Primary exercises
- Create and investigate a list.
Three students received different sets of grades (Amy: 1,6,7,9,10; Bob: 6,7,4,3,5,2,2,1,4; Dan: 9,9,10).
In a variable scores create a list (the names of the list elements should be the names of the students and the values should be the corresponding grades).
Print the list, its class, length and structure (str) of scores.
scores <- list(
Amy = c( 1,6,7,9,10 ),
Bob = c( 6,7,4,3,5,2,2,1,4 ),
Dan = c( 9,9,10 )
)
scores
$Amy
[1] 1 6 7 9 10
$Bob
[1] 6 7 4 3 5 2 2 1 4
$Dan
[1] 9 9 10
class( scores )
[1] "list"
length( scores )
[1] 3
str( scores )
List of 3
$ Amy: num [1:5] 1 6 7 9 10
$ Bob: num [1:9] 6 7 4 3 5 2 2 1 4
$ Dan: num [1:3] 9 9 10
- Add an element, change an element.
Reuse scores from the previous exercise.
Add there grades for Eve (7,3,5,8,8,9) and print the list.
Then, for Dan merge new grades (8,8,6,7) with the existing grades (hint: use the combine function c to combine existing Dan’s grades with the new grades then put the result back to scores; do not type again 9,9,10).
scores[[ 'Eve' ]] <- c(7,3,5,8,8,9)
scores
$Amy
[1] 1 6 7 9 10
$Bob
[1] 6 7 4 3 5 2 2 1 4
$Dan
[1] 9 9 10
$Eve
[1] 7 3 5 8 8 9
scores[[ "Dan" ]] <- c( scores[[ "Dan" ]], c(8,8,6,7) )
scores
$Amy
[1] 1 6 7 9 10
$Bob
[1] 6 7 4 3 5 2 2 1 4
$Dan
[1] 9 9 10 8 8 6 7
$Eve
[1] 7 3 5 8 8 9
- Single and double bracket operators.
Reuse scores from the previous exercises.
Investigate the difference between scores[[ "Bob" ]] and scores[ "Bob" ].
Look at what is printed and what is the class of each result.
Then compare scores[[ c( "Amy", "Bob" ) ]] with scores[ c( "Amy", "Bob" ) ].
Understand, why the error is reported.
scores[[ "Bob" ]] # Returns the value of Bob element (vector)
[1] 6 7 4 3 5 2 2 1 4
scores[ "Bob" ] # Creates a new list with only Bob there (list)
$Bob
[1] 6 7 4 3 5 2 2 1 4
class( scores[[ "Bob" ]] )
[1] "numeric"
class( scores[ "Bob" ] )
[1] "list"
scores[[ c( "Amy", "Bob" ) ]] # A list is needed to return two elements
Error in scores[[c("Amy", "Bob")]]: subscript out of bounds
scores[ c( "Amy", "Bob" ) ] # This creates a list, so many elements are ok
$Amy
[1] 1 6 7 9 10
$Bob
[1] 6 7 4 3 5 2 2 1 4
- Dollar operator.
Reuse scores from the previous exercises.
Investigate the (lack of) difference between scores$Bob and scores[[ "Bob" ]].
Look at what is printed and what is the class of each result.
Then compare scores$Bo with scores[[ "Bo" ]].
Understand, why the NULL is returned.
scores$Bob # another way to access Bob
[1] 6 7 4 3 5 2 2 1 4
scores[[ "Bob" ]] # get an element with exact name Bob
[1] 6 7 4 3 5 2 2 1 4
class( scores$Bob )
[1] "numeric"
class( scores[[ "Bob" ]] )
[1] "numeric"
scores$Bo # strange matching of names, it still finds Bob
[1] 6 7 4 3 5 2 2 1 4
scores[[ "Bo" ]] # there is no "Bo" so NULL is returned
NULL
Multitopic exercises
- (ADV) Split a table into list of tables by a column factor; merge back.
Some functions might require an input to be provided as a list of tables.
Let’s assume that the pulse table should be split into a list of table parts based on the exercise argument.
Load the pulse.csv data to variable pulse.
Try l <- pulse %>% split( .$exercise ) and investigate the class, length and names of the result l.
Use double square bracket to extract the part for exercise being low.
Finally, check that with bind_rows applied to l you can recreate the pulse table (but with a different order of rows).
l <- pulse %>% split( .$exercise ) # . represents the object on the left side of %>%
class( l )
[1] "list"
length( l )
[1] 3
names( l )
[1] "high" "low" "moderate"
l[[ "low" ]]
# A tibble: 37 × 13
id name height weight age gender smokes alcohol exercise ran pulse1 pulse2 year
<chr> <chr> <dbl> <dbl> <dbl> <chr> <chr> <chr> <chr> <chr> <dbl> <dbl> <dbl>
1 1993_E Lauri 173 64 18 female no yes low sat 90 88 1993
2 1993_F George 184 74 22 male no yes low ran 78 141 1993
3 1993_L Frederick 178 58 19 male no no low sat 74 76 1993
4 1993_P Mathew 185 110 22 male no yes low sat 77 73 1993
5 1993_Q Leslie 170 56 19 male no no low sat 64 63 1993
6 1993_U Jerome 175 60 19 male no no low sat 88 86 1993
7 1993_V Arlene 140 50 34 female no no low ran 70 98 1993
8 1993_W Glenna 163 55 20 female no no low sat 78 74 1993
9 1995_B Olga 172 60 21 female no no low sat 81 79 1995
10 1995_H Eliza 164 66 23 female no no low ran 74 168 1995
# … with 27 more rows
recreatedPulse <- bind_rows( l )
dim( pulse )
[1] 110 13
dim( recreatedPulse )
[1] 110 13
- (ADV) Split a table by a column and write each part to a different file.
Continue with the setup of the previous exercise.
Study/type/exectute the following example.
Find the newly created files in your filesystem.
l <- pulse %>% split( .$exercise )
exercises <- names( l ) # name in l of each table chunk
for( exercise in exercises ) { # exercise will be a name of a single chunk
fileName <- paste0( "pulse_", exercise, ".csv" ) # name of the file for the chunk
message( "Writing file '", fileName, "'..." )
write_csv( l[[ exercise ]], file = fileName )
}
