一种简单的方法是使用rle：
以下是您的示例数据：

x <- scan(what = character(), text = "a a a b c c d e a a b b b e e d d")
# Read 17 items

rle返回具有两个值的list：游程长度（"lengths"），以及为该游程重复的值（"values"）。

rle(x)$values
# [1] "a" "b" "c" "d" "e" "a" "b" "e" "d"

更新：对于data.frame

如果您使用的是data.frame，请尝试以下操作：

## Sample data
mydf <- data.frame(
  V1 = c("a", "a", "a", "b", "c", "c", "d", "e", 
         "a", "a", "b", "b", "e", "e", "d", "d"),
  V2 = c(1, 2, 3, 2, 4, 1, 3, 9, 
         4, 8, 10, 199, 2, 5, 4, 10)
)

## Use rle, as before
X <- rle(mydf$V1)
## Identify the rows you want to keep
Y <- cumsum(c(1, X$lengths[-length(X$lengths)]))
Y
# [1]  1  4  5  7  8  9 11 13 15
mydf[Y, ]
#    V1 V2
# 1   a  1
# 4   b  2
# 5   c  4
# 7   d  3
# 8   e  9
# 9   a  4
# 11  b 10
# 13  e  2
# 15  d  4

更新2

"data.table"包中有一个函数rleid，可以让你很容易地完成这个任务。使用上面的mydf，尝试：

library(data.table)
as.data.table(mydf)[, .SD[1], by = rleid(V1)]
#    rleid V2
# 1:     1  1
# 2:     2  2
# 3:     3  4
# 4:     4  3
# 5:     5  9
# 6:     6  4
# 7:     7 10
# 8:     8  2
# 9:     9  4

library(dplyr)
x <- c("a", "a", "a", "b", "c", "c", "d", "e", "a", "a", "b", "b", "b", "e", "e", "d", "d")
x[x!=lag(x, default=1)]
#[1] "a" "b" "c" "d" "e" "a" "b" "e" "d"

编辑：对于data.frame

mydf <- data.frame(
    V1 = c("a", "a", "a", "b", "c", "c", "d", "e", 
         "a", "a", "b", "b", "e", "e", "d", "d"),
    V2 = c(1, 2, 3, 2, 4, 1, 3, 9, 
         4, 8, 10, 199, 2, 5, 4, 10),
   stringsAsFactors=FALSE)

DPLYR溶液是一种线性：

mydf %>% filter(V1!= lag(V1, default="1"))
#  V1 V2
#1  a  1
#2  b  2
#3  c  4
#4  d  3
#5  e  9
#6  a  4
#7  b 10
#8  e  2
#9  d  4

事后脚本

@Carl Witthoft建议的lead(x,1)以相反的顺序迭代。

leadit<-function(x) x!=lead(x, default="what")
rows <- leadit(mydf[ ,1])
mydf[rows, ]

#   V1  V2
#3   a   3
#4   b   2
#6   c   1
#7   d   3
#8   e   9
#10  a   8
#12  b 199
#14  e   5
#16  d  10

以R为底，我喜欢有趣的算法：

x <- c("a", "a", "a", "b", "c", "c", "d", "e", "a", "a", "b", "b", "b", "e", "e", "d", "d")

x[x!=c(x[-1], FALSE)]
#[1] "a" "b" "c" "d" "e" "a" "b" "e" "d"

虽然我很喜欢，......呃，* 爱 * rle，这里有一个枪战：
编辑：不能弄清楚dplyr到底是怎么回事，所以我用了dplyr::lead。我在OSX，R3.1.2，和最新的dplyr从CRAN。

xlet<-sample(letters,1e5,rep=T)
rleit<-function(x) rle(x)$values
lagit<-function(x) x[x!=lead(x, default=1)]
tailit<-function(x) x[x!=c(tail(x,-1), tail(x,1))]


  microbenchmark(rleit(xlet),lagit(xlet),tailit(xlet),times=20)
Unit: milliseconds
         expr      min       lq   median       uq      max neval
  rleit(xlet) 27.43996 30.02569 30.20385 30.92817 37.10657    20
  lagit(xlet) 12.44794 15.00687 15.14051 15.80254 46.66940    20
 tailit(xlet) 12.48968 14.66588 14.78383 15.32276 55.59840    20

Tidyverse解决方案：

x <- scan(what = character(), text = "a a a b c c d e a a b b b e e d d")
x <- tibble(x)
x |> 
 mutate(id = consecutive_id(x)) |> 
 distinct(x, id)

此外，如果存在与连续值列相关联的另一列y，则此解决方案允许一些灵活性：

x <- scan(what = character(), text = "a a a b c c d e a a b b b e e d d")
x <- tibble(x, y = runif(length(x)))
x |> 
    group_by(id = consecutive_id(x)) |> 
    slice_min(y)

我们可以选择不同的切片函数，如slice_max、slice_min、slice_head和slice_tail。
这个堆栈溢出线程出现在R4DS的第二版中，在书中的数字章节。