R for Data Scienceの例題を解く- Chapter 5 データ変換

神Hadley R for Data Science の例題たちとその解答を書き残します。
今回はChapter 5 Data transformationです。

この章では下記のライブラリを使う。

library(dplyr)
library(nycflights13)

Chapter 5

5.2 Filter rows with `filter()`

5.2.4. Exercises

1. Find all flights that

Had an arrival delay of two or more hours

Flew to Houston (IAH or HOU)

Were operated by United, American, or Delta

Departed in summer (July, August, and September)

Arrived more than two hours late, but didn’t leave late

Were delayed by at least an hour, but made up over 30 minutes in flight

Departed between midnight and 6am (inclusive)

#1.
filter(flights, arr_delay > 120)
#2.
filter(flights, dest %in% c("IAH", "HOU"))
#3.
filter(flights, carrier %in% c("UA", "DL"))
#4.
filter(flights, month %in% 7:9)
#5.
filter(flights, arr_delay > 120, dep_delay < 1)
#6.
filter(flights, arr_delay > 60, dep_delay  > 30)
#7.
filter(flights, dep_time <= 600)

2. Another useful dplyr filtering helper is between(). What does it do? Can you use it to simplify the code needed to answer the previous challenges?

between(x, left, right)でベクトルxの元それぞれに対してleft以上でかつright以下であればTRUE, そうでなければFALSEを返す。
これを使って上の例題の4番を次のように書き換えることができる。

filter(flights, between(month, 7, 9))

3. How many flights have a missing dep_time? What other variables are missing? What might these rows represent?

8255行。dep_timeだけでなく、dep_delay, arr_time, arr_delay, air_timeもNAになっている。
air_timeがNAであることから、便が欠航になったと推測できる。

4. Why is NA ^ 0 not missing? Why is NA | TRUE not missing? Why is FALSE & NA not missing? Can you figure out the general rule? (NA * 0 is a tricky counterexample!)

NAにどのような値が入っても演算の結果が一定に定まるとき、結果はNAでなく値を返す。
ではNA * 0 は0ではないかと思ってしまうのだが、NAがInfであったとき、Inf * 0 = NaN となり0でなくなる。
そのため値は一意に決定しないので計算結果はNAとなる。

5.3. Arrange rows with `arrange()`?

5.3.1. Exercises

1. How could you use arrange() to sort all missing values to the start? (Hint: use is.na()).

arrange(data, !is.na(variable1))

2. Sort flights to find the most delayed flights. Find the flights that left earliest.

arrange(flights, desc(arr_delay))
arrange(flights, dep_delay)

3. Sort flights to find the fastest flights.

arrange(flights, distance/air_time)

4. Which flights travelled the longest? Which travelled the shortest?

arrange(flights, desc(air_time))
arrange(flights, air_time)

5.4 Select column with `select()`

5.4.1. Exercises

1. Brainstorm as many ways as possible to select dep_time, dep_delay, arr_time, and arr_delay from flights.

select(flights, starts_with("dep"), starts_with("arr"))
select(flights, matches("^(dep|arr)"))
select(flights, num_range("dep_", c("time", "delay")), num_range("arr_", c("time", "delay")))

など

2. What happens if you include the name of a variable multiple times in a select() call?

その変数が一度だけ現れるテーブルが返される。
つまり変数が二回selectされたからといって二回現れるわけではない。

3. What does the one_of() function do? Why might it be helpful in conjunction with this vector?
vars <- c("year", "month", "day", "dep_delay", "arr_delay")

select(flights, one_of(vars))

で文字列から変数を指定できる。
ということだと思うが、次でも普通にselectできた。

select(flights, vars)

となるとone_of関数の使いみちがよくわからない。はて。

4. Does the result of running the following code surprise you? How do the select helpers deal with case by default? How can you change that default?
select(flights, contains("TIME"))

文字列timeを含んだ列たち(dep_time, sched_dep_time,…)がselectされる。
次の式だとエラーを返す。

select(flights, TIME)
select(flights, "TIME")

select helper関数たちは引数としてignore.caseを持ち、デフォルトはTRUEになっている。
これをFALSEにすれば大文字小文字を区別して変数名を指定できる。

select(flights, contains("TIME", ignore.case = FALSE))
## A tibble: 336,776 x 0

5.5. Add new variables with `mutate()`

5.5.1. Exercises

1. Currently dep_time and sched_dep_time are convenient to look at, but hard to compute with because they’re not really continuous numbers. Convert them to a more convenient representation of number of minutes since midnight.

 flights2 <- mutate(flights, 
        dep_time2 = dep_time %/% 100 * 60 + dep_time %% 100, 
        sched_dep_time2 = sched_dep_time %/% 100 * 60 + sched_dep_time %% 100
       )

2. Compare air_time with arr_time - dep_time. What do you expect to see? What do you see? What do you need to do to fix it?

## A tibble: 6 x 5
#   year month   day air_time  diff
#  <int> <int> <int>    <dbl> <dbl>
#1  2013     1     1      227   193
#2  2013     1     1      227   197
#3  2013     1     1      160   221
#4  2013     1     1      183   260
#5  2013     1     1      116   138
#6  2013     1     1      150   106

diffはarr_time - dep_time (in minutes)で計算した変数だが、air_timeと微妙にずれている。
dep_time, arr_time は共に現地時刻なので飛行時間との差は時差が影響していると考えられる。
その場合arr_timeを現地時間ではなく東海岸時刻に調整すればよい。

3. Compare dep_time, sched_dep_time, and dep_delay. How would you expect those three numbers to be related?

flights3 <- mutate(flights2, 
    cal_depdel = dep_time2 - sched_dep_tim2,
    dif_depdel = cal_depdel - dep_delay
)
filter(flights3, dif_depdel != 0)
## A tibble: 1,207 x 5
#   dep_time sched_dep_time dep_delay cal_depdel dif_depdel
#      <dbl>          <dbl>     <dbl>      <dbl>      <dbl>
# 1      528           1115       853       -587      -1440
# 2       42           1439        43      -1397      -1440
# 3       86           1370       156      -1284      -1440
# 4       32           1439        33      -1407      -1440
# 5       50           1305       185      -1255      -1440
# 6      155           1439       156      -1284      -1440
# 7       25           1439        26      -1414      -1440
# 8       66           1365       141      -1299      -1440
# 9       14           1439        15      -1425      -1440
#10       37           1350       127      -1313      -1440

1440 = 60 * 24ということで、日付をまたいだ場合に計算した遅延時間と一日分のズレが発生して一致しなくなる。

4. Find the 10 most delayed flights using a ranking function. How do you want to handle ties? Carefully read the documentation for min_rank().

flights %>% 
   mutate(delay_rank = min_rank(desc(dep_delay))) %>% 
   filter(delay_rank <= 10)

上記の例ではmin_rank関数の仕様により１０位タイが複数あった場合にはすべてが抽出される。
１０位タイが複数あった場合に１０位をすべて除くためにはmin_rankを単にrankとする。

5. What does 1:3 + 1:10 return? Why?

Rにおいて長さの異なるベクトルの四則演算は、長さが短いベクトルを繰り返して補完することになっている。
この例題の場合は短いのは1:3の方なのでこれが繰り返されて長さ10のベクトルとなる。
その結果は下記に示す要素を持つベクトルになる。

[1] 2 = 1 + 1
[2] 4 = 2 + 2
[3] 6 = 3 + 3
[4] 5 = 1 + 4
[5] 7 = 2 + 5
[6] 9 = 3 + 6
[7] 8 = 1 + 7
[8] 10 = 2 + 8
[9] 12 = 3 + 9
[10] 11 = 1 + 10

6. What trigonometric functions does R provide?

サイン: sin(x)
コサイン: cos(x)
タンジェント: tan(x)
アークサイン: asin(x)
アークコサイン: acos(x)
アークタンジェント: atan(x)
サイン（単位π）: sinpi(x)
コサイン（単位π）: cospi(x)
タンジェント（単位π）: tanpi(x)

sinpi, cospi, tanpiはそれぞれsin(pi*x), cos(pi*x), tan(pi*x)と等価な関数。
その他atan2(x, y)という関数があるが使いみちは良くわからない。

5.6 Grouped summaries with `summarise()`

5.6.7. Exercises

1. Brainstorm at least 5 different ways to assess the typical delay characteristics of a group of flights. Consider the following scenarios:

A flight is 15 minutes early 50% of the time, and 15 minutes late 50% of the time.

A flight is always 10 minutes late.

A flight is 30 minutes early 50% of the time, and 30 minutes late 50% of the time.

99% of the time a flight is on time. 1% of the time it’s 2 hours late.

Which is more important: arrival delay or departure delay?

# フライトのグループ化
gf <- flights %>%
   filter(!is.na(dep_delay), !is.na(arr_delay)) %>%
   group_by(flight)

# １つめ
gf %>%
  summarise(l15 = sum(arr_delay >= 15), e15 = sum(arr_delay <= -15), n = n()) %>%
  filter(l15/n >= 0.5, e15/n >= 0.5)
gf %>%
  summarise(l15ratio = mean(arr_delay >= 15), e15ratio = mean(arr_delay <= -15)) %>%
  filter(l15ratio >= 0.5, e15ratio >= 0.5)
gf %>%
  summarise(l15 = sum(arr_delay >= 15), e15 = sum(arr_delay <= -15), abs15 = sum(abs(arr_delay) >= 15), n = n()) %>%
  filter(abs15 == n, l15 == e15)
gf %>%
  filter(all(abs(arr_delay)>=15)) %>%
  summarise(l15 = sum(arr_delay > 15), n = n()) %>%
  filter(l15 == n/2)
gf %>%
  filter(all(abs(arr_delay)>=15)) %>%
  count(e15 = arr_delay > 15) %>%
  filter(n == sum(n)/2) %>%
  count()

# ２つめ
gf %>%
  summarise(nl10 = sum(arr_delay < 10)) %>%
  filter(nl10 == 0)
gf %>%
  count(wt = arr_delay < 10) %>%
  filter(n == 0)
gf %>%
  filter(all(arr_delay >= 10)) %>%
  summarise()
gf %>%
  filter(!any(arr_delay < 10)) %>%
  count()
gf %>%
  count(e10 = arr_delay >= 10) %>%
  filter(all(e10 == TRUE)) %>%
  count

# ３つめ
gf %>%
  filter(mean(arr_delay < -30) >= 0.5) %>%
  filter(mean(arr_delay > 30) >= 0.5) %>%
  summarise()
gf %>%
  summarise(l30 = sum(arr_delay >= 30), e15 = sum(arr_delay <= -30), n = n()) %>%
  filter(l30/n >= 0.5, e30/n >= 0.5)
gf %>%
  filter(all(abs(arr_delay) >= 30)) %>%
  count(delay = arr_delay < 0) %>%
  filter(n == sum(n)/2) %>%
  summarise
gf %>%
  mutate(l30 = mean(arr_delay >= 30), e30 = mean(arr_delay <= -30)) %>%
  filter(l30 >= 0.5, e30 >= 0.5) %>%
  summarise
gf %>%
  summarise(l30l = mean(arr_delay >= 30) >= 0.5, e30l = mean(arr_delay <= -30) >= 0.5) %>%
  filter(l30l, e30l)

# ４つめ
# 10分以上遅れないことを"on time"の定義とする。
# 下記の条件に当てはまるflightは存在しないような気がするが・・・
# 99%を95%などと条件を緩めれば見つかる。
gf %>%
  summarise(on_time_ratio = mean(arr_delay <= 10), max_delay = max(arr_delay)) %>%
  filter(on_time_ratio >= 0.99, max_delay >= 120)
gf %>%
  filter(mean(arr_delay <= 10) >= 0.99) %>%
  filter(max(arr_delay) >= 120) %>%
  summarise()
gf %>%
  filter(any(arr_delay >= 120)) %>%
  summarise(on_time_ratio = mean(arr_delay <= 10)) %>%
  filter(on_time_ratio >= 0.99)  
gf %>%
  filter(mean(arr_delay <= 10) >= 0.99) %>%
  count(wt = arr_delay >= 120) %>%
  filter(n > 0)  
gf %>%
  filter(!all(arr_delay < 120)) %>%
  group_by(flight, ontime = arr_delay <= 10) %>%
  summarise(n = n()) %>%
  filter(ontime == TRUE, n >= sum(n)*0.99)

データ分析とは関係無いが、出発時刻が遅れたとしても到着時刻に影響がなければ問題ないように思う。

2. Come up with another approach that will give you the same output as not_cancelled %>% count(dest) and not_cancelled %>% count(tailnum, wt = distance) (without using count()).

not_cancelled %>% group_by(dest) %>% summarise(n = n())
not_cancelled %>% group_by(tailnum) %>% summarise(dist = sum(distance))

3. Our definition of cancelled flights (is.na(dep_delay) | is.na(arr_delay) ) is slightly suboptimal. Why? Which is the most important column?

flights %>% count(is.na(dep_delay), is.na(arr_delay))
## A tibble: 3 x 3
#  `is.na(dep_delay)` `is.na(arr_delay)`      n
#               <lgl>              <lgl>  <int>
#1              FALSE              FALSE 327346
#2              FALSE               TRUE   1175
#3               TRUE               TRUE   8255

dep_delayはNAではないが、arr_delayがNAのフライトが存在する。
出発したはよいが引き返したのかもしれない。
何にせよ出発しなければ到着もしないので、フライトキャンセルの条件はia.na(arr_delay)だけで良さそう。

4. Look at the number of cancelled flights per day. Is there a pattern? Is the proportion of cancelled flights related to the average delay?

flights %>% group_by(day) %>% summarise(cancelled.num = sum(is.na(arr_delay))) %>% ggplot(aes(factor(day), cancelled.num)) + geom_col()
flights %>% group_by(day) %>% summarise(cancelled.ratio = mean(is.na(arr_delay))) %>% ggplot(aes(factor(day), cancelled.ratio)) + geom_col()

８日がダントツにdelayが多い。
４日、１６日、２０日が少ない、など４の倍数に何かヒミツがあるような気がしたり。

5. Which carrier has the worst delays? Challenge: can you disentangle the effects of bad airports vs. bad carriers? Why/why not? (Hint: think about flights %>% group_by(carrier, dest) %>% summarise(n()))

flights %>% filter(arr_delay == max(arr_delay, na.rm=TRUE))
## A tibble: 1 x 19
#   year month   day dep_time sched_dep_time dep_delay arr_time sched_arr_time arr_delay carrier flight tailnum origin  dest air_time
#  <int> <int> <int>    <int>          <int>     <dbl>    <int>      <int>     <dbl>   <chr>  <int>   <chr>  <chr> <chr>    <dbl>
#1  2013     1     9      641            900      1301     1242       1530      1272      HA     51  N384HA    JFK   HNL      640
## ... with 4 more variables: distance <dbl>, hour <dbl>, minute <dbl>, time_hour <dttm>

ホノルル行きは長距離なので遅れが大きくなるのは当然、つまりハワイアン航空が遅れやすい航空会社だと結論するのは時期尚早です。
遅延時間は航路に大きく左右されるので、航空会社が良いか悪いかを判断するためには航路の影響を除かなければなりません。

flights %>%
   group_by(carrier, dest) %>%
   summarise(delay = mean(arr_delay[arr_delay > 0], na.rm=TRUE), n = n()) %>%
   ggplot(aes(carrier, dest)) +
     geom_count(aes(size = n, color = delay), alpha = 0.9) +
     scale_size_continuous(range = c(3,10)) +
     scale_color_gradient(low = "#56B1F7", high = "red")

横軸が航空会社、縦軸が行き先で、色は赤が遅延（平均値）が大きいことを意味しています。灰色は遅延が無いです。
横向きに色を比較すれば同じ行き先について優秀な航空会社、悪い航空会社を見つけることができます。
こう見るとOO(SkyWest Airline)はORD, DTW行きで遅れがひどいことがわかります。
データ数は少ないですが少ないデータのうちに二つもひどい遅れを出している以上、『遅れる航空会社』のレッテルを貼られるのは避けられないでしょう。
データ数が少ないカテゴリを除くと景色が変わります。

flights %>%
   group_by(carrier, dest) %>%
   summarise(delay = mean(arr_delay[arr_delay > 0], na.rm=TRUE), n = n()) %>%
   filter(n > 2) %>%
   ggplot(aes(carrier, dest)) +
     geom_count(aes(size = n, color = delay), alpha = 0.9) +
     scale_size_continuous(range = c(3,10)) +
     scale_color_gradient(low = "#56B1F7", high = "red")

航空会社、目的地別の平均遅延時間。ただしデータが3以上無い系列は表示しないようにした。灰色は遅延無し。大きさはデータ数を表す。

これを見ればEV(エクスプレスジェット航空)は大手航空会社ですが相対的に赤色が多く、遅延が大きい航空会社であることが見て取れます。

6. What does the sort argument to count() do. When might you use it?

sort = TRUEで大きい順に並び替えをする。

5.7 Grouped mutate (and filters)

5.7.1 Exercises

1. Refer back to the lists of useful mutate and filtering functions. Describe how each operation changes when you combine it with grouping.

関数arrangeを使えばグループ化された内部のみで整列することができる。

2. Which plane (tailnum) has the worst on-time record?

flights %>%
   group_by(tailnum) %>%
   filter(arr_delay == max(arr_delay, na.rm=TRUE)) %>%
   arrange(desc(arr_delay))
## A tibble: 2,210 x 19
## Groups:   tailnum [2,189]
#    year month   day dep_time sched_dep_time dep_delay arr_time sched_arr_time arr_delay carrier flight tailnum origin  dest air_time
#   <int> <int> <int>    <int>          <int>     <dbl>    <int>          <int>     <dbl>   <chr>  <int>   <chr>  <chr> <chr>    <dbl>
# 1  2013     1     9      641            900      1301     1242           1530      1272      HA     51  N384HA    JFK   HNL      640
# 2  2013     7    22      845           1600      1005     1044           1815       989      MQ   3075  N665MQ    JFK   CVG       96
# 3  2013     4    10     1100           1900       960     1342           2211       931      DL   2391  N959DL    JFK   TPA      139
# 4  2013     3    17     2321            810       911      135           1020       915      DL   2119  N927DA    LGA   MSP      167
# 5  2013     7    22     2257            759       898      121           1026       895      DL   2047  N6716C    LGA   ATL      109
# 6  2013    12     5      756           1700       896     1058           2020       878      AA    172  N5DMAA    EWR   MIA      149
# 7  2013    12    19      734           1725       849     1046           2039       847      DL   1223  N375NC    EWR   SLC      290
# 8  2013     2    10     2243            830       853      100           1106       834      F9    835  N203FR    LGA   DEN      233
# 9  2013     4    19      617           1700       797      858           1955       783      AA    257  N3GJAA    JFK   LAS      313
#10  2013     2    19     2324           1016       788      114           1227       767      DL   2319  N324US    LGA   MSP      136
## ... with 2,200 more rows, and 4 more variables: distance <dbl>, hour <dbl>, minute <dbl>, time_hour <dttm>

3. What time of day should you fly if you want to avoid delays as much as possible?

flights %>%
   group_by(day) %>%
   summarise(ave.delay = mean(arr_delay, na.rm=TRUE)) %>%
   arrange(desc(ave.delay))
## A tibble: 31 x 2
#     day ave.delay
#   <int>     <dbl>
# 1     8  19.07218
# 2    22  17.40492
# 3    23  16.74046
# 4    10  14.73601
# 5    12  11.13897
# 6    11  10.51807
# 7    24  10.47845
# 8    19  10.19267
# 9    18  10.10772
#10    17  10.07892
## ... with 21 more rows

質問の意図がよく分からなかった。この節のテクニックをどう使えばよいのか分からない。

4. For each destination, compute the total minutes of delay. For each, flight, compute the proportion of the total delay for its destination.

flights %>%
   filter(arr_delay > 0) %>%
   group_by(dest) %>%
   mutate(total_delay = sum(arr_delay, na.rm = TRUE))
flights %>%
   filter(arr_delay > 0) %>%
   group_by(flight) %>%
   mutate(total_delay = sum(arr_delay, na.rm = TRUE)) %>%
   group_by(flight, dest) %>%
   summarise(sum(arr_delay)/mean(total_delay))

5. Delays are typically temporally correlated: even once the problem that caused the initial delay has been resolved, later flights are delayed to allow earlier flights to leave. Using lag() explore how the delay of a flight is related to the delay of the immediately preceding flight.

flights %>%
  arrange(time_hour) %>%
  group_by(origin) %>%
  mutate(lag_delay = lag(dep_delay)) %>%
  ggplot(aes(dep_delay, lag_delay)) +
    geom_point(aes(color = origin), alpha = 0.1) +
    geom_abline(intercept = 0, slope = 1)

出発遅れと次のフライトの遅れの相関。左側で縦長に寄っているフライトは実際に何か問題が発生したフライトで、右側の相関のあるフライトが玉突き事故の結果と考えられる。

6. Look at each destination. Can you find flights that are suspiciously fast? (i.e. flights that represent a potential data entry error). Compute the air time a flight relative to the shortest flight to that destination. Which flights were most delayed in the air?

flights %>%
   group_by(dest) %>%
   mutate(min_air = min(air_time, na.rm = TRUE)) %>%
   mutate(diff = air_time - min_air) %>%
   arrange(desc(diff)) %>%
   select(flight, dep_time, arr_time, air_time, min_air, diff)
#Adding missing grouping variables: `dest`
## A tibble: 336,776 x 7
## Groups:   dest [105]
#    dest flight dep_time arr_time air_time min_air  diff
#   <chr>  <int>    <int>    <int>    <dbl>   <dbl> <dbl>
# 1   SFO    841     1727     2242      490     295   195
# 2   LAX    426     1812     2302      440     275   165
# 3   EGE    575     1806     2253      382     219   163
# 4   DEN    745     1513     1914      331     182   149
# 5   LAX     17     1814     2240      422     275   147
# 6   LAS    587     2142      143      399     256   143
# 7   SFO    434     1727     2206      438     295   143
# 8   SAN     89     1646     2107      413     279   134
# 9   HNL     15     1337     1937      695     562   133
#10   SFO    177     1746     2225      426     295   131
## ... with 336,766 more rows

SFO行きは最速295分で到達している実績があるにもかかわらず、flight 841はair_timeが490分もかかっています。

最速との差分を見るとSFO(San Francisco International Airport)など長距離のフライトばかりが候補に上がってしまいますので、データの入力エラーを見つけるのには適していないです。
データの入力エラーを見つけるのであれば我々に最も馴染みのある統計値の１つである偏差値を使うのが良いと思います。

flights %>%
   group_by(dest) %>%
   mutate(min_air = mean_air = mean(air_time, na.rm = TRUE), sd_air = sd(air_time, na.rm = TRUE)) %>%
   mutate(hensachi = (air_time - mean_air) * 10 / sd_air + 50) %>%
   arrange(desc(hensachi)) %>%
   select(dest, flight, air_time, mean_air, sd_air, hensachi)
## A tibble: 336,776 x 6
## Groups:   dest [105]
#    dest flight air_time mean_air   sd_air hensachi
#   <chr>  <int>    <dbl>    <dbl>    <dbl>    <dbl>
# 1   BOS   1703      112 38.95300 4.948552 197.6129
# 2   BOS   2136      107 38.95300 4.948552 187.5089
# 3   DCA   2175      131 45.85655 6.460277 181.7954
# 4   ACK   1491      141 42.06818 8.127495 171.7249
# 5   BOS   2480       99 38.95300 4.948552 171.3426
# 6   SYR   1516       97 43.03984 4.530559 169.1027
# 7   BOS   1750       96 38.95300 4.948552 165.2802
# 8   DTW   1131      170 84.83202 7.853928 158.4400
# 9   DTW   1026      170 84.83202 7.853928 158.4400
#10   RDU   1185      138 70.88533 6.201977 158.2150

BOS(Boston Logan International Airport)行き1703便は通常のBOS行きが平均38.9でいけているところを112分もかかっています。
偏差値にして197(!)ということで、私はこれが一番ひどい遅れであるように思います。

7. Find all destinations that are flown by at least two carriers. Use that information to rank the carriers.

flights %>%
   group_by(dest) %>%
   filter(n_distinct(carrier) > 1) %>%
   summarise(car_num = n_distinct(carrier))
## A tibble: 76 x 2
#    dest car_num
#   <chr>   <int>
# 1   ATL       7
# 2   AUS       6
# 3   AVL       2
# 4   BDL       2
# 5   BGR       2
# 6   BNA       5
# 7   BOS       7
# 8   BQN       2
# 9   BTV       3
#10   BUF       4

上記の条件を使って航空会社をランク付けせよという質問の意図がよく理解できなかったが、２つ以上が競合している空港にたくさん便を持っている航空会社が高順位だとしてみました。

flights %>% 
  group_by(dest) %>% 
  filter(n_distinct(carrier) > 1) %>% 
  ungroup() %>% 
  group_by(carrier) %>% 
  summarise(dest_num = n_distinct(dest)) %>% 
  arrange(desc(dest_num))
## A tibble: 16 x 2
#   carrier dest_num
#     <chr>    <int>
# 1      EV       51
# 2      9E       48
# 3      UA       42
# 4      DL       39
# 5      B6       35
# 6      AA       19
# 7      MQ       19
# 8      WN       10
# 9      OO        5
#10      US        5
#11      VX        4
#12      YV        3
#13      FL        2
#14      AS        1
#15      F9        1
#16      HA        1

8. For each plane, count the number of flights before the first delay of greater than 1 hour.

flights %>%
   mutate(delay_1hour = arr_delay > 60) %>%
   group_by(tailnum) %>%
   filter(any(delay_1hour)) %>%
   group_by(tailnum, delay_1hour) %>%
   mutate(first_delay = (time_hour == min(time_hour)) & delay_1hour)  %>%
   group_by(tailnum) %>%
   mutate(first_delay_timehour = min(time_hour[first_delay], na.rm=TRUE)) %>%
   summarise(answer = sum(time_hour < first_delay_timehour))
## A tibble: 3,371 x 2
#   tailnum answer
#     <chr>  <int>
# 1  D942DN      0
# 2  N0EGMQ      0
# 3  N10156      9
# 4  N102UW     25
# 5  N104UW      3
# 6  N10575      0
# 7  N105UW     22
# 8  N107US     20
# 9  N108UW     36
#10  N109UW     28
## ... with 3,361 more rows

参考サイト

R for Data Science Chapter 5 Data transformation

西陣に住むデータ分析屋のブログ

R for Data Scienceの例題を解く- Chapter 5 データ変換

Chapter 5

5.2 Filter rows with `filter()`

5.2.4. Exercises

5.3. Arrange rows with `arrange()`?

5.3.1. Exercises

5.4 Select column with `select()`

5.4.1. Exercises

5.5. Add new variables with `mutate()`

5.5.1. Exercises

5.6 Grouped summaries with `summarise()`

5.6.7. Exercises

5.7 Grouped mutate (and filters)

5.7.1 Exercises

参考サイト

関連記事

Chapter 5

5.2 Filter rows with filter()

5.2.4. Exercises

5.3. Arrange rows with arrange()?

5.3.1. Exercises

5.4 Select column with select()

5.4.1. Exercises

5.5. Add new variables with mutate()

5.5.1. Exercises

5.6 Grouped summaries with summarise()

5.6.7. Exercises

5.7 Grouped mutate (and filters)

5.7.1 Exercises

参考サイト

関連記事

5.2 Filter rows with `filter()`

5.3. Arrange rows with `arrange()`?

5.4 Select column with `select()`

5.5. Add new variables with `mutate()`

5.6 Grouped summaries with `summarise()`