18 R Tidyverse Exercise

18.1 Load Libraries

Load the tidyverse packages

 library(tidyverse)
# library(tidylog)

18.2 Untidy data

Let’s use the World Health Organization TB data set from the tidyr package

who <- tidyr::who
dim(who)

[1] 7240   60

head(who[,1:6] %>% filter(!is.na(new_sp_m014)))

# A tibble: 6 × 6
  country     iso2  iso3   year new_sp_m014 new_sp_m1524
  <chr>       <chr> <chr> <dbl>       <dbl>        <dbl>
1 Afghanistan AF    AFG    1997           0           10
2 Afghanistan AF    AFG    1998          30          129
3 Afghanistan AF    AFG    1999           8           55
4 Afghanistan AF    AFG    2000          52          228
5 Afghanistan AF    AFG    2001         129          379
6 Afghanistan AF    AFG    2002          90          476

See the help page for who for more information about this data set.

In particular, note this description:

“The data uses the original codes given by the World Health Organization. The column names for columns five through 60 are made by combining new_ to a code for method of diagnosis (rel = relapse, sn = negative pulmonary smear, sp = positive pulmonary smear, ep = extrapulmonary) to a code for gender (f = female, m = male) to a code for age group (014 = 0-14 yrs of age, 1524 = 15-24 years of age, 2534 = 25 to 34 years of age, 3544 = 35 to 44 years of age, 4554 = 45 to 54 years of age, 5564 = 55 to 64 years of age, 65 = 65 years of age or older).”

So new_sp_m014 represents the counts of new TB cases detected by a positive pulmonary smear in males in the 0-14 age group.

18.3 Tidy data

Tidy data: Have each variable in a column.

Question: Are these data tidy?

Expand to see solution

No these data are not tidy because aspects of the data that should be variables are encoded in the name of the variables.

These aspects are

test type.
sex of the subjects.
age range of the subjects.

Question: How would we make these data tidy?

Consider this portion of the data:

head(who[,1:5] %>% filter(!is.na(new_sp_m014) & new_sp_m014>0), 1)

# A tibble: 1 × 5
  country     iso2  iso3   year new_sp_m014
  <chr>       <chr> <chr> <dbl>       <dbl>
1 Afghanistan AF    AFG    1998          30

Expand to see solution

We would replace the new_sp_m014 with the following four columns:

type  sex   age   n
sp    m     014  30

This would place each variable in its own column.

18.4 Gather

stocks <- tibble(
  time = as.Date('2009-01-01') + 0:9,
  X = rnorm(10, 0, 1),
  Y = rnorm(10, 0, 2),
  Z = rnorm(10, 0, 4)
)

head(stocks)

# A tibble: 6 × 4
  time            X       Y        Z
  <date>      <dbl>   <dbl>    <dbl>
1 2009-01-01  0.244  3.46    4.60   
2 2009-01-02 -1.65   2.03    0.0932 
3 2009-01-03 -1.38   2.44   -0.00714
4 2009-01-04 -2.35   0.472   3.98   
5 2009-01-05  1.19   0.0770  6.72   
6 2009-01-06  1.17  -0.0612  6.48

stocks %>% gather("stock", "price", -time) %>% head()

# A tibble: 6 × 3
  time       stock  price
  <date>     <chr>  <dbl>
1 2009-01-01 X      0.244
2 2009-01-02 X     -1.65 
3 2009-01-03 X     -1.38 
4 2009-01-04 X     -2.35 
5 2009-01-05 X      1.19 
6 2009-01-06 X      1.17

18.5 Pivot_longer

stocks %>% pivot_longer(c(X,Y,Z), names_to= "stock", values_to = "price") %>% 
  head()

# A tibble: 6 × 3
  time       stock   price
  <date>     <chr>   <dbl>
1 2009-01-01 X      0.244 
2 2009-01-01 Y      3.46  
3 2009-01-01 Z      4.60  
4 2009-01-02 X     -1.65  
5 2009-01-02 Y      2.03  
6 2009-01-02 Z      0.0932

18.6 WHO TB data

Question: How would we convert this to tidy form?

head(who[,1:6] %>% filter(!is.na(new_sp_m014)))

# A tibble: 6 × 6
  country     iso2  iso3   year new_sp_m014 new_sp_m1524
  <chr>       <chr> <chr> <dbl>       <dbl>        <dbl>
1 Afghanistan AF    AFG    1997           0           10
2 Afghanistan AF    AFG    1998          30          129
3 Afghanistan AF    AFG    1999           8           55
4 Afghanistan AF    AFG    2000          52          228
5 Afghanistan AF    AFG    2001         129          379
6 Afghanistan AF    AFG    2002          90          476

Expand to see solution

who.long <- who %>% pivot_longer(starts_with("new"), names_to = "demo", values_to = "n") %>%  filter(!is.na(n))
head(who.long)

# A tibble: 6 × 6
  country     iso2  iso3   year demo             n
  <chr>       <chr> <chr> <dbl> <chr>        <dbl>
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

Question: How would we split demo into variables?

head(who.long)

# A tibble: 6 × 6
  country     iso2  iso3   year demo             n
  <chr>       <chr> <chr> <dbl> <chr>        <dbl>
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

Look at the variable naming scheme:

names(who) %>% grep("m014",., value=TRUE)

[1] "new_sp_m014" "new_sn_m014" "new_ep_m014" "newrel_m014"

Question: How should we adjust the demo strings so as to be able to easily split all of them into the desired variables?

Expand to see solution

who.long <- who.long %>%  
  mutate(demo = str_replace(demo, "newrel", "new_rel"))
grep("m014",who.long$demo, value=TRUE) %>%  unique()

[1] "new_sp_m014"  "new_sn_m014"  "new_ep_m014"  "new_rel_m014"

Question: After adjusting the demo strings, how would we then separate them into the desired variables?

Hint: Use separate_wider_position() and separate_wider_delim().

Expand to see solution

who.long.v1 <- who.long %>% 
  separate(demo, into = c("new", "type", "sexagerange"), sep="_") %>% 
  separate(sexagerange, into=c("sex","age_range"), sep=1) %>%
  select(-new)
head(who.long.v1)

# A tibble: 6 × 8
  country     iso2  iso3   year type  sex   age_range     n
  <chr>       <chr> <chr> <dbl> <chr> <chr> <chr>     <dbl>
1 Afghanistan AF    AFG    1997 sp    m     014           0
2 Afghanistan AF    AFG    1997 sp    m     1524         10
3 Afghanistan AF    AFG    1997 sp    m     2534          6
4 Afghanistan AF    AFG    1997 sp    m     3544          3
5 Afghanistan AF    AFG    1997 sp    m     4554          5
6 Afghanistan AF    AFG    1997 sp    m     5564          2

Note that separate() has been superseded in favour of separate_wider_position() and separate_wider_delim(). So here we use those two functions instead of separate():

who.long.v2 <- who.long %>% 
  separate_wider_delim(demo, names = c("new", "type", "sexagerange"), delim="_") %>% 
  separate_wider_position(sexagerange, widths=c("sex"=1,"age_range"=4), too_few="align_start" ) %>%
  select(-new)
head(who.long.v2)

# A tibble: 6 × 8
  country     iso2  iso3   year type  sex   age_range     n
  <chr>       <chr> <chr> <dbl> <chr> <chr> <chr>     <dbl>
1 Afghanistan AF    AFG    1997 sp    m     014           0
2 Afghanistan AF    AFG    1997 sp    m     1524         10
3 Afghanistan AF    AFG    1997 sp    m     2534          6
4 Afghanistan AF    AFG    1997 sp    m     3544          3
5 Afghanistan AF    AFG    1997 sp    m     4554          5
6 Afghanistan AF    AFG    1997 sp    m     5564          2

18.7 Conclusion

Now our untidy data are tidy.

head(who.long)

# A tibble: 6 × 6
  country     iso2  iso3   year demo             n
  <chr>       <chr> <chr> <dbl> <chr>        <dbl>
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

18.8 Acknowledgment

This exercise was modeled, in part, on this exercise:

https://people.duke.edu/\~ccc14/cfar-data-workshop-2018/CFAR_R_Workshop_2018_Exercisees.html