问题
I have the following dataframe:
a a a b c c d e a a b b b e e d d
The required result should be
a b c d e a b e d
It means no two consecutive rows should have same value. How it can be done without using loop.
As my data set is quite huge, looping is taking lot of time to execute.
The dataframe structure is like the following
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
Result:
a 1
b 2
c 4
d 3
e 9
a 4
b 10
e 2
d 4
Its should delete the entire row.
回答1:
One easy way is to use rle:
Here's your sample data:
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 returns a list with two values: the run length ("lengths"), and the value that is repeated for that run ("values").
rle(x)$values
# [1] "a" "b" "c" "d" "e" "a" "b" "e" "d"
Update: For a data.frame
If you are working with a data.frame, try something like the following:
## Sample data
mydf <- data.frame(
V1 = c("a", "a", "a", "b", "c", "c", "d", "e",
"a", "a", "b", "b", "e", "e", "d", "d"),
V2 = c(1, 2, 3, 2, 4, 1, 3, 9,
4, 8, 10, 199, 2, 5, 4, 10)
)
## Use rle, as before
X <- rle(mydf$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
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
回答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"
EDIT: For 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 solution is one liner:
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
post scriptum
lead(x,1) suggested by @Carl Witthoft iterates in reverse order.
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
回答3:
With base R, I like funny algorithmics:
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"
回答4:
Much as I like,... errr, love rle , here's a shootoff:
EDIT: Can't figure out exactly what's up with dplyr so I used dplyr::lead . I'm on OSX, R3.1.2, and latest dplyr from 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
来源：https://stackoverflow.com/questions/27482712/remove-collapse-consecutive-duplicate-values-in-sequence