2534     36
# ... with 76,036 more rows

Pipeline version

who %>% 
  pivot_longer(cols = new_sp_m014:newrel_f65,
               names_to = "key",
               values_to = "cases",
               values_drop_na = TRUE) %>% 
  mutate(key = stringr::str_replace(key, "newrel", "new_rel")) %>% 
  separate(key, c("new", "type", "sexage"),
                  sep = "_") %>% 
  select(-new, -iso2, -iso3) %>% 
  separate(sexage, c("sex", "age"),
           # first pos = sex, rest = age group
           sep = 1)

# A tibble: 76,046 x 6
   country      year type  sex   age   cases
   <chr>       <int> <chr> <chr> <chr> <int>
 1 Afghanistan  1997 sp    m     014       0
 2 Afghanistan  1997 sp    m     1524     10
 3 Afghanistan  1997 sp    m     2534      6
 4 Afghanistan  1997 sp    m     3544      3
 5 Afghanistan  1997 sp    m     4554      5
 6 Afghanistan  1997 sp    m     5564      2
 7 Afghanistan  1997 sp    m     65        0
 8 Afghanistan  1997 sp    f     014       5
 9 Afghanistan  1997 sp    f     1524     38
10 Afghanistan  1997 sp    f     2534     36
# ... with 76,036 more rows

Exercises

1. In this case study I set values_drop_na = TRUE just to make it easier to check that we had the correct values. Is this reasonable? Think about how missing values are represented in this dataset. Are there implicit missing values? What’s the difference between an NA and zero?

   If we consider the missing values in this data set, it does seem reasonable.

   who %>% 
     select(new_sp_m014:newrel_f65) %>% 
     skimr::skim()

   Data summary

   Name	Piped data
   Number of rows	7240
   Number of columns	56
   _______________________
   Column type frequency:
   numeric	56
   ________________________
   Group variables	None

   Variable type: numeric

   skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
   new_sp_m014	4067	0.44	83.71	316.14	0	0.00	5.0	37.00	5001	▇▁▁▁▁
   new_sp_m1524	4031	0.44	1015.66	4885.38	0	9.00	90.0	502.00	78278	▇▁▁▁▁
   new_sp_m2534	4034	0.44	1403.80	5718.39	0	14.00	150.0	715.50	84003	▇▁▁▁▁
   new_sp_m3544	4021	0.44	1315.88	6003.26	0	13.00	130.0	583.50	90830	▇▁▁▁▁
   new_sp_m4554	4017	0.45	1103.86	5441.06	0	12.00	102.0	440.00	82921	▇▁▁▁▁
   new_sp_m5564	4022	0.44	800.70	4418.31	0	8.00	63.0	279.00	63814	▇▁▁▁▁
   new_sp_m65	4031	0.44	682.82	4089.14	0	8.00	53.0	232.00	70376	▇▁▁▁▁
   new_sp_f014	4066	0.44	114.33	504.63	0	1.00	7.0	50.75	8576	▇▁▁▁▁
   new_sp_f1524	4046	0.44	826.11	3552.02	0	7.00	66.0	421.00	53975	▇▁▁▁▁
   new_sp_f2534	4040	0.44	917.30	3580.15	0	9.00	84.0	476.25	49887	▇▁▁▁▁
   new_sp_f3544	4041	0.44	640.43	2542.51	0	6.00	57.0	308.00	34698	▇▁▁▁▁
   new_sp_f4554	4036	0.44	445.78	1799.23	0	4.00	38.0	211.00	23977	▇▁▁▁▁
   new_sp_f5564	4045	0.44	313.87	1381.25	0	3.00	25.0	146.50	18203	▇▁▁▁▁
   new_sp_f65	4043	0.44	283.93	1267.94	0	4.00	30.0	129.00	21339	▇▁▁▁▁
   new_sn_m014	6195	0.14	308.75	1727.25	0	1.00	9.0	61.00	22355	▇▁▁▁▁
   new_sn_m1524	6210	0.14	513.02	3643.27	0	2.00	15.5	102.00	60246	▇▁▁▁▁
   new_sn_m2534	6218	0.14	653.69	3430.03	0	2.00	23.0	135.50	50282	▇▁▁▁▁
   new_sn_m3544	6215	0.14	837.87	8524.53	0	2.00	19.0	132.00	250051	▇▁▁▁▁
   new_sn_m4554	6213	0.14	520.79	3301.70	0	2.00	19.0	127.50	57181	▇▁▁▁▁
   new_sn_m5564	6219	0.14	448.62	3488.68	0	2.00	16.0	101.00	64972	▇▁▁▁▁
   new_sn_m65	6220	0.14	460.36	3991.90	0	2.00	20.5	111.75	74282	▇▁▁▁▁
   new_sn_f014	6200	0.14	291.95	1647.30	0	1.00	8.0	58.00	21406	▇▁▁▁▁
   new_sn_f1524	6218	0.14	407.90	2379.13	0	1.00	12.0	89.00	35518	▇▁▁▁▁
   new_sn_f2534	6224	0.14	466.26	2272.86	0	2.00	18.0	103.25	28753	▇▁▁▁▁
   new_sn_f3544	6220	0.14	506.59	5013.53	0	1.00	11.0	82.25	148811	▇▁▁▁▁
   new_sn_f4554	6222	0.14	271.16	1511.72	0	1.00	10.0	76.75	23869	▇▁▁▁▁
   new_sn_f5564	6223	0.14	213.39	1468.62	0	1.00	8.0	56.00	26085	▇▁▁▁▁
   new_sn_f65	6221	0.14	230.75	1597.70	0	1.00	13.0	74.00	29630	▇▁▁▁▁
   new_ep_m014	6202	0.14	128.61	460.14	0	0.00	6.0	55.00	7869	▇▁▁▁▁
   new_ep_m1524	6214	0.14	158.30	537.74	0	1.00	11.0	88.00	8558	▇▁▁▁▁
   new_ep_m2534	6220	0.14	201.23	764.05	0	1.00	13.0	124.00	11843	▇▁▁▁▁
   new_ep_m3544	6216	0.14	272.72	3381.41	0	1.00	10.5	91.25	105825	▇▁▁▁▁
   new_ep_m4554	6220	0.14	108.11	380.61	0	1.00	8.5	63.25	5875	▇▁▁▁▁
   new_ep_m5564	6225	0.14	72.17	234.55	0	1.00	7.0	46.00	3957	▇▁▁▁▁
   new_ep_m65	6222	0.14	78.94	227.34	0	1.00	10.0	55.00	3061	▇▁▁▁▁
   new_ep_f014	6208	0.14	112.89	446.55	0	0.00	5.0	50.00	6960	▇▁▁▁▁
   new_ep_f1524	6219	0.14	149.17	543.89	0	1.00	9.0	78.00	7866	▇▁▁▁▁
   new_ep_f2534	6219	0.14	189.52	761.79	0	1.00	12.0	95.00	10759	▇▁▁▁▁
   new_ep_f3544	6219	0.14	241.70	3218.50	0	1.00	9.0	77.00	101015	▇▁▁▁▁
   new_ep_f4554	6223	0.14	93.77	339.33	0	1.00	8.0	56.00	6759	▇▁▁▁▁
   new_ep_f5564	6223	0.14	63.04	212.95	0	1.00	6.0	42.00	4684	▇▁▁▁▁
   new_ep_f65	6226	0.14	72.31	202.72	0	0.00	10.0	51.00	2548	▇▁▁▁▁
   newrel_m014	7050	0.03	538.18	2082.18	0	5.00	32.5	210.00	18617	▇▁▁▁▁
   newrel_m1524	7058	0.03	1489.51	6848.18	0	17.50	171.0	684.25	84785	▇▁▁▁▁
   newrel_m2534	7057	0.03	2139.72	7539.87	0	25.00	217.0	1091.00	76917	▇▁▁▁▁
   newrel_m3544	7056	0.03	2036.40	7847.94	0	24.75	208.0	851.25	84565	▇▁▁▁▁
   newrel_m4554	7056	0.03	1835.07	8324.28	0	19.00	175.0	688.50	100297	▇▁▁▁▁
   newrel_m5564	7055	0.03	1525.30	8760.27	0	13.00	136.0	536.00	112558	▇▁▁▁▁
   newrel_m65	7058	0.03	1426.00	9431.99	0	17.00	117.0	453.50	124476	▇▁▁▁▁
   newrel_f014	7050	0.03	532.84	2117.78	0	5.00	32.5	226.00	18054	▇▁▁▁▁
   newrel_f1524	7056	0.03	1161.85	4606.76	0	10.75	123.0	587.75	49491	▇▁▁▁▁
   newrel_f2534	7058	0.03	1472.80	5259.59	0	18.00	161.0	762.50	44985	▇▁▁▁▁
   newrel_f3544	7057	0.03	1125.01	4210.58	0	12.50	125.0	544.50	38804	▇▁▁▁▁
   newrel_f4554	7057	0.03	877.27	3556.18	0	10.00	92.0	400.50	37138	▇▁▁▁▁
   newrel_f5564	7057	0.03	686.41	3379.33	0	8.00	69.0	269.00	40892	▇▁▁▁▁
   newrel_f65	7055	0.03	683.76	3618.47	0	9.00	69.0	339.00	47438	▇▁▁▁▁

   who %>% 
     select(new_sp_m014) %>% 
     filter(new_sp_m014 == 0 |
            is.na(new_sp_m014)) %>% 
     skimr::skim()   

   Data summary

   Name	Piped data
   Number of rows	4929
   Number of columns	1
   _______________________
   Column type frequency:
   numeric	1
   ________________________
   Group variables	None

   Variable type: numeric

   skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
   new_sp_m014	4067	0.17	0	0	0	0	0	0	0	▁▁▇▁▁

   who %>% 
     select(new_sp_m014) %>% 
     filter(new_sp_m014 == 0) %>% 
     skimr::skim()   

   Data summary

   Name	Piped data
   Number of rows	862
   Number of columns	1
   _______________________
   Column type frequency:
   numeric	1
   ________________________
   Group variables	None

   Variable type: numeric

   skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
   new_sp_m014	0	1	0	0	0	0	0	0	0	▁▁▇▁▁

   who %>% 
     select(new_sp_m014) %>% 
     filter(is.na(new_sp_m014)) %>% 
     skimr::skim()   

   Data summary

   Name	Piped data
   Number of rows	4067
   Number of columns	1
   _______________________
   Column type frequency:
   numeric	1
   ________________________
   Group variables	None

   Variable type: numeric

   skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
   new_sp_m014	4067	0	NaN	NA	NA	NA	NA	NA	NA

   There may be implicit values if a year is missing for a country.

   0 means I received a measure for that variable and it is no cases. NA could mean I received no information for that variable so cases may or may not exist.

2. What happens if you neglect the mutate() step? (mutate(names_from = stringr::str_replace(key, "newrel", "new_rel")))

   For some of the steps we will have misaligned data when we separate these later in the pipeline.

   Let’s take for example newrel_m3544

   • in the separate(key, …) step new = newrel, type = m3544
   • this will complicate the dropping of “new”
   • it will also mean that type would be m3544.

3. I claimed that iso2 and iso3 were redundant with country. Confirm this claim.

   who %>% 
     count(country, sort = TRUE)

   # A tibble: 219 x 2
      country                 n
      <chr>               <int>
    1 Afghanistan            34
    2 Albania                34
    3 Algeria                34
    4 American Samoa         34
    5 Andorra                34
    6 Angola                 34
    7 Anguilla               34
    8 Antigua and Barbuda    34
    9 Argentina              34
   10 Armenia                34
   # ... with 209 more rows

   who %>% 
     count(iso2, sort = TRUE)

   # A tibble: 219 x 2
      iso2      n
      <chr> <int>
    1 AD       34
    2 AE       34
    3 AF       34
    4 AG       34
    5 AI       34
    6 AL       34
    7 AM       34
    8 AO       34
    9 AR       34
   10 AS       34
   # ... with 209 more rows

   who %>% 
     count(iso3, sort = TRUE)

   # A tibble: 219 x 2
      iso3      n
      <chr> <int>
    1 ABW      34
    2 AFG      34
    3 AGO      34
    4 AIA      34
    5 ALB      34
    6 AND      34
    7 ARE      34
    8 ARG      34
    9 ARM      34
   10 ASM      34
   # ... with 209 more rows

   who %>% 
     select(country, iso2, iso3) %>% 
     skimr::skim()

   Data summary

   Name	Piped data
   Number of rows	7240
   Number of columns	3
   _______________________
   Column type frequency:
   character	3
   ________________________
   Group variables	None

   Variable type: character

   skim_variable	n_missing	complete_rate	min	max	empty	n_unique	whitespace
   country	0	1	4	52	0	219	0
   iso2	0	1	2	2	0	219	0
   iso3	0	1	3	3	0	219	0

   In the skim for character variables the min tells you the min chars in a observation, and max is the max chars in a variable.

4. For each country, year, and sex compute the total number of cases of TB. Make an informative visualisation of the data.

   who5 %>% 
     group_by(country, year, sex) %>% 
     summarise(total_cases = sum(cases)) %>% 
     ungroup() %>% 
     arrange(country, year, sex) %>% 
     filter(country %in% c("South Africa", "Botswana",
                          "Zimbabwe", "Namibia", "Swaziland",
                          "Lesotho", "Mozambique")) %>% 
     ggplot(aes(x = year, y = total_cases, colour = sex)) +
     geom_point(alpha = 0.5) +
     facet_wrap(~ country, scales = "free_y") +
     labs(title = "Total Cases Across South Africa 
          and neighbouring countries over time",
          subtitle = "The number of cases is growing quite a bit over time",
          x = "", 
          y = "Cases Per Year") +
     scale_colour_tq()

   

   who5 %>% 
     group_by(country, year, sex) %>% 
     summarise(total_cases = sum(cases)) %>% 
     ungroup() %>% 
     arrange(country, year, sex) %>% 
     group_by(year, sex) %>% 
     summarise(cases_per_year = sum(total_cases)) %>% 
     ungroup() %>% 
     ggplot(aes(x = year, y = cases_per_year, fill = sex)) +
     geom_col(alpha = 0.5) +
     scale_fill_tq() +
     scale_y_log10() +
     labs(title = "Total Cases Across all countries over time",
          subtitle = "The number of cases is growing quite a bit over time",
          x = "", 
          y = "Cases Per Year")

   

Session information

sessionInfo()

R version 3.6.3 (2020-02-29)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 18363)

Matrix products: default

locale:
[1] LC_COLLATE=English_South Africa.1252  LC_CTYPE=English_South Africa.1252   
[3] LC_MONETARY=English_South Africa.1252 LC_NUMERIC=C                         
[5] LC_TIME=English_South Africa.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
 [1] tidyquant_1.0.0            quantmod_0.4.17           
 [3] TTR_0.23-6                 PerformanceAnalytics_2.0.4
 [5] xts_0.12-0                 zoo_1.8-7                 
 [7] magrittr_1.5               lubridate_1.7.8           
 [9] emo_0.0.0.9000             flair_0.0.2               
[11] forcats_0.5.0              stringr_1.4.0             
[13] dplyr_1.0.0                purrr_0.3.4               
[15] readr_1.3.1                tidyr_1.1.0               
[17] tibble_3.0.3               ggplot2_3.3.0             
[19] tidyverse_1.3.0            workflowr_1.6.2           

loaded via a namespace (and not attached):
 [1] httr_1.4.2        sass_0.2.0        jsonlite_1.7.0    modelr_0.1.6     
 [5] assertthat_0.2.1  highr_0.8         cellranger_1.1.0  yaml_2.2.1       
 [9] pillar_1.4.6      backports_1.1.6   lattice_0.20-38   glue_1.4.1       
[13] quadprog_1.5-8    digest_0.6.25     promises_1.1.0    checkmate_2.0.0  
[17] rvest_0.3.5       snakecase_0.11.0  colorspace_1.4-1  htmltools_0.5.0  
[21] httpuv_1.5.2      pkgconfig_2.0.3   broom_0.5.6       haven_2.2.0      
[25] scales_1.1.0      whisker_0.4       later_1.0.0       git2r_0.26.1     
[29] generics_0.0.2    farver_2.0.3      DT_0.13           ellipsis_0.3.1   
[33] repr_1.1.0        withr_2.2.0       janitor_2.0.1     skimr_2.1.1      
[37] cli_2.0.2         crayon_1.3.4      readxl_1.3.1      evaluate_0.14    
[41] fs_1.4.1          fansi_0.4.1       nlme_3.1-144      xml2_1.3.2       
[45] tools_3.6.3       hms_0.5.3         lifecycle_0.2.0   munsell_0.5.0    
[49] reprex_0.3.0      compiler_3.6.3    rlang_0.4.7       grid_3.6.3       
[53] gt_0.2.2          rstudioapi_0.11   htmlwidgets_1.5.1 crosstalk_1.1.0.1
[57] base64enc_0.1-3   labeling_0.3      rmarkdown_2.4     gtable_0.3.0     
[61] DBI_1.1.0         curl_4.3          R6_2.4.1          knitr_1.28       
[65] utf8_1.1.4        rprojroot_1.3-2   Quandl_2.10.0     stringi_1.4.6    
[69] Rcpp_1.0.4.6      vctrs_0.3.2       dbplyr_1.4.3      tidyselect_1.1.0 
[73] xfun_0.13