CODEWITHSUNDEEP
rif-statementrecursionvectorinteger
user3375672
user3375672

Reputation: 3768

R: recursive function to give groups of consecutive numbers

Given a sorted vector x:

x <- c(1,2,4,6,7,10,11,12,15)

I am trying to write a small function that will yield a similar sized vector y giving the last consecutive integer in order to group consecutive numbers. In my case it is (defining groups 2, 4, 7, 12 and 15):

> y
[1]  2  2  4  7  7 12 12 12 15

I tried this this recursive idea (were x is the vector, and i an index that would start by 1 in most cases: if the content of the next index is one larger than the current i, then call the function with i+1; else return the content):

fun <- function(x,i){
  ifelse(x[i]+1 == x[i+1],
         fun(x,i+1),
         return(x[i]))
}

However:

> sapply(x,fun,1)
[1] NA NA NA NA NA NA NA NA NA

How to get this to work.

Upvotes: 0

Views: 228

Answers (2)

RHertel
RHertel

Reputation: 23798

Maybe this helps:

find_non_consec <- function(x){ c(x[which(as.logical(diff(x)-1))],x[length(x)]) }
x <- c(1,2,4,6,7,10,11,12,15)
res <- find_non_consec(x)

The result is:

> res
[1]  2  4  7 12 15

This function identifies the numbers where the series ceases to be consecutive.

Upvotes: 2

C_Z_
C_Z_

Reputation: 7806

Your sapply call is applying fun across all values of x, when you really want it to be applying across all values of i. To get the sapply to do what I assume you want to do, you can do the following:

sapply(X = 1:length(x), FUN = fun, x = x)

[1]  2  2  4  7  7 12 12 12 NA

Although it returns NA as the last value instead of 15. This is because I don't think your function is set up to handle the last value of a vector (there is no x[10], so it returns NA). You can probably edit your function to handle this fairly easily.

Upvotes: 3

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