In-class worksheet 6
Feb 6, 2020
In this worksheet, we will continue to work with the tidyverse libraries:
library(tidyverse)1. The msleep dataset
The msleep dataset, provided with ggplot2, contains information about sleep and awake times of different mammals:
msleep## # A tibble: 83 x 11
## name genus vore order conservation sleep_total sleep_rem sleep_cycle awake
## <chr> <chr> <chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
## 1 Chee… Acin… carni Carn… lc 12.1 NA NA 11.9
## 2 Owl … Aotus omni Prim… <NA> 17 1.8 NA 7
## 3 Moun… Aplo… herbi Rode… nt 14.4 2.4 NA 9.6
## 4 Grea… Blar… omni Sori… lc 14.9 2.3 0.133 9.1
## 5 Cow Bos herbi Arti… domesticated 4 0.7 0.667 20
## 6 Thre… Brad… herbi Pilo… <NA> 14.4 2.2 0.767 9.6
## 7 Nort… Call… carni Carn… vu 8.7 1.4 0.383 15.3
## 8 Vesp… Calo… <NA> Rode… <NA> 7 NA NA 17
## 9 Dog Canis carni Carn… domesticated 10.1 2.9 0.333 13.9
## 10 Roe … Capr… herbi Arti… lc 3 NA NA 21
## # … with 73 more rows, and 2 more variables: brainwt <dbl>, bodywt <dbl>Verify that the sum of total sleep time (column sleep_total) and total awake time (column awake) adds up to 24h for all animals in the msleep dataset.
(msleep$sleep_total + msleep$awake) == 24## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [25] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
## [37] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [49] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
## [61] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [73] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUEThere are two cases where the sum is not equal to exactly 24 hours.
msleep %>%
mutate(day_total = sleep_total + awake) %>%
filter(day_total != 24) %>%
select(name, vore, sleep_total, awake, day_total)## # A tibble: 2 x 5
## name vore sleep_total awake day_total
## <chr> <chr> <dbl> <dbl> <dbl>
## 1 Pilot whale carni 2.7 21.4 24.0
## 2 Common porpoise carni 5.6 18.4 24.0In the pilot whale and the common porpoise, the total sleep and awake times add up to 24.05 hours.
Make a list of all the domesticated species in the msleep dataset, in alphabetical order. Hint: Domesticated species have the entry “domesticated” in the column conservation.
msleep %>%
filter(conservation == "domesticated") %>%
select(name) %>%
arrange(name)## # A tibble: 10 x 1
## name
## <chr>
## 1 Chinchilla
## 2 Cow
## 3 Dog
## 4 Domestic cat
## 5 Donkey
## 6 Guinea pig
## 7 Horse
## 8 Pig
## 9 Rabbit
## 10 SheepFor the different vore classifications, tally how many species are awake for at least 18 hours. Hint: Use the function tally().
msleep %>%
filter(awake >= 18) %>%
group_by(vore) %>%
tally()## # A tibble: 3 x 2
## vore n
## <chr> <int>
## 1 carni 4
## 2 herbi 11
## 3 <NA> 1Using the function top_n(), identify the top-10 least-awake animals and list them from least awake to most awake. Explain why this analysis gives you 11 results instead of 10. Hint: Before calling top_n(), use the function select() to extract the two columns name and sleep_total, in that order.
msleep %>%
select(name, sleep_total) %>%
top_n(10) %>%
arrange(desc(sleep_total))## Selecting by sleep_total## # A tibble: 11 x 2
## name sleep_total
## <chr> <dbl>
## 1 Little brown bat 19.9
## 2 Big brown bat 19.7
## 3 Thick-tailed opposum 19.4
## 4 Giant armadillo 18.1
## 5 North American Opossum 18
## 6 Long-nosed armadillo 17.4
## 7 Owl monkey 17
## 8 Arctic ground squirrel 16.6
## 9 Golden-mantled ground squirrel 15.9
## 10 Tiger 15.8
## 11 Eastern american chipmunk 15.8There are 11 results because there is a tie. Both the Tiger and the Eastern american chipmunk have a total sleep time of 15.8h, and they are in positions 10 and 11 of the list. Note that by default, top_n() orders based on the last variable in the table. Since we selected sleep_total as the last column before we called top_n(), we get the desired result.
Considering only carnivores and herbivores, make a plot of the percent of time each animal is in REM sleep (out of the total sleep time) vs. the animal’s total sleep time. Hint: Use the operator | to indicate logical OR in the filter() function.
msleep %>%
filter(vore == "carni" | vore == "herbi") %>%
mutate(perc_rem = sleep_rem / sleep_total) %>%
ggplot(aes(x = sleep_total, y = perc_rem, color = vore)) +
geom_point()## Warning: Removed 17 rows containing missing values (geom_point).
2. The diamonds dataset
The diamonds dataset provided by ggplot2 provides information about quality and price of 53940 diamonds:
head(diamonds)## # A tibble: 6 x 10
## carat cut color clarity depth table price x y z
## <dbl> <ord> <ord> <ord> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
## 1 0.23 Ideal E SI2 61.5 55 326 3.95 3.98 2.43
## 2 0.21 Premium E SI1 59.8 61 326 3.89 3.84 2.31
## 3 0.23 Good E VS1 56.9 65 327 4.05 4.07 2.31
## 4 0.290 Premium I VS2 62.4 58 334 4.2 4.23 2.63
## 5 0.31 Good J SI2 63.3 58 335 4.34 4.35 2.75
## 6 0.24 Very Good J VVS2 62.8 57 336 3.94 3.96 2.48The best cuts of diamonds are “Very Good”, “Premium”, and “Ideal”. Make a table that selects only those diamonds, and find the minimum, median, and maximum price for each cut. Hint: The operator %in% is helpful for selecting the diamond cuts.
diamonds %>%
filter(cut %in% c("Very Good", "Premium", "Ideal")) %>%
group_by(cut) %>%
summarize(
min_price = min(price),
med_price = median(price),
max_price = max(price)
)## # A tibble: 3 x 4
## cut min_price med_price max_price
## <ord> <int> <dbl> <int>
## 1 Very Good 336 2648 18818
## 2 Premium 326 3185 18823
## 3 Ideal 326 1810 18806For each of the different diamond cuts, calculate the mean carat level among the diamonds whose price falls within 10% of the most expensive diamond for that cut.
diamonds %>%
group_by(cut) %>%
filter(price > 0.9 * max(price)) %>%
summarize(mean_carat = mean(carat))## # A tibble: 5 x 2
## cut mean_carat
## <ord> <dbl>
## 1 Fair 2.73
## 2 Good 2.08
## 3 Very Good 1.97
## 4 Premium 2.08
## 5 Ideal 1.99For each of the different diamond cuts, calculate the mean carat level among the top-10% most expensive diamonds.
diamonds %>%
group_by(cut) %>%
mutate(price_rank = rank(desc(price))) %>% # rank diamonds by price, in descending order
filter(price_rank < 0.1 * max(price_rank)) %>% # pick the top 10%
summarize(mean_carat = mean(carat))## # A tibble: 5 x 2
## cut mean_carat
## <ord> <dbl>
## 1 Fair 2.04
## 2 Good 1.76
## 3 Very Good 1.71
## 4 Premium 1.88
## 5 Ideal 1.58Make a table that contains the median price for each combination of cut and clarity, and arrange the final table in descending order of median price.
diamonds %>%
group_by(cut, clarity) %>%
summarize(med_price = median(price)) %>%
arrange(desc(med_price))## # A tibble: 40 x 3
## # Groups: cut [5]
## cut clarity med_price
## <ord> <ord> <dbl>
## 1 Premium SI2 4291
## 2 Ideal SI2 4060.
## 3 Very Good SI2 4042
## 4 Good SI2 3770
## 5 Fair SI2 3681
## 6 Ideal I1 3674.
## 7 Premium SI1 3618
## 8 Fair SI1 3528.
## 9 Very Good I1 3283
## 10 Premium I1 3261
## # … with 30 more rowsNow arrange the same table first by cut and then within each cut group by median price.
diamonds %>%
group_by(cut, clarity) %>%
summarize(med_price = median(price)) %>%
arrange(desc(cut), desc(med_price))## # A tibble: 40 x 3
## # Groups: cut [5]
## cut clarity med_price
## <ord> <ord> <dbl>
## 1 Ideal SI2 4060.
## 2 Ideal I1 3674.
## 3 Ideal SI1 2537
## 4 Ideal VS1 1813
## 5 Ideal VS2 1689
## 6 Ideal VVS2 1330
## 7 Ideal VVS1 1114
## 8 Ideal IF 1022.
## 9 Premium SI2 4291
## 10 Premium SI1 3618
## # … with 30 more rows3. If this was easy
For the diamonds data set, separately for each diamond cut, calculate the percentage of diamonds with a price above $10,000, and the median carat value for diamonds priced $10,000 or more.
diamonds %>%
group_by(cut) %>%
summarize(
percent_above_10k = round(100 * sum(price >= 10000) / n(), 1),
median_carat_above_10k = median(carat[price >= 10000])
)## # A tibble: 5 x 3
## cut percent_above_10k median_carat_above_10k
## <ord> <dbl> <dbl>
## 1 Fair 9.1 2.01
## 2 Good 7.6 2
## 3 Very Good 9.3 1.7
## 4 Premium 13.1 1.72
## 5 Ideal 8.2 1.57