按顺序删除/折叠连续的重复值 [英] Remove/collapse consecutive duplicate values in sequence

问题描述

我有以下 dataframe :

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

所需结果应为

a b c d e a b e d 

这意味着任何两个连续的行都不应具有相同的值.不使用循环怎么办?

It means no two consecutive rows should have same value. How it can be done without using loop.

由于我的数据集非常庞大，因此执行循环需要大量时间.

As my data set is quite huge, looping is taking lot of time to execute.

数据框结构如下

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.

推荐答案

一种简单的方法是使用rle:

One easy way is to use 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:行程长度("lengths")和该行程重复的值("values").

rle returns a list with two values: the run length ("lengths"), and the value that is repeated for that run ("values").

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

---

更新:对于data.frame

如果您正在使用data.frame，请尝试以下操作:

---

Update: For a data.frame

If you are working with a data.frame, try something like the following:

## 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"包具有功能rleid，使您可以轻松地执行此操作.从上方使用mydf，请尝试:

---

Update 2

The "data.table" package has a function rleid that lets you do this quite easily. Using mydf from above, try:

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

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