第一题 编写代码
利用nycflights13包的flights数据集是2013年从纽约三大机场(JFK、LGA、EWR)起飞的所有航班的准点数据,共336776条记录。
计算纽约三大机场2013起飞航班数和平均延误时间(可使用group_by, summarise函数)
flights %>%
group_by (origin) %>%
summarise (n= n (),delay_m= mean (dep_delay),na.rm= T)
# A tibble: 3 × 4
origin n delay_m na.rm
<chr> <int> <dbl> <lgl>
1 EWR 120835 NA TRUE
2 JFK 111279 NA TRUE
3 LGA 104662 NA TRUE
计算不同航空公司2013从纽约起飞航班数和平均延误时间
flights %>%
group_by (carrier) %>%
summarise (n= n (),delay_m= mean (dep_delay)) %>%
arrange (desc (n))
# A tibble: 16 × 3
carrier n delay_m
<chr> <int> <dbl>
1 UA 58665 NA
2 B6 54635 NA
3 EV 54173 NA
4 DL 48110 NA
5 AA 32729 NA
6 MQ 26397 NA
7 US 20536 NA
8 9E 18460 NA
9 WN 12275 NA
10 VX 5162 NA
11 FL 3260 NA
12 AS 714 NA
13 F9 685 NA
14 YV 601 NA
15 HA 342 4.90
16 OO 32 NA
计算纽约三大机场排名前三个目的地和平均飞行距离(可使用group_by, summarise, arrange, slice_max函数)
flights %>%
group_by (origin,dest) %>%
summarise (n= n (),dist_m= mean (distance)) %>%
slice_max (n,n= 3 )
`summarise()` has grouped output by 'origin'. You can override using the
`.groups` argument.
# A tibble: 9 × 4
# Groups: origin [3]
origin dest n dist_m
<chr> <chr> <int> <dbl>
1 EWR ORD 6100 719
2 EWR BOS 5327 200
3 EWR SFO 5127 2565
4 JFK LAX 11262 2475
5 JFK SFO 8204 2586
6 JFK BOS 5898 187
7 LGA ATL 10263 762
8 LGA ORD 8857 733
9 LGA CLT 6168 544
第二题 解释代码
代码含义:使用tibble函数将数据集iris转化成tibble格式,它提供了更友好的打印输出和一些其他特性;将数据集通过管道符传递给下个函数,增加代码的可读性;arrange对该数据集下的变量Species进行排序,across函数对在多个列上应用相同操作。starts_with(“Sepal”)用于选择所有以Sepal开头的列。desc按降序进行排列。
这段代码的功能是将iris数据集转换为tibble格式,然后按照鸢尾花的种类进行分组,并在每个组内将Sepal相关的列按照降序排列。
tibble (iris) %>%
arrange (Species,across (starts_with ("Sepal" ), desc))
# A tibble: 150 × 5
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
<dbl> <dbl> <dbl> <dbl> <fct>
1 5.8 4 1.2 0.2 setosa
2 5.7 4.4 1.5 0.4 setosa
3 5.7 3.8 1.7 0.3 setosa
4 5.5 4.2 1.4 0.2 setosa
5 5.5 3.5 1.3 0.2 setosa
6 5.4 3.9 1.7 0.4 setosa
7 5.4 3.9 1.3 0.4 setosa
8 5.4 3.7 1.5 0.2 setosa
9 5.4 3.4 1.7 0.2 setosa
10 5.4 3.4 1.5 0.4 setosa
# ℹ 140 more rows
代码含义:starwars是一个数据集,它包含了《星球大战》系列中角色的相关信息。group_by(gender)按变量gender进行分组。filter()用于筛选满足特定条件的行。mass > mean(mass, na.rm = TRUE)是筛选条件。其中, mean(mass, na.rm = TRUE)计算了每个gender组内mass的平均值,na.rm = TRUE表示在计算平均值时忽略缺失值(NA)。
从starwars数据集中,按照角色的性别进行分组,然后在每个性别组内筛选出体重超过该组平均体重的角色数据。最
starwars %>%
group_by (gender) %>%
filter (mass > mean (mass, na.rm = TRUE ))
# A tibble: 15 × 14
# Groups: gender [3]
name height mass hair_color skin_color eye_color birth_year sex gender
<chr> <int> <dbl> <chr> <chr> <chr> <dbl> <chr> <chr>
1 Darth … 202 136 none white yellow 41.9 male mascu…
2 Owen L… 178 120 brown, gr… light blue 52 male mascu…
3 Beru W… 165 75 brown light blue 47 fema… femin…
4 Chewba… 228 112 brown unknown blue 200 male mascu…
5 Jabba … 175 1358 <NA> green-tan… orange 600 herm… mascu…
6 Jek To… 180 110 brown fair blue NA <NA> <NA>
7 IG-88 200 140 none metal red 15 none mascu…
8 Bossk 190 113 none green red 53 male mascu…
9 Ayla S… 178 55 none blue hazel 48 fema… femin…
10 Gregar… 185 85 black dark brown NA <NA> <NA>
11 Lumina… 170 56.2 black yellow blue 58 fema… femin…
12 Zam We… 168 55 blonde fair, gre… yellow NA fema… femin…
13 Shaak … 178 57 none red, blue… black NA fema… femin…
14 Grievo… 216 159 none brown, wh… green, y… NA male mascu…
15 Tarfful 234 136 brown brown blue NA male mascu…
# ℹ 5 more variables: homeworld <chr>, species <chr>, films <list>,
# vehicles <list>, starships <list>
代码含义:首先传入starwars数据集,利用select函数选择变量名为:name, homeworld, species的数据列,mutate用于对数据集的列进行变换,对以上变量除了变量name以外的其余两者转换成因子类型的数据。
starwars %>%
select (name, homeworld, species) %>%
mutate (across (! name, as.factor))
# A tibble: 87 × 3
name homeworld species
<chr> <fct> <fct>
1 Luke Skywalker Tatooine Human
2 C-3PO Tatooine Droid
3 R2-D2 Naboo Droid
4 Darth Vader Tatooine Human
5 Leia Organa Alderaan Human
6 Owen Lars Tatooine Human
7 Beru Whitesun Lars Tatooine Human
8 R5-D4 Tatooine Droid
9 Biggs Darklighter Tatooine Human
10 Obi-Wan Kenobi Stewjon Human
# ℹ 77 more rows
代码含义:使用tibble函数将数据集mtcars转化成tibble格式,依照vs分组,mutate 函数用于在数据集中添加新的列或修改现有列的值。cut 函数用于将连续的数值变量 hp分割成离散的区间,并且划分为 3 个等宽的区间。将划分后的区间结果存储在一个新的列 hp_cut中。最后再按hp_cut分组。
tibble (mtcars) %>%
group_by (vs) %>%
mutate (hp_cut = cut (hp, 3 )) %>%
group_by (hp_cut)
# A tibble: 32 × 12
# Groups: hp_cut [6]
mpg cyl disp hp drat wt qsec vs am gear carb hp_cut
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4 (90.8,172]
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4 (90.8,172]
3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1 (75.7,99.3]
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1 (99.3,123]
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2 (172,254]
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1 (99.3,123]
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4 (172,254]
8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2 (51.9,75.7]
9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2 (75.7,99.3]
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4 (99.3,123]
# ℹ 22 more rows
第三题 查找帮助理解函数
阅读 https://dplyr.tidyverse.org/reference/mutate-joins.html 内容,说明4个数据集链接函数函数的作用。分别举一个实际例子演示并解释其输出结果。
inner_join() :对两个数据集内具有相同变量名的数据进行匹配,从而合并数据。
grades <- tibble (
ID = c (1 , 2 , 3 ),
Grade = c (98 , 100 , 77 )
)
names<- tibble (
ID = c (1 , 2 , 4 ),
Names = c ("Lihua" ,"Jack" , "Mark" )
)
inner_join (grades, names, by = "ID" )
# A tibble: 2 × 3
ID Grade Names
<dbl> <dbl> <chr>
1 1 98 Lihua
2 2 100 Jack
left_join() :保留左侧数据集的所有行,右侧无匹配时填充 NA
grades <- tibble (
ID = c (1 , 2 , 3 ),
Grade = c (98 , 100 , 77 )
)
names<- tibble (
ID = c (1 , 2 , 4 ),
Names = c ("Lihua" ,"Jack" , "Mark" )
)
left_join (grades, names, by = "ID" )
# A tibble: 3 × 3
ID Grade Names
<dbl> <dbl> <chr>
1 1 98 Lihua
2 2 100 Jack
3 3 77 <NA>
right_join() :保留右侧数据集的所有行,左侧无匹配时填充 NA
grades <- tibble (
ID = c (1 , 2 , 3 ),
Grade = c (98 , 100 , 77 )
)
names<- tibble (
ID = c (1 , 2 , 4 ),
Names = c ("Lihua" ,"Jack" , "Mark" )
)
right_join (grades, names, by = "ID" )
# A tibble: 3 × 3
ID Grade Names
<dbl> <dbl> <chr>
1 1 98 Lihua
2 2 100 Jack
3 4 NA Mark
full_join() :保留两个数据集的所有行(并集),无匹配时填充 NA
grades <- tibble (
ID = c (1 , 2 , 3 ),
Grade = c (98 , 100 , 77 )
)
names<- tibble (
ID = c (1 , 2 , 4 ),
Names = c ("Lihua" ,"Jack" , "Mark" )
)
full_join (grades, names, by = "ID" )
# A tibble: 4 × 3
ID Grade Names
<dbl> <dbl> <chr>
1 1 98 Lihua
2 2 100 Jack
3 3 77 <NA>
4 4 NA Mark