CODEWITHSUNDEEP
rpattern-matchingconditional-statements
user3324491
user3324491

Reputation: 559

Finding numeric element patterns using conditionals R

I have x

x <- c(2, 28, 1, 31, 1, 10, 2, 1, 17, 4, 2, 5, 19, 6)

I want to find those numbers which oscillates three or more times between being the same or more than 7 followed by a 2 or 1 and convert x into 0s and 1s to reflect when this is true. 

oscillates_more_than_once
[1]  0 1 0 1 0 1 0 0 0 0 0 0 0 0

For a number of 7 and above that is flanked by a 4 or less I also want to convert x into 0s and 1s to reflect when this is true. 

oscillates_once
[1]  0 0 0 0 0 0 0 0 1 0 0 0 0 0

However, if the number is above 7, but is flanked by 5 or above I wish to also find it separately. 

flanked
[1]  0 0 0 0 0 0 0 0 0 0 0 1 1 1

Would pattern matching be the best way to solve this?

Attempt at oscillates_more_than_once that does not work as it does not go through each element.

if (length(x[x >= 7]) > 3) {
test <- as.numeric(x >=7)
lag.test <- c(tail(test,-1),NA)
 for (w in 1:length(test)) {
  if (!(as.numeric(test[w]==lag.test[w])) && as.numeric(test[w]==1)) {
    as.numeric(x >= 7) ###converts all of x not just those elements that oscillate
  }
  else {
    as.numeric(!(x))
  }
 }
}

Upvotes: 0

Views: 54

Answers (1)

talat
talat

Reputation: 70336

You could try the following:

library(dplyr) # for lead()

#1 
y1 <- as.integer(x > 7 & lead(x) %in% 1:2)
#[1] 0 1 0 1 0 1 0 0 0 0 0 0 0 0

#2
y2 <- as.integer(x >= 7 & (lead(x) <= 4 | lag(x) <= 4) & !y1)
#[1] 0 0 0 0 0 0 0 0 1 0 0 0 0 0

#3
y3 <- as.integer((x >= 5 & (lead(x) > 7 | lag(x) > 7) )| (x > 7 & lead(x) >= 5 & lag(x) >= 5))
#[1] 0 0 0 0 0 0 0 0 0 0 0 1 1 1

Upvotes: 1

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