Homework #2
Jerry Chan
2026-03-08
1. Data Transformation and Visualization with the flights data set
Working with the data set flights in the package nycflights13, answer the following questions by performing necessary data transformation/visualization.
- Create a histogram of arrival delays (excluding NAs) for all flights in June and July. Summarize your findings.
flights%>%
filter(month %in% c(6,7))%>%
filter(!is.na(arr_delay))%>%
ggplot(aes(x=arr_delay))+
geom_histogram(bins=50,
fill="blue",
color="black")+
xlim(c(-30,200))+
labs(
title="arrival delays (excluding NAs) for all flights in June and July",
x="arrival delay (minutes)",
y="number of flights"
)
The histogram shows a right skew, suggesting flights that are on time is common, and delays are not common.
- Create a smooth line graph of arrival delays vs departure delays for all flights departing from EWR on the first day of each month. Summarize your findings.
flights%>%
filter(origin == "EWR",
day == 1,
!is.na(arr_delay),
!is.na(dep_delay))%>%
ggplot(aes(x=arr_delay, y=dep_delay))+
geom_point()+
geom_smooth(method = "lm", se=F, color="red")+
labs(
title="arrival delays vs departure delays for all flights departing from EWR on the first day of each month",
x="arrival delays (minutes)",
y="departure delays (minutes)"
)
The plot shows a trend where arrival and departure delays are closely related, suggesting delays in on arrival will delay departure times.
- Find the flights that actually departed with the shortest travel distance. What is its origin and destination airport?
flights%>%
filter(!is.na(dep_time),
!is.na(distance))%>%
arrange(distance)%>%
select(origin, dest, distance)## # A tibble: 328,521 × 3
## origin dest distance
## <chr> <chr> <dbl>
## 1 EWR PHL 80
## 2 EWR PHL 80
## 3 EWR PHL 80
## 4 EWR PHL 80
## 5 EWR PHL 80
## 6 EWR PHL 80
## 7 EWR PHL 80
## 8 EWR PHL 80
## 9 EWR PHL 80
## 10 EWR PHL 80
## # ℹ 328,511 more rowsShortest distance: from EWR to PHL, distance of 80 miles.
- Create a new categorical variable with two labels. Flights with a travel distance shorter than 500 miles are marked as “short-distance”, and otherwise “long-distance”. Create a bar plot to compare the number of flights in each category. Summarize your findings.
flight_dist <-flights%>%
mutate(flight_dist = if_else(distance<500,
"short-distance",
"long-distance"
))
flight_dist%>%
ggplot(aes(x=flight_dist, fill=flight_dist))+
geom_bar()+
labs(
title = "Number of Short- vs Long-Distance Flights",
x = "Flight distance category",
y = "Number of flights"
)
long distance flights are more common compared to short distance flights.
- Find the destination airport that has the longest average departure delay by creating a graph.
flights%>%
group_by(dest)%>%
summarise(avg_dep_delay=mean(dep_delay, na.rm=T))%>%
ggplot(aes(x=avg_dep_delay, y=reorder(dest, avg_dep_delay), fill=avg_dep_delay))+
geom_col()+
labs(
title = "destination airport that has the longest average departure delay",
x="average departure delay (minutes)",
y="destination airport"
)
- Answer the question in (e) without creating a graph.
flights%>%
group_by(dest)%>%
summarise(avg_dep_delay=mean(dep_delay, na.rm=T))%>%
arrange(desc(avg_dep_delay), dest)## # A tibble: 105 × 2
## dest avg_dep_delay
## <chr> <dbl>
## 1 CAE 35.6
## 2 TUL 34.9
## 3 OKC 30.6
## 4 BHM 29.7
## 5 TYS 28.5
## 6 JAC 26.5
## 7 DSM 26.2
## 8 RIC 23.6
## 9 ALB 23.6
## 10 MSN 23.6
## # ℹ 95 more rows- Find the carriers with the highest and the lowest average flight speed for all their flights in the data set.
carrier_speed <- flights %>%
filter(!is.na(distance), !is.na(air_time)) %>%
mutate(speed = distance/air_time) %>%
group_by(carrier) %>%
summarise(
avg_speed = mean(speed, na.rm = TRUE),
n_flights = n(),
)# Highest average speed carriers
carrier_speed %>%
slice_max(avg_speed, n = 1)## # A tibble: 1 × 3
## carrier avg_speed n_flights
## <chr> <dbl> <int>
## 1 HA 8.01 342# Lowest average speed carriers
carrier_speed %>%
slice_min(avg_speed, n = 1)## # A tibble: 1 × 3
## carrier avg_speed n_flights
## <chr> <dbl> <int>
## 1 YV 5.53 5442. Analyzing the seattlepets data set
For the following questions, analyze the data set seattlepets in the package openintro. Read the help document and make sure that you understand the basic information about the data set before analysis.
- How many species are there in the data set? What are they?
table(seattlepets$species)##
## Cat Dog Goat Pig
## 17294 35181 38 6there are 4 species, they are: cat, dog, goat, and pig.
- What are the most popular primary breeds for cats and dogs, respectively?
seattlepets%>%
filter(species %in% c("Cat", "Dog"))%>%
group_by(species, primary_breed)%>%
summarise(n=n())%>%
arrange(species)%>%
slice_max(n)## `summarise()` has grouped output by 'species'. You can override using the
## `.groups` argument.## # A tibble: 2 × 3
## # Groups: species [2]
## species primary_breed n
## <chr> <chr> <int>
## 1 Cat Domestic Shorthair 10086
## 2 Dog Retriever, Labrador 4867- What are the three most common pet names in Seattle?
seattlepets%>%
filter(!is.na(animal_name))%>%
group_by(animal_name)%>%
summarise(n=n(), .groups = "drop")%>%
arrange(desc(n))%>%
slice_head(n=3)## # A tibble: 3 × 2
## animal_name n
## <chr> <int>
## 1 Lucy 439
## 2 Charlie 387
## 3 Luna 355three most common pet names are: Lucy, Charlie, Luna.
- What are the ten most common pet names for cats? What are the ten most common pet names for dogs? Write a code to print the result and their frequencies.
# ten most common pet names for cats
seattlepets%>%
filter(!is.na(animal_name))%>%
filter(species == "Cat")%>%
group_by(species, animal_name)%>%
summarise(n=n(), .groups="drop")%>%
slice_max(n, n=10)%>%
arrange(species, desc(n), animal_name)## # A tibble: 11 × 3
## species animal_name n
## <chr> <chr> <int>
## 1 Cat Luna 111
## 2 Cat Lucy 102
## 3 Cat Lily 86
## 4 Cat Max 83
## 5 Cat Bella 82
## 6 Cat Charlie 81
## 7 Cat Oliver 73
## 8 Cat Jack 65
## 9 Cat Sophie 59
## 10 Cat Leo 54
## 11 Cat Molly 54# ten most common pet names for dogs
seattlepets%>%
filter(!is.na(animal_name))%>%
filter(species == "Dog")%>%
group_by(species, animal_name)%>%
summarise(n=n(), .groups="drop")%>%
slice_max(n, n=10)%>%
arrange(species, desc(n), animal_name)## # A tibble: 10 × 3
## species animal_name n
## <chr> <chr> <int>
## 1 Dog Lucy 337
## 2 Dog Charlie 306
## 3 Dog Bella 249
## 4 Dog Luna 244
## 5 Dog Daisy 221
## 6 Dog Cooper 189
## 7 Dog Lola 187
## 8 Dog Max 186
## 9 Dog Molly 186
## 10 Dog Stella 185- How many names appear more than 100 times in the data set excluding “NA”?
seattlepets%>%
filter(!is.na(animal_name))%>%
group_by(animal_name)%>%
summarise(n=n(), .groups="drop")%>%
filter(n>100)%>%
summarise(num_names=n())## # A tibble: 1 × 1
## num_names
## <int>
## 1 56- For all names that appear more than 100 times in the data set, which has the highest “cat_to_dog” ratio? Which has the lowest? The “cat_to_dog” ratio can be computed this way - if a name appears 200 times, in which 150 are for cats and 50 are for dogs, the ratio is 150/50 = 3.
cat_dog_ratio <- seattlepets%>%
filter(!is.na(animal_name), species %in% c("Cat", "Dog"))%>%
group_by(species, animal_name)%>%
summarise(n=n(), .groups="drop")%>%
pivot_wider(
names_from = species,
values_from = n,
)%>%
mutate(total=Cat+Dog)%>%
filter(total>100, Dog>0)%>%
mutate(cat_to_dog = Cat/Dog)# highest cat to dog ratio
cat_dog_ratio%>%
arrange(desc(cat_to_dog))%>%
slice(1)## # A tibble: 1 × 5
## animal_name Cat Dog total cat_to_dog
## <chr> <int> <int> <int> <dbl>
## 1 Shadow 53 79 132 0.671# lowest cat to dog ratio
cat_dog_ratio%>%
arrange(cat_to_dog)%>%
slice(1)## # A tibble: 1 × 5
## animal_name Cat Dog total cat_to_dog
## <chr> <int> <int> <int> <dbl>
## 1 Riley 9 117 126 0.0769