Question

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？

Answer 1

可能有更快的方法：

df %>%
  group_by(id) %>%
  arrange(stopSequence) %>%
  filter(row_number()==1 | row_number()==n())

Answer 2

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

Answer 3

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

Answer 4

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。

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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()

Answer 9

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)])))), ]

从分组数据中选择第一行和最后一行

9 个答案: