R set outliers in time series NA in series already containing NA

Question

I have a time series containing NAs and some sudden jumps like this:

input=c(1:5, NA, 6:7,0,9:12)

In which 7,0,9 would be considered a jump for which 0 should be replaced by NA.

I would like to remove the very first value in which a sudden jump (with set value of what qualifies as a jump, in the example a change > 1) occurs and set it NA

the output for the example should look like this:

output=c(1:5,NA,6:7,NA,9:12)

I only want to set outliers NA, I do not want to overwrite the remaining values. The jump can be both negative and positive.

Problems I encountered:

1. The value after an existing NA value being counted as a jump
2. The "jump back" after the outlier being counted as jump

both of which resulted in more than necessary NAs, I try to keep as much original data as possible.

Any ideas? I have been stuck for a while. Thanks in advance!

Answer 1

There are three situations that are very similar but require different degrees of difficulties in terms of exceptions:

Situation 1

If the pattern always jumps back to 1-increase with a couple of interruptions, I would create vector_check which resembles the perfect vector. Everything in input that deviates from this should be set to NA :

vector_check <- min(input):max(input)
inds         <- vector_check != input
input[inds]  <- NA

Situation 2

If the pattern is less predictable and you basically wish to look for 'irregular' pattern, you'll get a more complicated situation. A possible solution would be to create a while -loop that checks which increments are larger than 2 (or whatever value seems sensible) and then replaces the problematic location bump_inds with an NA . Here I assume that an outlier creates two large increments: one because the value suddenly drops (increases) and one because it rises back up (drops back down) to its old value. This process proceeds until no problematic locations remain:

bump_ind <- rep(0, 3)

while(length(bump_ind) > 1){
  bump_ind        <- which( abs(diff(input)) > 2 )
  input[bump_ind[2]] <- NA
}

input
# [1]  1  2  3  4  5 NA  6  7 NA  9 10 11 12

Situation3

A third option, based on your real data sensor shows that the data does not have to jump back to a the previous level:

input    <- c(20.2,20.2,20.2,20.2,20.1,20.2,20.2,20.1,20.2, 20.2,20.2,20.2,17.7,
              18.9,19.3,19.4,19.4,19.4,19.5,19.5,19.5)
bump_ind <- rep(0, 3)

while(length(bump_ind) > 1){
  bump_ind        <- which( abs(diff(input)) > 2 )
  if(length(bump_ind) > 2){
    bump_ind <- bump_ind[1:2]
  }
  if( length(bump_ind) == 1 ){
      input[bump_ind[1] + 1] <- NA
  } else if( diff(bump_ind > 1) ){
      input[bump_ind[1] + 1] <- NA
  } else{
      input[bump_ind[2]] <- NA
  }
}

input
# [1] 20.2 20.2 20.2 20.2 20.1 20.2 20.2 20.1 20.2 20.2 20.2 20.2   NA 18.9 19.3
# [16] 19.4 19.4 19.4 19.5 19.5 19.5

Answer 2

This may be a more robust solution since you could modify the linear model of your data below if necessary:

Your data:

 input <- c(1:5, NA, 6:7,0,9:12)

A sequence of numbers:

x <- seq_len(length(input))

Select some threshold value for the residual of a linear model:

threshhold = 2

Calculate the linear model of your data and the residuals and select the outliers:

select <- abs((predict(lm(input ~ x), newdata = data.frame(x = x)) -input)) >= threshhold

Replace the outliers with 'NA'

input[select] <- NA
input
 [1]  1  2  3  4  5 NA  6  7 NA  9 10 11 12

EDIT: With your data:

input=c(20.2, 20.2, 20.2, 20.2,
        20.1, 20.2, 20.2, 20.1,
        20.2, 20.2, 20.2, 20.2,
        17.7, 18.9, 19.3, 19.4,
        19.4, 19.4, 19.5, 19.5,
        19.5)

x <- seq_len(length(input))
threshhold = 0.7
select <- abs((predict(lm(input ~ x), newdata = data.frame(x = x)) - input)) >= threshhold

inputnew <- input
inputnew[select] <- NA

input
 [1] 20.2 20.2 20.2 20.2 20.1 20.2 20.2 20.1 20.2 20.2 20.2 20.2 17.7 18.9 19.3
 [16] 19.4 19.4 19.4 19.5 19.5 19.5

inputnew
 [1] 20.2 20.2 20.2 20.2 20.1 20.2 20.2 20.1 20.2 20.2 20.2 20.2   NA 18.9 19.3
 [16] 19.4 19.4 19.4 19.5 19.5 19.5