CODEWITHSUNDEEP
rperformanceloopsindexingcounting
Rens
Rens

Reputation: 143

Create group number for contiguous runs of equal values

Is there is a faster way to make a counter index than using a loop? For each contiguous run of equal values, the index should be the same. I find the looping very slow especially when the data is so big.

For illustration, here is the input and desired output

x <- c(2, 3, 9, 2, 4, 4, 3, 4, 4, 5, 5, 5, 1)

Desired resulting counter:

c(1, 2, 3, 4, 5, 5, 6, 7, 7, 8, 8, 8, 9)

Note that non-contiguous runs have different indexes. E.g. see the desired indexes of the values 2 and 4

My inefficient code is this:

group[1]<-1
counter<-1
for (i in 2:n){
if (x[i]==x[i-1]){
    group[i]<-counter
}else{
    counter<-counter+1
    group[1]<-counter}
}

Upvotes: 14

Views: 844

Answers (5)

Ma&#235;l
Ma&#235;l

Reputation: 52239

With dplyr, you can use consecutive_id:

library(dplyr) #1.1.0+
consecutive_id(x)
# [1] 1 2 3 4 5 5 6 7 7 8 8 8 9

Upvotes: 1

AnilGoyal
AnilGoyal

Reputation: 26238

Above answer by Jota can be further simplified to, which will be even faster

with(rle(x), rep(1:length(lengths), lengths))

 [1] 1 2 3 4 5 5 6 7 7 8 8 8 9

Upvotes: 3

Arun
Arun

Reputation: 118849

Using data.table, which has the function rleid():

require(data.table) # v1.9.5+
rleid(x)
#  [1] 1 2 3 4 5 5 6 7 7 8 8 8 9

Upvotes: 13

Jota
Jota

Reputation: 17611

This will work with numeric of character values:

rep(1:length(rle(x)$values), times = rle(x)$lengths)
#[1] 1 2 3 4 5 5 6 7 7 8 8 8 9

You can also be a bit more efficient by calling rle just once (about 2x faster) and a very slight speed improvement can be made using rep.int instead of rep:

y <- rle(x)
rep.int(1:length(y$values), times = y$lengths)

Upvotes: 7

MrFlick
MrFlick

Reputation: 206411

If you have numeric values like this, you can use diff and cumsum to add up changes in values

x <- c(2,3,9,2,4,4,3,4,4,5,5,5,1)
cumsum(c(1,diff(x)!=0))
# [1] 1 2 3 4 5 5 6 7 7 8 8 8 9

Upvotes: 13

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