Tidying Data with tidyr
R Markdown
Paul Jozefek
2020-07-31
The tidyr package (part of the tidyverse) helps to organize or restructure messy or inconvenient datasets.
The Datasets
The following datasets will be used to demonstrate some of tidyr’s functionality. Each dataset shows the same four variables in a different way: country, year, population, and cases
> library(tidyverse) #load tidyr and datasets
> library(knitr) #load library for table formatting
> library(kableExtra) #load library for table formatting
>
> # table1
> kable(table1) %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| 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 |
> # table2
> kable(table2) %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | type | count |
|---|---|---|---|
| Afghanistan | 1999 | cases | 745 |
| Afghanistan | 1999 | population | 19987071 |
| Afghanistan | 2000 | cases | 2666 |
| Afghanistan | 2000 | population | 20595360 |
| Brazil | 1999 | cases | 37737 |
| Brazil | 1999 | population | 172006362 |
| Brazil | 2000 | cases | 80488 |
| Brazil | 2000 | population | 174504898 |
| China | 1999 | cases | 212258 |
| China | 1999 | population | 1272915272 |
| China | 2000 | cases | 213766 |
| China | 2000 | population | 1280428583 |
> # table3
> kable(table3) %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | rate |
|---|---|---|
| Afghanistan | 1999 | 745/19987071 |
| Afghanistan | 2000 | 2666/20595360 |
| Brazil | 1999 | 37737/172006362 |
| Brazil | 2000 | 80488/174504898 |
| China | 1999 | 212258/1272915272 |
| China | 2000 | 213766/1280428583 |
> # table4a
> kable(table4a) %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | 1999 | 2000 |
|---|---|---|
| Afghanistan | 745 | 2666 |
| Brazil | 37737 | 80488 |
| China | 212258 | 213766 |
> # table4b
> kable(table4b) %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | 1999 | 2000 |
|---|---|---|
| Afghanistan | 19987071 | 20595360 |
| Brazil | 172006362 | 174504898 |
| China | 1272915272 | 1280428583 |
There are three rules (interrelated) that make a datset tidy:
- Each variable must have its own column.
- Each observation must have its own row.
- Each value must have its own cell.
In this case only table1 is tidy.
The Datasets - Exercises
- Compute the
ratefortable2andtable4a+table4b. You will need to perform four operations:
Extract the number of TB cases per country per year.
Extract the matching population per country per year.
Divide cases by population and multiply by 10,000.
Store back in the appropriate place.
- Which representation is easiest to work with?
Table 4a and 4b were easier to work with.
Table2
# A tibble: 6 x 4
country year type count
<chr> <int> <chr> <int>
1 Afghanistan 1999 cases 745
2 Afghanistan 2000 cases 2666
3 Brazil 1999 cases 37737
4 Brazil 2000 cases 80488
5 China 1999 cases 212258
6 China 2000 cases 213766# A tibble: 6 x 4
country year type count
<chr> <int> <chr> <int>
1 Afghanistan 1999 population 19987071
2 Afghanistan 2000 population 20595360
3 Brazil 1999 population 172006362
4 Brazil 2000 population 174504898
5 China 1999 population 1272915272
6 China 2000 population 1280428583> (Rate_yr <- TB_yr %>% inner_join(Pop_yr, by=c("country","year")) %>%
+ mutate(rate=count.x/count.y*10000, type.z="rate") %>%
+ select(country,year,type=type.z,count=rate)) # 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 # A tibble: 18 x 4
country year type count
<chr> <int> <chr> <dbl>
1 Afghanistan 1999 cases 7.45e+2
2 Afghanistan 1999 population 2.00e+7
3 Afghanistan 1999 rate 3.73e-1
4 Afghanistan 2000 cases 2.67e+3
5 Afghanistan 2000 population 2.06e+7
6 Afghanistan 2000 rate 1.29e+0
7 Brazil 1999 cases 3.77e+4
8 Brazil 1999 population 1.72e+8
9 Brazil 1999 rate 2.19e+0
10 Brazil 2000 cases 8.05e+4
11 Brazil 2000 population 1.75e+8
12 Brazil 2000 rate 4.61e+0
13 China 1999 cases 2.12e+5
14 China 1999 population 1.27e+9
15 China 1999 rate 1.67e+0
16 China 2000 cases 2.14e+5
17 China 2000 population 1.28e+9
18 China 2000 rate 1.67e+0Table4a + Table4b
> table4a %>% inner_join(table4b,by="country",
+ suffix=c("_cases","_pop")) %>%
+ mutate(`1999_rate`=`1999_cases`/`1999_pop`*10000,
+ `2000_rate`=`2000_cases`/`2000_pop`*10000) %>%
+ select(country,`1999`=`1999_rate`,`2000`=`2000_rate`)# 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- Re-create the plot showing change in cases over time using
table2instead oftable1.
> table2 %>%
+ filter(type == "cases") %>%
+ ggplot(aes(year, count)) +
+ geom_line(aes(group = country), color = "grey50") +
+ geom_point(aes(color = country)) +
+ scale_x_continuous(breaks = unique(table2$year)) +
+ ylab("cases")
Pivoting Longer (Gathering)
Sometimes columns would be better represented as values of a variable. In the case of table4a, 1999 and 2000 should be gathered and placed into the variable year.
This can be accomplished with the function gather()
- key = The gathered column’s name
- value = The column name for the values spread over those cells.
> #using gather
> table4a %>%
+ gather(`1999`,`2000`,key="year",value="cases") %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | cases |
|---|---|---|
| Afghanistan | 1999 | 745 |
| Brazil | 1999 | 37737 |
| China | 1999 | 212258 |
| Afghanistan | 2000 | 2666 |
| Brazil | 2000 | 80488 |
| China | 2000 | 213766 |
However, the gather() function has be sunsetted and replaced with pivot_longer().
pivot_longer() “lengthens” data, increasing the number of rows and decreasing the number of columns.
Below you can see the same output with the newer function. Note that instead of selecting 1999 and 2000 I just de-selected country instead.
> #using pivot longer
> table4a %>%
+ pivot_longer(-country,names_to="year",values_to="cases") %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | cases |
|---|---|---|
| Afghanistan | 1999 | 745 |
| Afghanistan | 2000 | 2666 |
| Brazil | 1999 | 37737 |
| Brazil | 2000 | 80488 |
| China | 1999 | 212258 |
| China | 2000 | 213766 |
We can do the same with table4b.
> #using pivot longer
> table4b %>%
+ pivot_longer(-country,names_to="year",values_to="population") %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | population |
|---|---|---|
| Afghanistan | 1999 | 19987071 |
| Afghanistan | 2000 | 20595360 |
| Brazil | 1999 | 172006362 |
| Brazil | 2000 | 174504898 |
| China | 1999 | 1272915272 |
| China | 2000 | 1280428583 |
We can then join table4a and table4b with a left join from dplyr. The output matches table1.
> tidy4a <- table4a %>%
+ pivot_longer(-country,names_to="year",values_to="cases")
>
> tidy4b <- table4b %>%
+ pivot_longer(-country,names_to="year",values_to="population")
>
> left_join(tidy4a,tidy4b) %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")Joining, by = c("country", "year")| 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 |
We can now plot case growth over time.
> CombinTbl <- left_join(tidy4a,tidy4b)
> ggplot(CombinTbl,aes(year,cases))+
+ geom_line(aes(group=country),color="grey50")+
+ geom_point(aes(color=country))+theme_light()
Pivoting Wider (Spreading)
The function spread() is the opposite of gather() and the function pivot_wider is the opposite of pivot_longer.
Like gather(), spread() has been sunsetted. Going forward it is best to use pivot_wider.
pivot_wider() “widens” data, increasing the number of columns and decreasing the number of rows.
> #table2
> table2 %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | type | count |
|---|---|---|---|
| Afghanistan | 1999 | cases | 745 |
| Afghanistan | 1999 | population | 19987071 |
| Afghanistan | 2000 | cases | 2666 |
| Afghanistan | 2000 | population | 20595360 |
| Brazil | 1999 | cases | 37737 |
| Brazil | 1999 | population | 172006362 |
| Brazil | 2000 | cases | 80488 |
| Brazil | 2000 | population | 174504898 |
| China | 1999 | cases | 212258 |
| China | 1999 | population | 1272915272 |
| China | 2000 | cases | 213766 |
| China | 2000 | population | 1280428583 |
> #table 2 spread
> table2 %>% spread(key=type, value=count) %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| 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 |
> #table2 pivot wider
> table2 %>% pivot_wider(names_from=type, values_from=count ) %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| 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 |
Pivoting - Exercises
- Why are
pivot_longer()andpivot_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)
+ )
> stocks %>%
+ pivot_wider(names_from = year, values_from = return) %>%
+ pivot_longer(`2015`:`2016`, names_to = "year", values_to = "return")- In the original tibble
yearwas<dbl>. When we went frompivot_widerback topivot_longerit became<chr>.
pivot_widerconverted the years into column names (characters).
> (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 0.17pivot_longer() has a names_ptypes argument, e.g. names_ptypes = list(year = double()). What does it do?
names_ptypesandnames_transformallow you to confirm or change the type of specific columns.
names_ptypes- A list of column name-prototype pairs. A prototype (or ptype for short) is a zero-length vector (like integer() or numeric()) that defines the type, class, and attributes of a vector. Use these arguments to confirm that the created columns are the types that you expectnames_transform- A list of column name-function pairs. Use these arguments if you need to change the type of specific columns.
> stocks %>%
+ pivot_wider(names_from = year, values_from = return)%>%
+ pivot_longer(`2015`:`2016`, names_to = "year", values_to = "return",
+ names_ptypes = list(year = double()))Error: Can't convert <character> to <double>.> stocks %>%
+ pivot_wider(names_from = year, values_from = return)%>%
+ pivot_longer(`2015`:`2016`, names_to = "year", values_to = "return",
+ names_transform = list(year = as.numeric))# 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- Why does this code fail?
Error: Can't subset columns that don't exist.
[31mx[39m Locations 1999 and 2000 don't exist.
[34mi[39m There are only 3 columns.- The code interprets
1999and2000as column numbers. They need to be surrounded by backticks.
# 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- What would happen if you widen this table? Why? How could you add a new column to uniquely identify each value?
> people <- tribble(
+ ~name, ~key, ~value,
+ #-----------------|--------|------
+ "Phillip Woods", "age", 45,
+ "Phillip Woods", "height", 186,
+ "Phillip Woods", "age", 50,
+ "Jessica Cordero", "age", 37,
+ "Jessica Cordero", "height", 156
+ )- The name and key don’t uniquely identify each value. You can group them and add a row number.
# A tibble: 5 x 4
# Groups: name, key [4]
name key value obs
<chr> <chr> <dbl> <int>
1 Phillip Woods age 45 1
2 Phillip Woods height 186 1
3 Phillip Woods age 50 2
4 Jessica Cordero age 37 1
5 Jessica Cordero height 156 1# A tibble: 3 x 4
# Groups: key [2]
key obs `Phillip Woods` `Jessica Cordero`
<chr> <int> <dbl> <dbl>
1 age 1 45 37
2 height 1 186 156
3 age 2 50 NA- Tidy the simple tibble below. Do you need to make it wider or longer?
pivot_longerworks well with this data.
> preg %>% pivot_longer(c(male, female), names_to = "sex",
+ values_to = "count", values_drop_na = TRUE)# A tibble: 3 x 3
pregnant sex count
<chr> <chr> <dbl>
1 yes female 10
2 no male 20
3 no female 12Separate
separate() pulls apart one column into multiple columns based on a separator character.
> table3 %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | year | rate |
|---|---|---|
| Afghanistan | 1999 | 745/19987071 |
| Afghanistan | 2000 | 2666/20595360 |
| Brazil | 1999 | 37737/172006362 |
| Brazil | 2000 | 80488/174504898 |
| China | 1999 | 212258/1272915272 |
| China | 2000 | 213766/1280428583 |
> table3 %>% separate(rate, into=c("cases","population")) %>%
+ kable %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| 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 |
By default it selects a non alphanumeric character on which to split the variable. However, you can select it:
Also, by default separate() leaves the type of column as is. In this instance both cases and population remain as character. However, by adding convert=TRUE the function will change them to integers, which is more appropriate.
> SepTbl3a <- table3 %>% separate(rate, into=c("cases","population"), sep="/")
>
> SepTbl3b <- table3 %>% separate(
+ rate, into=c("cases","population"),
+ convert=TRUE
+ )
>
> str(SepTbl3a)tibble [6 x 4] (S3: tbl_df/tbl/data.frame)
$ country : chr [1:6] "Afghanistan" "Afghanistan" "Brazil" "Brazil" ...
$ year : int [1:6] 1999 2000 1999 2000 1999 2000
$ cases : chr [1:6] "745" "2666" "37737" "80488" ...
$ population: chr [1:6] "19987071" "20595360" "172006362" "174504898" ...tibble [6 x 4] (S3: tbl_df/tbl/data.frame)
$ country : chr [1:6] "Afghanistan" "Afghanistan" "Brazil" "Brazil" ...
$ year : int [1:6] 1999 2000 1999 2000 1999 2000
$ cases : int [1:6] 745 2666 37737 80488 212258 213766
$ population: int [1:6] 19987071 20595360 172006362 174504898 1272915272 1280428583You can also separate variables based on integers. Positive values start from the left and negative values start from the right.
> table3 %>%
+ separate(year, into=c("century","year"), sep=2) %>%
+ kable %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | century | year | rate |
|---|---|---|---|
| Afghanistan | 19 | 99 | 745/19987071 |
| Afghanistan | 20 | 00 | 2666/20595360 |
| Brazil | 19 | 99 | 37737/172006362 |
| Brazil | 20 | 00 | 80488/174504898 |
| China | 19 | 99 | 212258/1272915272 |
| China | 20 | 00 | 213766/1280428583 |
Unite
unite() is the inverse of separate().
> table5 %>%
+ kable %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | century | year | rate |
|---|---|---|---|
| Afghanistan | 19 | 99 | 745/19987071 |
| Afghanistan | 20 | 00 | 2666/20595360 |
| Brazil | 19 | 99 | 37737/172006362 |
| Brazil | 20 | 00 | 80488/174504898 |
| China | 19 | 99 | 212258/1272915272 |
| China | 20 | 00 | 213766/1280428583 |
By default it will place an underscore _ between values.
> table5 %>% unite(new,century,year) %>%
+ kable %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | new | rate |
|---|---|---|
| Afghanistan | 19_99 | 745/19987071 |
| Afghanistan | 20_00 | 2666/20595360 |
| Brazil | 19_99 | 37737/172006362 |
| Brazil | 20_00 | 80488/174504898 |
| China | 19_99 | 212258/1272915272 |
| China | 20_00 | 213766/1280428583 |
However, you can specify "" for no separator.
> table5 %>% unite(new,century,year, sep="") %>%
+ kable %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| country | new | rate |
|---|---|---|
| Afghanistan | 1999 | 745/19987071 |
| Afghanistan | 2000 | 2666/20595360 |
| Brazil | 1999 | 37737/172006362 |
| Brazil | 2000 | 80488/174504898 |
| China | 1999 | 212258/1272915272 |
| China | 2000 | 213766/1280428583 |
Separating and Uniting Exercises
- What do the extra and fill arguments do in
separate()? Experiment with the various options for the following two toy datasets.
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 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 - extra - instructions for too many pieces
- fill - instructions for too few pieces
> 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 > 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 - Both
unite()andseparate()have a remove argument. What does it do? Why would you set it to FALSE?
- remove discards the original columns. Setting it to FALSE keeps them.
> tibble(x = c("a,b,c", "d,e,f", "h,i,j")) %>%
+ separate(x, c("one", "two", "three"),
+ remove = FALSE)# A tibble: 3 x 4
x one two three
<chr> <chr> <chr> <chr>
1 a,b,c a b c
2 d,e,f d e f
3 h,i,j h i j - Compare and contrast
separate()andextract(). Why are there three variations of separation (by position, by separator, and with groups), but only one unite?
extract()is more flexible because it can handle non-standard patterns.
# 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# 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# 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 1280428583unite()only has one option because you are just merging entire columns.
# A tibble: 6 x 3
country year 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 213766/1280428583Missing Values
There are two types of missing values
- Explicitly - recorded as NA
- Implicitly - excluded from the dataset
In the following dataset Q4 2015 is recorded as NA and Q1 2016 is not present.
> 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)
+ )
> stocks %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| year | qtr | return |
|---|---|---|
| 2015 | 1 | 1.88 |
| 2015 | 2 | 0.59 |
| 2015 | 3 | 0.35 |
| 2015 | 4 | NA |
| 2016 | 2 | 0.92 |
| 2016 | 3 | 0.17 |
| 2016 | 4 | 2.66 |
We can first remove the implicit missing value by giving each year its own column.
> stocks %>% pivot_wider(names_from = year, values_from=return) %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| qtr | 2015 | 2016 |
|---|---|---|
| 1 | 1.88 | NA |
| 2 | 0.59 | 0.92 |
| 3 | 0.35 | 0.17 |
| 4 | NA | 2.66 |
If we want to eliminate the NA values we could add values_drop_na = TRUE.
> stocks %>% pivot_wider(names_from = year, values_from=return) %>%
+ pivot_longer(-qtr,names_to="year",values_to="return",
+ values_drop_na = TRUE) %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| qtr | year | return |
|---|---|---|
| 1 | 2015 | 1.88 |
| 2 | 2015 | 0.59 |
| 2 | 2016 | 0.92 |
| 3 | 2015 | 0.35 |
| 3 | 2016 | 0.17 |
| 4 | 2016 | 2.66 |
The function complete() finds all unique combinations from a set of columns and fills in missing values with NA.
> stocks %>%
+ complete(year,qtr) %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| year | qtr | return |
|---|---|---|
| 2015 | 1 | 1.88 |
| 2015 | 2 | 0.59 |
| 2015 | 3 | 0.35 |
| 2015 | 4 | NA |
| 2016 | 1 | NA |
| 2016 | 2 | 0.92 |
| 2016 | 3 | 0.17 |
| 2016 | 4 | 2.66 |
Sometimes it makes sense for a value to be carried forward. We can do so with the fill() function.
> treatment <- tribble(
+ ~person, ~treatment, ~response,
+ "Derrick Whitmore", 1, 7,
+ NA, 2, 10,
+ NA, 3, 9,
+ "Katherine Burke", 1, 4
+ )
>
> treatment %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| person | treatment | response |
|---|---|---|
| Derrick Whitmore | 1 | 7 |
| NA | 2 | 10 |
| NA | 3 | 9 |
| Katherine Burke | 1 | 4 |
> treatment %>%
+ fill(person) %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| person | treatment | response |
|---|---|---|
| Derrick Whitmore | 1 | 7 |
| Derrick Whitmore | 2 | 10 |
| Derrick Whitmore | 3 | 9 |
| Katherine Burke | 1 | 4 |
The function coalesce() will replace missing values.
> nonalcoholic = tibble(
+ name=c("g&t","mimosa","rum and coke","calimocho","polar"),
+ mixer=c("tonic","orange juice", "coke", "coke","seltzer"),
+ price=c(3, 2, 2, 2, 5))
>
> alcoholic = tibble(
+ name=c("g&t","mimosa","rum and coke","calimocho","IPA"),
+ spirit=c("gin","sparkling wine", "rum", "red wine","beer"),
+ price=c(5, 4, 5, 3, 5))
>
> full_join(nonalcoholic, alcoholic, by='name') %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| name | mixer | price.x | spirit | price.y |
|---|---|---|---|---|
| g&t | tonic | 3 | gin | 5 |
| mimosa | orange juice | 2 | sparkling wine | 4 |
| rum and coke | coke | 2 | rum | 5 |
| calimocho | coke | 2 | red wine | 3 |
| polar | seltzer | 5 | NA | NA |
| IPA | NA | NA | beer | 5 |
We can first change the NA values in price.x and price.y to 0s and combine the prices for a total.
> drinks <- full_join(nonalcoholic, alcoholic, by='name')%>%
+ mutate(price.x=coalesce(price.x,0), price.y=coalesce(price.y,0)
+ ,total=price.x+price.y)
>
> drinks %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| name | mixer | price.x | spirit | price.y | total |
|---|---|---|---|---|---|
| g&t | tonic | 3 | gin | 5 | 8 |
| mimosa | orange juice | 2 | sparkling wine | 4 | 6 |
| rum and coke | coke | 2 | rum | 5 | 7 |
| calimocho | coke | 2 | red wine | 3 | 5 |
| polar | seltzer | 5 | NA | 0 | 5 |
| IPA | NA | 0 | beer | 5 | 5 |
> drinks %>% select(spirit, mixer, total) %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| spirit | mixer | total |
|---|---|---|
| gin | tonic | 8 |
| sparkling wine | orange juice | 6 |
| rum | coke | 7 |
| red wine | coke | 5 |
| NA | seltzer | 5 |
| beer | NA | 5 |
We can then create a new column that merges the information in spirit and mixer
> drinks <- drinks %>% select(spirit, mixer, total)
> drinks %>% unite(name, spirit, mixer, sep=' & ') %>%
+ kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| name | total |
|---|---|
| gin & tonic | 8 |
| sparkling wine & orange juice | 6 |
| rum & coke | 7 |
| red wine & coke | 5 |
| NA & seltzer | 5 |
| beer & NA | 5 |
Unfortunately, it doesn’t look great because there are NAs. You can substitute the NAs by nothing using coalesce to make the output look a little nicer:
> new <- drinks %>%
+ mutate(spirit=coalesce(spirit, "nothing"), mixer=coalesce(mixer, "nothing")) %>%
+ unite(name, spirit, mixer, sep=' & ') %>%
+ mutate(name=trimws(name))
> new %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| name | total |
|---|---|
| gin & tonic | 8 |
| sparkling wine & orange juice | 6 |
| rum & coke | 7 |
| red wine & coke | 5 |
| nothing & seltzer | 5 |
| beer & nothing | 5 |
The function trimws got rid of some empty spaces at the beginning and end of name.
You can make the new variable look even nicer with the function gsub by substituting & nothing and nothing & by (literally) nothing.
> final <- new %>% mutate(name1 = gsub("& nothing","",name),
+ name = trimws(gsub("nothing &","",name1))) %>%
+ select(-name1)
> final %>% kable() %>%
+ kable_styling(bootstrap_options = c("striped","condensed"),
+ full_width = F, font_size = 12,position="left")%>%
+ row_spec(0,background="lightpink")| name | total |
|---|---|
| gin & tonic | 8 |
| sparkling wine & orange juice | 6 |
| rum & coke | 7 |
| red wine & coke | 5 |
| seltzer | 5 |
| beer | 5 |
Missing Values - Exercises
- Compare and contrast the fill arguments to
pivot_wider()andcomplete().
values_fillreplaces implicitly missing values .fillreplaces all NA values, implicit and explicit.
> 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)
+ )
> stocks %>%
+ pivot_wider(names_from = year, values_from = return,
+ values_fill = 0)# A tibble: 4 x 3
qtr `2015` `2016`
<dbl> <dbl> <dbl>
1 1 1.88 0
2 2 0.59 0.92
3 3 0.35 0.17
4 4 NA 2.66# 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 0
5 2016 1 0
6 2016 2 0.92
7 2016 3 0.17
8 2016 4 2.66- What does the direction argument to
fill()do?.
- Direction in which to fill missing values. Currently either “down” (the default), “up”, “downup” (i.e. first down and then up) or “updown” (first up and then 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# 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# A tibble: 4 x 3
person treatment response
<chr> <dbl> <dbl>
1 Derrick Whitmore 1 7
2 Katherine Burke 2 10
3 Katherine Burke 3 9
4 Katherine Burke 1 4Case Study
We can apply some of the previous methods to the who dataset, which contains tuberculosis cases broken down by year, country, age, gender, and diagnosis method.
# 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>- country, iso2, and iso3 all identify the country
- year is a variable
- The other columns, new_sp_m014 to newrel_f65, should be variables
We can begin tidying the data by using the pivot_longer() function.
> who1 <- who %>%
+ pivot_longer(-(country:year),names_to="key",values_to="cases",
+ values_drop_na=TRUE)
> who1# 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 rowsWe can examine it further by evaluating the count.
# A tibble: 56 x 2
key n
<chr> <int>
1 new_ep_f014 1032
2 new_ep_f1524 1021
3 new_ep_f2534 1021
4 new_ep_f3544 1021
5 new_ep_f4554 1017
6 new_ep_f5564 1017
7 new_ep_f65 1014
8 new_ep_m014 1038
9 new_ep_m1524 1026
10 new_ep_m2534 1020
# ... with 46 more rowsThe new key column can be parsed further:
- The first three letters denote whether it is a new or old case. In this dataset they are all new.
- The next two letters describe the type of TB. The values are rel, ep, sn, sp
- The sixth letter represent’s the patients sex, M,F.
- The remaining letter represent an age group. ex 014 = 0 to 14.
Most follow the format new_ep etc., but instead of new_rel we have newrel. We can change that with str_replace().
# 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 rowsThe we can separate key into more useful columns.
# 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 rowsSince iso2 and iso3 are redundant (we already have country) and since all cases are “new” we can drop those three columns.
Finally, we can split sexage into sex and age.
# 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 rowsThe resulting dataset is now tidy.
Case Study - Exercises
> (who1 <- who %>%
+ 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# A tibble: 76,046 x 7
country iso2 iso3 year key cases names_from
<chr> <chr> <chr> <int> <chr> <int> <chr>
1 Afghanistan AF AFG 1997 new_sp_m014 0 new_sp_m014
2 Afghanistan AF AFG 1997 new_sp_m1524 10 new_sp_m1524
3 Afghanistan AF AFG 1997 new_sp_m2534 6 new_sp_m2534
4 Afghanistan AF AFG 1997 new_sp_m3544 3 new_sp_m3544
5 Afghanistan AF AFG 1997 new_sp_m4554 5 new_sp_m4554
6 Afghanistan AF AFG 1997 new_sp_m5564 2 new_sp_m5564
7 Afghanistan AF AFG 1997 new_sp_m65 0 new_sp_m65
8 Afghanistan AF AFG 1997 new_sp_f014 5 new_sp_f014
9 Afghanistan AF AFG 1997 new_sp_f1524 38 new_sp_f1524
10 Afghanistan AF AFG 1997 new_sp_f2534 36 new_sp_f2534
# ... with 76,036 more rows# A tibble: 76,046 x 9
country iso2 iso3 year new type sexage cases names_from
<chr> <chr> <chr> <int> <chr> <chr> <chr> <int> <chr>
1 Afghanistan AF AFG 1997 new sp m014 0 new_sp_m014
2 Afghanistan AF AFG 1997 new sp m1524 10 new_sp_m1524
3 Afghanistan AF AFG 1997 new sp m2534 6 new_sp_m2534
4 Afghanistan AF AFG 1997 new sp m3544 3 new_sp_m3544
5 Afghanistan AF AFG 1997 new sp m4554 5 new_sp_m4554
6 Afghanistan AF AFG 1997 new sp m5564 2 new_sp_m5564
7 Afghanistan AF AFG 1997 new sp m65 0 new_sp_m65
8 Afghanistan AF AFG 1997 new sp f014 5 new_sp_f014
9 Afghanistan AF AFG 1997 new sp f1524 38 new_sp_f1524
10 Afghanistan AF AFG 1997 new sp f2534 36 new_sp_f2534
# ... with 76,036 more rows# A tibble: 2 x 2
new n
<chr> <int>
1 new 73466
2 newrel 2580# A tibble: 76,046 x 6
country year type sexage cases names_from
<chr> <int> <chr> <chr> <int> <chr>
1 Afghanistan 1997 sp m014 0 new_sp_m014
2 Afghanistan 1997 sp m1524 10 new_sp_m1524
3 Afghanistan 1997 sp m2534 6 new_sp_m2534
4 Afghanistan 1997 sp m3544 3 new_sp_m3544
5 Afghanistan 1997 sp m4554 5 new_sp_m4554
6 Afghanistan 1997 sp m5564 2 new_sp_m5564
7 Afghanistan 1997 sp m65 0 new_sp_m65
8 Afghanistan 1997 sp f014 5 new_sp_f014
9 Afghanistan 1997 sp f1524 38 new_sp_f1524
10 Afghanistan 1997 sp f2534 36 new_sp_f2534
# ... with 76,036 more rows# A tibble: 76,046 x 7
country year type sex age cases names_from
<chr> <int> <chr> <chr> <chr> <int> <chr>
1 Afghanistan 1997 sp m 014 0 new_sp_m014
2 Afghanistan 1997 sp m 1524 10 new_sp_m1524
3 Afghanistan 1997 sp m 2534 6 new_sp_m2534
4 Afghanistan 1997 sp m 3544 3 new_sp_m3544
5 Afghanistan 1997 sp m 4554 5 new_sp_m4554
6 Afghanistan 1997 sp m 5564 2 new_sp_m5564
7 Afghanistan 1997 sp m 65 0 new_sp_m65
8 Afghanistan 1997 sp f 014 5 new_sp_f014
9 Afghanistan 1997 sp f 1524 38 new_sp_f1524
10 Afghanistan 1997 sp f 2534 36 new_sp_f2534
# ... with 76,036 more rows- In this case study, I set
values_drop_na = TRUEjust 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?
- An NA is an explicitly missing value and a 0 is a recorded value of no TB cases.
- There are some implicit missing values, where a county,year combination is not present.
[1] 11080[1] 7240[1] 7446- What happens if you neglect the
mutate()step?(mutate(key = str_replace(key, "newrel", "new_rel"))
- It’s expecting 3 columns but will only have 2, so the data is wrong and some values are also missing.
Warning: Expected 3 pieces. Missing pieces filled with `NA` in 2580 rows [243,
244, 679, 680, 681, 682, 683, 684, 685, 686, 687, 688, 689, 690, 691, 692, 903,
904, 905, 906, ...].# A tibble: 6 x 8
country iso2 iso3 year new type sexage cases
<chr> <chr> <chr> <int> <chr> <chr> <chr> <int>
1 Afghanistan AF AFG 2013 newrel m014 <NA> 1705
2 Afghanistan AF AFG 2013 newrel f014 <NA> 1749
3 Albania AL ALB 2013 newrel m014 <NA> 14
4 Albania AL ALB 2013 newrel m1524 <NA> 60
5 Albania AL ALB 2013 newrel m2534 <NA> 61
6 Albania AL ALB 2013 newrel m3544 <NA> 32- I claimed that iso2 and iso3 were redundant with country. Confirm this claim.
- This is a correct assumption
> who3 %>% select(country, iso2, iso3) %>%
+ distinct() %>%
+ group_by(country) %>%
+ filter(n() > 1)# A tibble: 0 x 3
# Groups: country [0]
# ... with 3 variables: country <chr>, iso2 <chr>, iso3 <chr>- For each country, year, and sex compute the total number of cases of TB. Make an informative visualization of the data.
> who5 %>%
+ group_by(country, year, sex) %>%
+ summarize(cases = sum(cases)) %>%
+ na.omit() %>%
+ unite(country_sex, country, sex, remove = FALSE) %>%
+ filter(year > 1995) %>%
+ ggplot(aes(x = year, y = cases, group = country_sex, color = sex)) +
+ geom_line()