Does the dataset contains 336,776 flights records and 19 fields?
library(nycflights13)dim(flights)## [1] 336776 19How could you know which flight has the largest departure delay?
library(dplyr)## Warning: package 'dplyr' was built under R version 4.4.3##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(nycflights13)sortf <- arrange(flights,desc(dep_delay))
select(sortf, carrier, flight, tailnum, everything())## # A tibble: 336,776 × 19
## carrier flight tailnum year month day dep_time sched_dep_time dep_delay
## <chr> <int> <chr> <int> <int> <int> <int> <int> <dbl>
## 1 HA 51 N384HA 2013 1 9 641 900 1301
## 2 MQ 3535 N504MQ 2013 6 15 1432 1935 1137
## 3 MQ 3695 N517MQ 2013 1 10 1121 1635 1126
## 4 AA 177 N338AA 2013 9 20 1139 1845 1014
## 5 MQ 3075 N665MQ 2013 7 22 845 1600 1005
## 6 DL 2391 N959DL 2013 4 10 1100 1900 960
## 7 DL 2119 N927DA 2013 3 17 2321 810 911
## 8 DL 2007 N3762Y 2013 6 27 959 1900 899
## 9 DL 2047 N6716C 2013 7 22 2257 759 898
## 10 AA 172 N5DMAA 2013 12 5 756 1700 896
## # ℹ 336,766 more rows
## # ℹ 10 more variables: arr_time <int>, sched_arr_time <int>, arr_delay <dbl>,
## # origin <chr>, dest <chr>, air_time <dbl>, distance <dbl>, hour <dbl>,
## # minute <dbl>, time_hour <dttm>maxdep <- max(flights$dep_delay, na.rm=TRUE)
maxdep_id <- which(flights$dep_delay==maxdep)
flights[maxdep_id, 10:12]## # A tibble: 1 × 3
## carrier flight tailnum
## <chr> <int> <chr>
## 1 HA 51 N384HAYour friend David tries to get the average departure delay per date, and he writes the following code but get outputs as NAs:
flights %>%
group_by(year, month, day) %>%
summarise(mean = mean(dep_delay))
Which of the following do you think will help solve his problem and provide the correct values?
Option 1:
flights %>%
group_by(year, month, day) %>%
summarise(mean_delay = mean(dep_delay, na.rm = TRUE))## `summarise()` has regrouped the output.
## ℹ Summaries were computed grouped by year, month, and day.
## ℹ Output is grouped by year and month.
## ℹ Use `summarise(.groups = "drop_last")` to silence this message.
## ℹ Use `summarise(.by = c(year, month, day))` for per-operation grouping
## (`?dplyr::dplyr_by`) instead.## # A tibble: 365 × 4
## # Groups: year, month [12]
## year month day mean_delay
## <int> <int> <int> <dbl>
## 1 2013 1 1 11.5
## 2 2013 1 2 13.9
## 3 2013 1 3 11.0
## 4 2013 1 4 8.95
## 5 2013 1 5 5.73
## 6 2013 1 6 7.15
## 7 2013 1 7 5.42
## 8 2013 1 8 2.55
## 9 2013 1 9 2.28
## 10 2013 1 10 2.84
## # ℹ 355 more rowsOption 2:
not_cancelled <- flights %>%
filter(!is.na(dep_delay))
not_cancelled %>%
group_by(year, month, day) %>%
summarise(mean_delay = mean(dep_delay))## `summarise()` has regrouped the output.
## ℹ Summaries were computed grouped by year, month, and day.
## ℹ Output is grouped by year and month.
## ℹ Use `summarise(.groups = "drop_last")` to silence this message.
## ℹ Use `summarise(.by = c(year, month, day))` for per-operation grouping
## (`?dplyr::dplyr_by`) instead.## # A tibble: 365 × 4
## # Groups: year, month [12]
## year month day mean_delay
## <int> <int> <int> <dbl>
## 1 2013 1 1 11.5
## 2 2013 1 2 13.9
## 3 2013 1 3 11.0
## 4 2013 1 4 8.95
## 5 2013 1 5 5.73
## 6 2013 1 6 7.15
## 7 2013 1 7 5.42
## 8 2013 1 8 2.55
## 9 2013 1 9 2.28
## 10 2013 1 10 2.84
## # ℹ 355 more rowsOption 3:
delays <- not_cancelled %>%
group_by(tailnum) %>%
summarise(delay = mean(arr_delay))This code runs, but does not answer the question being asked. This code groups by tailnum (aircraft), not by date. Gives wrong analysis/output for this question.
Option 4:
not_cancelled <- flights %>%
filter(!is.na(dep_delay), !is.na(arr_delay))
not_cancelled %>%
group_by(year, month, day) %>%
summarise(mean_delay = mean(dep_delay))## `summarise()` has regrouped the output.
## ℹ Summaries were computed grouped by year, month, and day.
## ℹ Output is grouped by year and month.
## ℹ Use `summarise(.groups = "drop_last")` to silence this message.
## ℹ Use `summarise(.by = c(year, month, day))` for per-operation grouping
## (`?dplyr::dplyr_by`) instead.## # A tibble: 365 × 4
## # Groups: year, month [12]
## year month day mean_delay
## <int> <int> <int> <dbl>
## 1 2013 1 1 11.4
## 2 2013 1 2 13.7
## 3 2013 1 3 10.9
## 4 2013 1 4 8.97
## 5 2013 1 5 5.73
## 6 2013 1 6 7.15
## 7 2013 1 7 5.42
## 8 2013 1 8 2.56
## 9 2013 1 9 2.30
## 10 2013 1 10 2.84
## # ℹ 355 more rowsWrite the code to get the tailnum that has the lowest average arrival delay?
avg_delay <- flights %>%
filter(!is.na(arr_delay)) %>%
group_by(tailnum) %>%
summarise(avg_arr_delay = mean(arr_delay))avg_delay %>%
arrange(avg_arr_delay) %>%
slice(1)## # A tibble: 1 × 2
## tailnum avg_arr_delay
## <chr> <dbl>
## 1 N560AS -53not_cancelled %>%
group_by(year, month, day) %>%
summarise(
first = min(dep_time),
last = max(dep_time)
)## `summarise()` has regrouped the output.
## ℹ Summaries were computed grouped by year, month, and day.
## ℹ Output is grouped by year and month.
## ℹ Use `summarise(.groups = "drop_last")` to silence this message.
## ℹ Use `summarise(.by = c(year, month, day))` for per-operation grouping
## (`?dplyr::dplyr_by`) instead.## # A tibble: 365 × 5
## # Groups: year, month [12]
## year month day first last
## <int> <int> <int> <int> <int>
## 1 2013 1 1 517 2356
## 2 2013 1 2 42 2354
## 3 2013 1 3 32 2349
## 4 2013 1 4 25 2358
## 5 2013 1 5 14 2357
## 6 2013 1 6 16 2355
## 7 2013 1 7 49 2359
## 8 2013 1 8 454 2351
## 9 2013 1 9 2 2252
## 10 2013 1 10 3 2320
## # ℹ 355 more rowsWrite the code to obtain what proportion of flights have departure delays of more than an hour for each months.
delay_prop <- flights %>%
filter(!is.na(dep_delay)) %>%
group_by(month) %>%
summarise(
prop_over_1hr = mean(dep_delay > 60)
)
delay_prop## # A tibble: 12 × 2
## month prop_over_1hr
## <int> <dbl>
## 1 1 0.0688
## 2 2 0.0698
## 3 3 0.0837
## 4 4 0.0916
## 5 5 0.0818
## 6 6 0.128
## 7 7 0.134
## 8 8 0.0796
## 9 9 0.0490
## 10 10 0.0469
## 11 11 0.0402
## 12 12 0.0942Months 6 & 7 (June & July) have the highest proportions out of any month.
Given that the n_distinct() function can count the number of dinstict (unique) values, could you find out which destinations have the most carriers?
dest_carriers <- flights %>%
group_by(dest) %>%
summarise(num_carriers = n_distinct(carrier)) %>%
arrange(desc(num_carriers))
dest_carriers## # A tibble: 105 × 2
## dest num_carriers
## <chr> <int>
## 1 ATL 7
## 2 BOS 7
## 3 CLT 7
## 4 ORD 7
## 5 TPA 7
## 6 AUS 6
## 7 DCA 6
## 8 DTW 6
## 9 IAD 6
## 10 MSP 6
## # ℹ 95 more rowsORD, CLT, BOS, ATL, & TPA are all tied for the most carriers with 7.
Does the following code achieve the similar goal with the above question except for missing the plotting step?
delays <- flights %>%
group_by(dest) %>%
summarise(
count = n(),
dist = mean(distance, na.rm = TRUE),
delay = mean(arr_delay, na.rm = TRUE)
) %>%
filter(count > 20, dest != "HNL")