I have a sequence of integers. What I would like to do is identify all sequences of 3's that is preceded AND followed by a 5. For example:

c(5,3,3,5,5,4,3,3,5)

The desired output would be:

c(F,T,T,F,F,F,F,F,F)

Explanation: The first sequence of 3's is preceded and followed by a 5. Hence True . The second sequence is preceded by a 4, hence False.

3 Answers

0
Ronak Shah On

Couldn't come up with a smarter solution so here is a for loop

x <- c(5,3,3,5,5,4,3,3,5) #Initial vector
current_inds <- numeric() #Variable to hold indices which need to be changed
saw_3 <- FALSE  #If 3 was seen before
output <- rep(FALSE, length(x))  #output vector
num_to_check <- 5   #Value to compare
last_val <- 0 #Last non-3 value

for (i in seq_along(x)) {
    #If the current value is not equal to 3
    if (x[i] != 3 ) {
      #Check if we previously saw 3 and if previous non_3 value was 5
      # and the next value is 5
      if(saw_3 & x[i + 1] == num_to_check & last_val == num_to_check) {
         #Change the group of 3 indices to TRUE
         output[current_inds] <- TRUE
         #Make the saw_3 flag as FALSE
         saw_3 <- FALSE
       }
      #Update the last seen non_3 value to current value
      last_val = x[i]
      }
    else {
     #If it is a 3 then append the indices in current_inds
     current_inds <- c(current_inds, i)
     #Make saw_3 flag TRUE
     saw_3 = TRUE
    }
}

output
#[1] FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE
0
user3343378 On

I have a very lengthy & ugly solution, but it works :p I hope someone can find a cleaner one :) I first create a matrix that contains 1 column which is every number in a non-repeated way (not unique, but without consecutives), and then 1 column with the number of times this number is repeated. Then I apply a logical function to see if a 3 is surrounded by 5s and in a final step, I unravel the vector back to its original length using the rep() function...

x <- c(5,3,3,5,5,4,3,3,5)

x_reduced <- x[x!=c(x[-1], FALSE)]
x_mat <- matrix(0, ncol = 3, nrow = length(x_reduced))
x_mat[ , 1] <- x_reduced

ctr = 1
x_ctr = 1
while (ctr < length(x)) {
  x_mat[x_ctr ,1] = x[ctr]
  x_mat[x_ctr, 2] = x_mat[x_ctr, 2] + 1 
  if(x[ctr+1] == x[ctr]){
    ctr = ctr + 1
  } else {
    x_ctr = x_ctr + 1
    ctr = ctr + 1
  }
}
x_mat[nrow(x_mat), 1] <- x[length(x)]
x_mat[nrow(x_mat), 2] <- x_mat[nrow(x_mat), 2] + 1

check_element <- function(pos) {
  if(pos == 1 | pos == nrow(x_mat)) return(FALSE)
  if(x_mat[pos+1, 1] == 5 & x_mat[pos-1, 1] == 5){
    return(TRUE)
  } else {
    return(FALSE)
  }
}

x_mat[,3] <- sapply(1:nrow(x_mat), check_element)
rep(x_mat[,3], x_mat[,2])

0
Roman On

There's room for optimization, but it's certainly possible with dplyr and rle().

> df_result
# A tibble: 9 x 1
  result
  <lgl> 
1 FALSE 
2 TRUE  
3 TRUE  
4 FALSE 
5 FALSE 
6 FALSE 
7 FALSE 
8 FALSE 
9 FALSE 

Code

df_result <- df %>%
    group_by(seq = {seq = rle(value); rep(seq_along(seq$lengths), seq$lengths)}) %>%
    ungroup() %>%
    mutate(last_3 = case_when(lag(seq) != seq ~ as.numeric(lag(value) == 5),
                              TRUE ~ NA_real_),
           next_5 = case_when(lead(seq) != seq ~ as.numeric(lead(value) == 5),
                              TRUE ~ NA_real_)) %>%
    group_by(seq, value) %>%
    mutate(result = case_when(sum(last_3, na.rm = TRUE) + sum(next_5, na.rm = TRUE) == 2 ~ TRUE,
                              TRUE ~ FALSE)) %>%
    ungroup() %>%
    select(result)

Data

library(dplyr)
df <- data.frame(value = c(5,3,3,5,5,4,3,3,5))