CODEWITHSUNDEEP
rrandomnumbers
Vanessa Almache
Vanessa Almache

Reputation: 15

Getting consecutive numbers without repetition

From a random sequence of numbers {"1", "2", "3", "4", "5"}, it is asked to determine the number of times that groups of 3 consecutive numbers appear, it means, the number of times that have been generated with the base = c ("1", "2", "3", "4", "5") any of the following groups {"123", "234", "345" }.

# I undertand that I have generate a sample with 5 numbers
a<-c(sample(1:5,5))
a
#I generated the list, as you can see I didn't fix a seed because I know that in every single sequence I will have differents grupos of 3 consecutive numbers, so I should obtain something like this

b<-c(2,3,4,5,1) #this example gives me just one that it would be {2,3,4}
b
#answer expected
1
#Then, I don't know how to obtain the sequence I have tried with permutations and combinations but I don't get it.

Upvotes: 0

Views: 89

Answers (2)

det
det

Reputation: 5232

Parameter k is for subsequence length (in your case 3):

f <- function(b, k = 3){

  if(k > length(b)) return(0)
  
  check_list <- lapply(k:length(b), function(x) all(diff(b[(x-k+1):x]) == 1)) 
  sum(unlist(check_list))
}

If you want to count increasing or decreasing subsequences replace == and 1 with appropriate relation and number.

Upvotes: 0

c0bra
c0bra

Reputation: 1080

This will count any occurrence of a three item increasing streak (e.g. 123).

countsequence <- function(x){
  if (length(x)<3){
    return(0)
  }
  count <- 0
  for(i in 3:length(x)){
    if(x[i-1]==x[i]-1 && x[i-2] == x[i]-2){
      count <- count + 1
    }
  }
  return(count)
} 

countsequence(1:5)
countsequence(c(2, 3, 4, 1, 5))

Upvotes: 1

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