Nrthugu

As inputs your function should take a vector of 0s and 1s;
Every time you see a sequence of 1s in the data you need to increase the number of children by 1;
Be careful with the two subsequent sequences of 1s, where the difference between them is less than 5 (i.e. when there are less than 5 0s in between them, then it is the same child and not a new child);
To help you social planner provides some examples of what your function should return:

#Input: c(1,1,1,1,0,0,0,0)



#Output: 1 1 1 1 1 1 1 1



#Input: c(0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1)



#Output: 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2



#Input: c(0,0,0,0,1,1,1,1,0,0,1,1,0,0,0,1,1,0,0,0,0,1,1,0,0,0,0,0,1)



#Output: 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2

Functions, which might be helpful:
diff()
cumsum()
which()
rle()

I dont quite understand how to approach the question, my thoughts on this are using diff function after the cumsum as it will help me to sustain a row of 1s but in this scenario i am loosing the length of vector (it obviously becomes shorter) also #rle$lenght seems to help me to detect gaps of length 5 or more to turn 1s into 2s. Sorry for this question I am only a beginner

I make use of which function in r (https://www.r-bloggers.com/which-function-in-r/) and run length encoding (http://www.cookbook-r.com/Manipulating_data/Finding_sequences_of_identical_values/). Here's my attempt:

vector_analyse <- function(sample_vector){

  # ----------------------------------------------------------------------------

  # Signature: vector --> vector

  # Author: kon_u

  # Description: Given a sample vector of 0s and 1s, return a sequence of 1s in 

  # the data you need to increase the number of children by 1 (when there are less 

  # 5 0s in between them, then it is the same child and not a new child)

  # ----------------------------------------------------------------------------



  # ----------------------------------------------------------------------------

  # Run Length Encoding gives a list of length and values

  # ----------------------------------------------------------------------------

  rle_object <- rle(sample_vector)

  x <- rle_object$lengths # original length 

  y <- rle_object$values # original values

  z <- which(y == 1) # index of 1 in vector y

  if (length(z) == 1){

    invisible()

  } else{

    for (i in 2:length(z)){

      if (x[z[i]-1] >= 5){

        y[z[i]] = y[z[i]]

      } else {

        y[z[i]] = y[z[i]] - 1

      }

    }

  }

  y_cumsum = cumsum(y)

  rle_object$values <- y_cumsum

  new_vector = inverse.rle(rle_object)

  return(new_vector)

}



vector_analyse(c(1,1,1,1,0,0,0,0)) # 1 1 1 1 1 1 1 1

vector_analyse(c(0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1)) # 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2

vector_analyse(c(0,0,0,0,1,1,1,1,0,0,1,1,0,0,0,1,1,0,0,0,0,1,1,0,0,0,0,0,1)) # 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2