Test syntax highlighting

Prologue

This quiz will test your understanding of the filter() function in the {dplyr} package. Good luck!

Question 1

df is a dataset from a smallpox outbreak in Nigeria.

df <- outbreaks::smallpox_abakaliki_1967 |> as_tibble() 
df

## # A tibble: 5 × 8
##   case_ID date_of_onset   age gender vaccinated vaccscar ftc   compound
##     <int> <date>        <int> <fct>  <fct>      <fct>    <fct> <fct>   
## 1       1 1967-04-05       10 f      n          n        y     1       
## 2       2 1967-04-18       25 f      n          n        y     1       
## 3       3 1967-04-25       35 m      n          n        y     1       
## 4       4 1967-04-27        4 f      n          n        y     1       
## 5       5 1967-04-30       11 m      n          n        y     1

Which of the following is an incorrect interpretation of a filter statement on the age column in df? {3}

1. df %>% filter(!age > 5) : Drop row(s) where age is above 5
2. df %>% filter(age < 5) : Keep row(s) where age is below 5
3. df %>% filter(age == 5) : Drop row(s) where age is equal to 5

Question 2

df is an excerpt of a dataset from an H7N9 outbreak in China.

df <- outbreaks::fluH7N9_china_2013[c(1,2,3,74, 75), c(6:8)]
df

##    gender age province
## 1       m  87 Shanghai
## 2       m  27 Shanghai
## 3       f  35    Anhui
## 74   <NA>   ? Shanghai
## 75   <NA>   ? Shanghai

To drop the single female patient, which of the following filter statements is correct {1}

1. df %>% filter(gender == "m" & is.na(gender))
2. df %>% filter(gender != "f")
3. df %>% filter(gender == "m")

Question 3

df <- 
  tribble(~id, ~state, 
           1,  "Kano",
           2,  "Lagos",
           3,  "Bauchi", 
           4,  "Kano", 
           5,  "FCT",
           6,  NA)

df

## # A tibble: 6 × 2
##      id state 
##   <dbl> <chr> 
## 1     1 Kano  
## 2     2 Lagos 
## 3     3 Bauchi
## 4     4 Kano  
## 5     5 FCT   
## 6     6 <NA>

From df, you would like to keep rows where state is either “FCT” or “Kano”. Which of the following uses correct syntax for this. {1}

1. df %>% filter(state %in% c("FCT", "Kano"))
2. df %>% filter(state == "FCT" & state == "Kano"))
3. df %>% filter(state == c("FCT", "Kano"))

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10