# load the necessary libraries
library(tidyverse)
library(babynames)Class Activity 7
Problem 1: Boolean Operators
Use Boolean operators to alter the code below to return only the rows that contain:
a. Girls named Rhea
Click for answer
filter(babynames, name == "Rhea", sex == "F")# A tibble: 136 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1882 F Rhea 7 0.0000605
2 1883 F Rhea 8 0.0000666
3 1884 F Rhea 13 0.0000945
4 1885 F Rhea 11 0.0000775
5 1886 F Rhea 13 0.0000846
6 1887 F Rhea 14 0.0000901
7 1888 F Rhea 20 0.000106
8 1889 F Rhea 31 0.000164
9 1890 F Rhea 39 0.000193
10 1891 F Rhea 24 0.000122
# ℹ 126 more rowsbabynames %>% filter(name == "Rhea", sex == "F")# A tibble: 136 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1882 F Rhea 7 0.0000605
2 1883 F Rhea 8 0.0000666
3 1884 F Rhea 13 0.0000945
4 1885 F Rhea 11 0.0000775
5 1886 F Rhea 13 0.0000846
6 1887 F Rhea 14 0.0000901
7 1888 F Rhea 20 0.000106
8 1889 F Rhea 31 0.000164
9 1890 F Rhea 39 0.000193
10 1891 F Rhea 24 0.000122
# ℹ 126 more rowsb. Names that were used by exactly 5 or 6 children in 1990
Click for answer
filter(babynames, year == 1990, n == 5 | n == 6)# A tibble: 6,144 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1990 F Aariel 6 0.00000292
2 1990 F Aarion 6 0.00000292
3 1990 F Abagael 6 0.00000292
4 1990 F Abbye 6 0.00000292
5 1990 F Abiola 6 0.00000292
6 1990 F Abreanna 6 0.00000292
7 1990 F Abygail 6 0.00000292
8 1990 F Acadia 6 0.00000292
9 1990 F Adilenne 6 0.00000292
10 1990 F Adriena 6 0.00000292
# ℹ 6,134 more rowsbabynames %>% filter(year == "1990", n == 5 | n == 6)# A tibble: 6,144 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1990 F Aariel 6 0.00000292
2 1990 F Aarion 6 0.00000292
3 1990 F Abagael 6 0.00000292
4 1990 F Abbye 6 0.00000292
5 1990 F Abiola 6 0.00000292
6 1990 F Abreanna 6 0.00000292
7 1990 F Abygail 6 0.00000292
8 1990 F Acadia 6 0.00000292
9 1990 F Adilenne 6 0.00000292
10 1990 F Adriena 6 0.00000292
# ℹ 6,134 more rowsc. Names that are one of Apple, Yoroi, Ada
Click for answer
filter(babynames, name == "Apple" | name == "Yoroi" | name == "Ada")# A tibble: 200 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1880 F Ada 652 0.00668
2 1881 F Ada 628 0.00635
3 1882 F Ada 689 0.00596
4 1883 F Ada 778 0.00648
5 1884 F Ada 854 0.00621
6 1885 F Ada 876 0.00617
7 1885 M Ada 5 0.0000431
8 1886 F Ada 915 0.00595
9 1886 M Ada 6 0.0000504
10 1887 F Ada 910 0.00586
# ℹ 190 more rowsd. Store the data tibble in part c into a new tibble and change all the character columns to upper case. Also, rename the n variable to count.
Click for answer
aya <- babynames %>% filter(name == "Apple" | name == "Yoroi" | name == "Ada")
aya %>% mutate_if(is.character, toupper)# A tibble: 200 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1880 F ADA 652 0.00668
2 1881 F ADA 628 0.00635
3 1882 F ADA 689 0.00596
4 1883 F ADA 778 0.00648
5 1884 F ADA 854 0.00621
6 1885 F ADA 876 0.00617
7 1885 M ADA 5 0.0000431
8 1886 F ADA 915 0.00595
9 1886 M ADA 6 0.0000504
10 1887 F ADA 910 0.00586
# ℹ 190 more rowsaya %>% mutate_at(vars(name), toupper)# A tibble: 200 × 5
year sex name n prop
<dbl> <chr> <chr> <int> <dbl>
1 1880 F ADA 652 0.00668
2 1881 F ADA 628 0.00635
3 1882 F ADA 689 0.00596
4 1883 F ADA 778 0.00648
5 1884 F ADA 854 0.00621
6 1885 F ADA 876 0.00617
7 1885 M ADA 5 0.0000431
8 1886 F ADA 915 0.00595
9 1886 M ADA 6 0.0000504
10 1887 F ADA 910 0.00586
# ℹ 190 more rowsaya %>% rename(count = n)# A tibble: 200 × 5
year sex name count prop
<dbl> <chr> <chr> <int> <dbl>
1 1880 F Ada 652 0.00668
2 1881 F Ada 628 0.00635
3 1882 F Ada 689 0.00596
4 1883 F Ada 778 0.00648
5 1884 F Ada 854 0.00621
6 1885 F Ada 876 0.00617
7 1885 M Ada 5 0.0000431
8 1886 F Ada 915 0.00595
9 1886 M Ada 6 0.0000504
10 1887 F Ada 910 0.00586
# ℹ 190 more rowse. Change all the column names to upper case in the previous problem.
Click for answer
aya %>% rename_at(vars(year:prop), toupper)# A tibble: 200 × 5
YEAR SEX NAME N PROP
<dbl> <chr> <chr> <int> <dbl>
1 1880 F Ada 652 0.00668
2 1881 F Ada 628 0.00635
3 1882 F Ada 689 0.00596
4 1883 F Ada 778 0.00648
5 1884 F Ada 854 0.00621
6 1885 F Ada 876 0.00617
7 1885 M Ada 5 0.0000431
8 1886 F Ada 915 0.00595
9 1886 M Ada 6 0.0000504
10 1887 F Ada 910 0.00586
# ℹ 190 more rowsf. What do these commands do?
polluted_cities <- tribble(
~city, ~size, ~amount,
"New York", "large", 23,
"New York", "small", 14,
"London", "large", 22,
"London", "small", 16,
"Beijing", "large", 121,
"Beijing", "small", 56
)
polluted_cities# A tibble: 6 × 3
city size amount
<chr> <chr> <dbl>
1 New York large 23
2 New York small 14
3 London large 22
4 London small 16
5 Beijing large 121
6 Beijing small 56polluted_cities %>% select_if(is.numeric) #1
polluted_cities %>% rename_all(toupper) #2
polluted_cities %>% rename_if(is.character, toupper) #3
polluted_cities %>% rename_at(vars(contains("it")), toupper) #4Click for answer
answer:
- Selects all numeric columns from the polluted_cities dataset.
- Renames all column names in the polluted_cities dataset to uppercase.
- Renames column names with character data type in the polluted_cities dataset to uppercase.
- Renames column names containing “it” in the polluted_cities dataset to uppercase.
polluted_cities %>% select_if(is.numeric) #1# A tibble: 6 × 1
amount
<dbl>
1 23
2 14
3 22
4 16
5 121
6 56polluted_cities %>% rename_all(toupper) #2# A tibble: 6 × 3
CITY SIZE AMOUNT
<chr> <chr> <dbl>
1 New York large 23
2 New York small 14
3 London large 22
4 London small 16
5 Beijing large 121
6 Beijing small 56polluted_cities %>% rename_if(is.character, toupper) #3# A tibble: 6 × 3
CITY SIZE amount
<chr> <chr> <dbl>
1 New York large 23
2 New York small 14
3 London large 22
4 London small 16
5 Beijing large 121
6 Beijing small 56polluted_cities %>% rename_at(vars(contains("it")), toupper) #4# A tibble: 6 × 3
CITY size amount
<chr> <chr> <dbl>
1 New York large 23
2 New York small 14
3 London large 22
4 London small 16
5 Beijing large 121
6 Beijing small 56Let’s look at an interesting example on how to join related information on various artists, bands, songs, and their labels.
artists <- tibble(first = c("Jimmy", "George", "Mick", "Tom", "Davy", "John",
"Paul", "Jimmy", "Joe", "Elvis", "Keith", "Paul",
"Ringo", "Joe", "Brian", "Nancy"),
last = c("Buffett", "Harrison", "Jagger", "Jones", "Jones",
"Lennon", "McCartney", "Page", "Perry", "Presley",
"Richards", "Simon", "Starr", "Walsh", "Wilson", "Wilson"),
instrument = c("Guitar", "Guitar", "Vocals", "Vocals", "Vocals",
"Guitar", "Bass", "Guitar", "Guitar", "Vocals", "Guitar",
"Guitar", "Drums", "Guitar", "Vocals", "Vocals"))
bands <- tibble(first = c("John", "John Paul", "Jimmy", "Robert", "George", "John",
"Paul", "Ringo", "Jimmy", "Mick", "Keith", "Charlie", "Ronnie"),
last = c("Bonham", "Jones", "Page", "Plant", "Harrison", "Lennon",
"McCartney", "Starr", "Buffett", "Jagger", "Richards", "Watts", "Wood"),
band = c("Led Zeppelin", "Led Zeppelin", "Led Zeppelin", "Led Zeppelin",
"The Beatles", "The Beatles", "The Beatles", "The Beatles",
"The Coral Reefers", "The Rolling Stones", "The Rolling Stones",
"The Rolling Stones", "The Rolling Stones"))
albums <- tibble(album = c("A Hard Day's Night", "Magical Mystery Tour", "Beggar's Banquet",
"Abbey Road", "Led Zeppelin IV", "The Dark Side of the Moon", "Aerosmith",
"Rumours", "Hotel California"),
band = c("The Beatles", "The Beatles", "The Rolling Stones", "The Beatles",
"Led Zeppelin", "Pink Floyd", "Aerosmith", "Fleetwood Mac", "Eagles"),
year = c(1964,1967,1968,1969,1971,1973,1973,1977,1982))
songs <- tibble(song = c("Come Together", "Dream On", "Hello, Goodbye", "It's Not Unusual"),
album = c("Abbey Road", "Aerosmith", "Magical Mystery Tour", "Along Came Jones"),
first = c("John", "Steven", "Paul", "Tom"),
last = c("Lennon", "Tyler", "McCartney", "Jones"))
labels <- tibble(album = c("Abbey Road", "A Hard Days Night", "Magical Mystery Tour",
"Led Zeppelin IV", "The Dark Side of the Moon", "Hotel California",
"Rumours", "Aerosmith", "Beggar's Banquet"),
label = c("Apple", "Parlophone", "Parlophone", "Atlantic", "Harvest",
"Asylum", "Warner Brothers", "Columbia", "Decca"))Let’s take a glimpse of the tibbles artists and bands. Notice that there are different number of rows in the dataset.
glimpse(artists)Rows: 16
Columns: 3
$ first <chr> "Jimmy", "George", "Mick", "Tom", "Davy", "John", "Paul", "…
$ last <chr> "Buffett", "Harrison", "Jagger", "Jones", "Jones", "Lennon"…
$ instrument <chr> "Guitar", "Guitar", "Vocals", "Vocals", "Vocals", "Guitar",…glimpse(bands)Rows: 13
Columns: 3
$ first <chr> "John", "John Paul", "Jimmy", "Robert", "George", "John", "Paul"…
$ last <chr> "Bonham", "Jones", "Page", "Plant", "Harrison", "Lennon", "McCar…
$ band <chr> "Led Zeppelin", "Led Zeppelin", "Led Zeppelin", "Led Zeppelin", …glimpse(albums)Rows: 9
Columns: 3
$ album <chr> "A Hard Day's Night", "Magical Mystery Tour", "Beggar's Banquet"…
$ band <chr> "The Beatles", "The Beatles", "The Rolling Stones", "The Beatles…
$ year <dbl> 1964, 1967, 1968, 1969, 1971, 1973, 1973, 1977, 1982glimpse(songs)Rows: 4
Columns: 4
$ song <chr> "Come Together", "Dream On", "Hello, Goodbye", "It's Not Unusual"
$ album <chr> "Abbey Road", "Aerosmith", "Magical Mystery Tour", "Along Came J…
$ first <chr> "John", "Steven", "Paul", "Tom"
$ last <chr> "Lennon", "Tyler", "McCartney", "Jones"glimpse(labels)Rows: 9
Columns: 2
$ album <chr> "Abbey Road", "A Hard Days Night", "Magical Mystery Tour", "Led …
$ label <chr> "Apple", "Parlophone", "Parlophone", "Atlantic", "Harvest", "Asy…Problem 2: Joining artists and bands data
a. Join the artists and bands tibbles using left_join(), right_join(), and full_join(). Verify that the datasets obtained from left_join() and right_join() are the same using setequal().
Click for answer
bands2 <- left_join(bands, artists, by = c("first", "last"))
bands3 <- right_join(artists, bands, by = c("first", "last"))
full_bands <- full_join(artists, bands, by = c("first", "last"))
# Check if the datasets are the same
setequal(bands2, bands3)[1] TRUEb. Use the pipe operator, %>%, to create one table that combines all information from artists, bands, albums, songs, and labels.
Click for answer
all_info <- artists %>%
full_join(bands, by = c("first", "last")) %>%
full_join(songs, by = c("first", "last")) %>%
full_join(albums, by = c("album", "band")) %>%
full_join(labels, by = c("album"))
all_info# A tibble: 30 × 8
first last instrument band song album year label
<chr> <chr> <chr> <chr> <chr> <chr> <dbl> <chr>
1 Jimmy Buffett Guitar The Coral Reefers <NA> <NA> NA <NA>
2 George Harrison Guitar The Beatles <NA> <NA> NA <NA>
3 Mick Jagger Vocals The Rolling Stones <NA> <NA> NA <NA>
4 Tom Jones Vocals <NA> It's Not Un… Alon… NA <NA>
5 Davy Jones Vocals <NA> <NA> <NA> NA <NA>
6 John Lennon Guitar The Beatles Come Togeth… Abbe… 1969 Apple
7 Paul McCartney Bass The Beatles Hello, Good… Magi… 1967 Parl…
8 Jimmy Page Guitar Led Zeppelin <NA> <NA> NA <NA>
9 Joe Perry Guitar <NA> <NA> <NA> NA <NA>
10 Elvis Presley Vocals <NA> <NA> <NA> NA <NA>
# ℹ 20 more rowsProblem 3: Filtering and counting rows in the data
a. Collect artists that have songs provided, and return rows of artists that don’t have bands info.
Click for answer
# Artists with songs
artists_with_songs <- artists %>%
semi_join(songs, by = c("first", "last"))
# Artists without bands info
artists_without_bands <- artists %>%
anti_join(bands, by = c("first","last"))
artists_with_songs# A tibble: 3 × 3
first last instrument
<chr> <chr> <chr>
1 Tom Jones Vocals
2 John Lennon Guitar
3 Paul McCartney Bass artists_without_bands# A tibble: 8 × 3
first last instrument
<chr> <chr> <chr>
1 Tom Jones Vocals
2 Davy Jones Vocals
3 Joe Perry Guitar
4 Elvis Presley Vocals
5 Paul Simon Guitar
6 Joe Walsh Guitar
7 Brian Wilson Vocals
8 Nancy Wilson Vocals b. Collect the albums made by a band, count the number of rows, find the rows of songs that match a row in labels, and count the number of rows.
Click for answer
# Albums made by a band
albums_by_band <- albums %>% semi_join(bands, by = "band")
n_albums_by_band <- nrow(albums_by_band)
# Rows of songs that match a row in labels
songs_with_labels <- songs %>% semi_join(labels, by = "album")
n_songs_with_labels <- nrow(songs_with_labels)
n_albums_by_band[1] 5n_songs_with_labels[1] 3