按顺序删除/折叠连续的重复值

Question

我有以下数据框：

a a a b c c d e a a b b b e e d d

要求的结果应该是

a b c d e a b e d 

这意味着没有两个连续的行应该具有相同的值。不使用循环如何完成。

由于我的数据集非常庞大，循环执行需要花费大量时间。

dataframe结构如下

a 1 
a 2
a 3
b 2
c 4
c 1
d 3
e 9
a 4
a 8
b 10
b 199
e 2
e 5
d 4
d 10

结果：

a 1 
b 2
c 4
d 3
e 9
a 4
b 10
e 2
d 4

它应该删除整行。

Answer 1

一种简单的方法是使用rle：

这是您的示例数据：

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：运行长度 ("lengths") 和为该运行重复的值 ("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”包有一个函数rleid，可以让你很容易地做到这一点。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

Answer 2

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 解决方案是一个班轮：

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

发布脚本

@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

Answer 3

使用 base R，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"

Answer 4

尽管我很喜欢，... errr，love rle ，这是一场枪战：

编辑：无法弄清楚 dplyr 到底是怎么回事，所以我使用了 dplyr::lead 。我在 OSX、R3.1.2 和来自 CRAN 的最新 dplyr 上。

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