Week10_Assignment
2022-07-24
16.1-16.2
3.Use the appropriate lubridate function to parse each of the following dates:
d1 <- "January 1, 2010"
d2 <- "2015-Mar-07"
d3 <- "06-Jun-2017"
d4 <- c("August 19 (2015)", "July 1 (2015)")
d5 <- "12/30/14"
mdy(d1)## [1] "2010-01-01"ymd(d2)## [1] "2015-03-07"dmy(d3)## [1] "2017-06-06"mdy(d4)## [1] "2015-08-19" "2015-07-01"mdy(d5)## [1] "2014-12-30"
16.3
2.Compare dep_time, sched_dep_time and dep_delay. Are they consistent? Explain your findings.
make_datetime_100 <- function(year, month, day, time) {
make_datetime(year, month, day, time %/% 100, time %% 100)
}
flights_dt <- flights %>%
filter(!is.na(dep_time), !is.na(arr_time)) %>%
mutate(
dep_time = make_datetime_100(year, month, day, dep_time),
arr_time = make_datetime_100(year, month, day, arr_time),
sched_dep_time = make_datetime_100(year, month, day, sched_dep_time),
sched_arr_time = make_datetime_100(year, month, day, sched_arr_time)
) %>%
select(origin, dest, ends_with("delay"), ends_with("time"))
#if consistent, dep_time should equal to sched_dep_time+dep_delay
flights_dt %>%
mutate(dep_time_ = sched_dep_time + dep_delay * 60) %>%
filter(dep_time_ != dep_time) %>%
select(dep_time_, dep_time, sched_dep_time, dep_delay)## # A tibble: 1,205 × 4
## dep_time_ dep_time sched_dep_time dep_delay
## <dttm> <dttm> <dttm> <dbl>
## 1 2013-01-02 08:48:00 2013-01-01 08:48:00 2013-01-01 18:35:00 853
## 2 2013-01-03 00:42:00 2013-01-02 00:42:00 2013-01-02 23:59:00 43
## 3 2013-01-03 01:26:00 2013-01-02 01:26:00 2013-01-02 22:50:00 156
## 4 2013-01-04 00:32:00 2013-01-03 00:32:00 2013-01-03 23:59:00 33
## 5 2013-01-04 00:50:00 2013-01-03 00:50:00 2013-01-03 21:45:00 185
## 6 2013-01-04 02:35:00 2013-01-03 02:35:00 2013-01-03 23:59:00 156
## 7 2013-01-05 00:25:00 2013-01-04 00:25:00 2013-01-04 23:59:00 26
## 8 2013-01-05 01:06:00 2013-01-04 01:06:00 2013-01-04 22:45:00 141
## 9 2013-01-06 00:14:00 2013-01-05 00:14:00 2013-01-05 23:59:00 15
## 10 2013-01-06 00:37:00 2013-01-05 00:37:00 2013-01-05 22:30:00 127
## # … with 1,195 more rowsThey are not always consistent, these might be flights of which the departure time is the next day of the schedule departure time.
3.Compare dep_time, sched_dep_time and dep_delay. Are they consistent? Explain your findings.
flights_dt %>%
mutate(
flight_duration = as.numeric(arr_time - dep_time),
air_time_mins = air_time,
diff = flight_duration - air_time_mins
) %>%
select(origin, dest, flight_duration, air_time_mins, diff)## # A tibble: 328,063 × 5
## origin dest flight_duration air_time_mins diff
## <chr> <chr> <dbl> <dbl> <dbl>
## 1 EWR IAH 193 227 -34
## 2 LGA IAH 197 227 -30
## 3 JFK MIA 221 160 61
## 4 JFK BQN 260 183 77
## 5 LGA ATL 138 116 22
## 6 EWR ORD 106 150 -44
## 7 EWR FLL 198 158 40
## 8 LGA IAD 72 53 19
## 9 JFK MCO 161 140 21
## 10 LGA ORD 115 138 -23
## # … with 328,053 more rowsI see that the air time doesn’t equal to arrival time - depature time, I think the reason of this difference is that the flight might not take off and land in the same time zone. To fix this, we might need more information to convert the original time data into the same time zone.
4.How does the average delay time change over the course of a day? Should you use dep_time or sched_dep_time? Why?
flights_dt %>%
mutate(sched_dep_hour = hour(sched_dep_time)) %>%
group_by(sched_dep_hour) %>%
summarise(dep_delay = mean(dep_delay)) %>%
ggplot(aes(y = dep_delay, x = sched_dep_hour)) +
geom_point() +
geom_smooth()## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
Use sched_dep_time because that is the relevant metric for someone scheduling a flight. Also, using dep_time will always bias delays to later in the day since delays will push flights later.
5.On what day of the week should you leave if you want to minimize the chance of a delay?
flights_dt %>%
mutate(dow = wday(sched_dep_time)) %>%
group_by(dow) %>%
summarise(
dep_delay = mean(dep_delay),
arr_delay = mean(arr_delay, na.rm = TRUE)
) %>%
print(n = Inf)## # A tibble: 7 × 3
## dow dep_delay arr_delay
## <dbl> <dbl> <dbl>
## 1 1 11.5 4.82
## 2 2 14.7 9.65
## 3 3 10.6 5.39
## 4 4 11.7 7.05
## 5 5 16.1 11.7
## 6 6 14.7 9.07
## 7 7 7.62 -1.45flights_dt %>%
mutate(wday = wday(dep_time, label = TRUE)) %>%
group_by(wday) %>%
summarize(ave_dep_delay = mean(dep_delay, na.rm = TRUE)) %>%
ggplot(aes(x = wday, y = ave_dep_delay)) +
geom_bar(stat = "identity")
flights_dt %>%
mutate(wday = wday(dep_time, label = TRUE)) %>%
group_by(wday) %>%
summarize(ave_arr_delay = mean(arr_delay, na.rm = TRUE)) %>%
ggplot(aes(x = wday, y = ave_arr_delay)) +
geom_bar(stat = "identity")
Flights on Saturday have the lowest departure delay and arrival delay.
16.4
3.Create a vector of dates giving the first day of every month in 2015. Create a vector of dates giving the first day of every month in the current year.
ymd("2015-01-01") + months(0:11)## [1] "2015-01-01" "2015-02-01" "2015-03-01" "2015-04-01" "2015-05-01"
## [6] "2015-06-01" "2015-07-01" "2015-08-01" "2015-09-01" "2015-10-01"
## [11] "2015-11-01" "2015-12-01"floor_date(today(), unit = "year") + months(0:11)## [1] "2022-01-01" "2022-02-01" "2022-03-01" "2022-04-01" "2022-05-01"
## [6] "2022-06-01" "2022-07-01" "2022-08-01" "2022-09-01" "2022-10-01"
## [11] "2022-11-01" "2022-12-01"
5.Why can’t (today() %–% (today() + years(1)) / months(1) work?
Because (today() %–% (today() + years(1)) is an interval, which has an exact number of seconds. The denominator of the expression, months(1), is a period.