问题
使用dplyr，如何在一个语句中选择分组数据的顶部和底部观察/行？
数据＆amp;实施例
给定数据框
df <- data.frame(id=c(1,1,1,2,2,2,3,3,3),
stopId=c("a","b","c","a","b","c","a","b","c"),
stopSequence=c(1,2,3,3,1,4,3,1,2))
我可以使用slice从每个组中获取顶部和底部观察结果，但使用两个单独的声明：
firstStop <- df %>%
group_by(id) %>%
arrange(stopSequence) %>%
slice(1) %>%
ungroup
lastStop <- df %>%
group_by(id) %>%
arrange(stopSequence) %>%
slice(n()) %>%
ungroup
我可以将这两个statmenet组合成一个选择两个顶部和底部观察的吗？
答案 0 :(得分：182)
可能有更快的方法：
df %>%
group_by(id) %>%
arrange(stopSequence) %>%
filter(row_number()==1 | row_number()==n())
答案 1 :(得分：87)
为了完整性：您可以传递slice索引向量：
df %>% arrange(stopSequence) %>% group_by(id) %>% slice(c(1,n()))
给出了
id stopId stopSequence
1 1 a 1
2 1 c 3
3 2 b 1
4 2 c 4
5 3 b 1
6 3 a 3
答案 2 :(得分：14)
不是dplyr，但使用data.table更直接：
library(data.table)
setDT(df)
df[ df[order(id, stopSequence), .I[c(1L,.N)], by=id]$V1 ]
# id stopId stopSequence
# 1: 1 a 1
# 2: 1 c 3
# 3: 2 b 1
# 4: 2 c 4
# 5: 3 b 1
# 6: 3 a 3
更详细的解释：
# 1) get row numbers of first/last observations from each group
# * basically, we sort the table by id/stopSequence, then,
# grouping by id, name the row numbers of the first/last
# observations for each id; since this operation produces
# a data.table
# * .I is data.table shorthand for the row number
# * here, to be maximally explicit, I've named the variable V1
# as row_num to give other readers of my code a clearer
# understanding of what operation is producing what variable
first_last = df[order(id, stopSequence), .(row_num = .I[c(1L,.N)]), by=id]
idx = first_last$row_num
# 2) extract rows by number
df[idx]
请务必查看Getting Started wiki，了解涵盖的data.table基础知识
答案 3 :(得分：6)
类似的东西：
library(dplyr)
df <- data.frame(id=c(1,1,1,2,2,2,3,3,3),
stopId=c("a","b","c","a","b","c","a","b","c"),
stopSequence=c(1,2,3,3,1,4,3,1,2))
first_last <- function(x) {
bind_rows(slice(x, 1), slice(x, n()))
}
df %>%
group_by(id) %>%
arrange(stopSequence) %>%
do(first_last(.)) %>%
ungroup
## Source: local data frame [6 x 3]
##
## id stopId stopSequence
## 1 1 a 1
## 2 1 c 3
## 3 2 b 1
## 4 2 c 4
## 5 3 b 1
## 6 3 a 3
使用do，您几乎可以对该组执行任意数量的操作，但@ jeremycg的答案更适合此任务。
答案 4 :(得分：6)
我知道指定的问题dplyr。但是，由于其他人已经使用其他软件包发布了解决方案，我也决定使用其他软件包：
基础套餐：
df <- df[with(df, order(id, stopSequence, stopId)), ]
merge(df[!duplicated(df$id), ],
df[!duplicated(df$id, fromLast = TRUE), ],
all = TRUE)
data.table：
df <- setDT(df)
df[order(id, stopSequence)][, .SD[c(1,.N)], by=id]
sqldf：
library(sqldf)
min <- sqldf("SELECT id, stopId, min(stopSequence) AS StopSequence
FROM df GROUP BY id
ORDER BY id, StopSequence, stopId")
max <- sqldf("SELECT id, stopId, max(stopSequence) AS StopSequence
FROM df GROUP BY id
ORDER BY id, StopSequence, stopId")
sqldf("SELECT * FROM min
UNION
SELECT * FROM max")
在一个查询中：
sqldf("SELECT *
FROM (SELECT id, stopId, min(stopSequence) AS StopSequence
FROM df GROUP BY id
ORDER BY id, StopSequence, stopId)
UNION
SELECT *
FROM (SELECT id, stopId, max(stopSequence) AS StopSequence
FROM df GROUP BY id
ORDER BY id, StopSequence, stopId)")
输出：
id stopId StopSequence
1 1 a 1
2 1 c 3
3 2 b 1
4 2 c 4
5 3 a 3
6 3 b 1
答案 5 :(得分：5)
使用which.min和which.max：
library(dplyr, warn.conflicts = F)
df %>%
group_by(id) %>%
slice(c(which.min(stopSequence), which.max(stopSequence)))
#> # A tibble: 6 x 3
#> # Groups: id [3]
#> id stopId stopSequence
#> <dbl> <fct> <dbl>
#> 1 1 a 1
#> 2 1 c 3
#> 3 2 b 1
#> 4 2 c 4
#> 5 3 b 1
#> 6 3 a 3
基准
它也比当前接受的答案快得多，因为我们按组查找最小值和最大值，而不是对整个stopSequence列进行排序。
# create a 100k times longer data frame
df2 <- bind_rows(replicate(1e5, df, F))
bench::mark(
mm =df2 %>%
group_by(id) %>%
slice(c(which.min(stopSequence), which.max(stopSequence))),
jeremy = df2 %>%
group_by(id) %>%
arrange(stopSequence) %>%
filter(row_number()==1 | row_number()==n()))
#> Warning: Some expressions had a GC in every iteration; so filtering is disabled.
#> # A tibble: 2 x 6
#> expression min median `itr/sec` mem_alloc `gc/sec`
#> <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl>
#> 1 mm 22.6ms 27ms 34.9 14.2MB 21.3
#> 2 jeremy 254.3ms 273ms 3.66 58.4MB 11.0
答案 6 :(得分：0)
在2018年使用Go to windows
-> choose Preference
-> Choose General
-> WebBrowser
-> it open an external Dilouge page choose there Use External Web Browser
-> then choose chrome/Firefox/Internet Explore
：
data.table答案 7 :(得分：0)
另一种使用lapply和dplyr语句的方法。我们可以将任意数量的摘要功能应用于同一条语句：
lapply(c(first, last),
function(x) df %>% group_by(id) %>% summarize_all(funs(x))) %>%
bind_rows()
例如，您可能也对具有最大stopSequence值的行感兴趣，然后这样做：
lapply(c(first, last, max("stopSequence")),
function(x) df %>% group_by(id) %>% summarize_all(funs(x))) %>%
bind_rows()
答案 8 :(得分：0)
一个不同的基数R替代方案是首先order和id stopSequence和split，id基于id并为每个{{1} }，我们仅选择第一个和最后一个索引，并使用这些索引对数据帧进行子集化。
df[sapply(with(df, split(order(id, stopSequence), id)), function(x)
c(x[1], x[length(x)])), ]
# id stopId stopSequence
#1 1 a 1
#3 1 c 3
#5 2 b 1
#6 2 c 4
#8 3 b 1
#7 3 a 3
或类似的使用by
df[unlist(with(df, by(order(id, stopSequence), id, function(x)
c(x[1], x[length(x)])))), ]