`fragmentation` doesn't handle NA values in `x` properly.

[etzkorn]

I have data where I have non-wear periods that are not at the end of the recording time.
I have elected to pre-screen the counts vector, and place NA values in the counts variable x where there is non-wear. Your function does not provide the same output as if there were 0s for NA values. I've compared running your function with NA's (row 2) vs. 0's (row 1) using the reproducible code below. I also tried pre-filtering my data for wear time before providing it to your function (row 3). These three methods produce different results. Which one is theoretically correct? Given your answer, I would recommend editing the function to make at least rows 1 and 2 consistent.

# simulate data
x <- c(sample(0:2, 600, prob = c(.75,.2,.05), replace = T), 
       rep(0,200),
       sample(0:2, 200, prob = c(.75,.2,.05), replace = T)) %>% 
  as.integer
wear <- c(rep(1,600), rep(0,200), rep(1,200))
x.NA <- x
x.NA[wear == 0] <- NA

# fragmentation metrics
bind_rows(
fragmentation(x, wear, thresh = 1, metrics = "all") %>% as.tibble,
fragmentation(x.NA, wear, thresh = 1, metrics = "all") %>% as.tibble,
fragmentation(x.NA[wear == 1], w = rep(1,800), thresh = 1, metrics = "all") %>% as.tibble
)

[junruidi]

See line 83 and 84 in the fragmentation function.

x = na.omit(as.integer(x))
w = na.omit(w)

The first step of my fragmentation function is to remove NA for both x and w. The current function requires that, if "x" has NA, "wear" needs to have NA at the same locations. In the NHANES data, there is no "NA" in the vectors, only 0. Therefore, only your row 1 is correct. 2 is not. In the next version, I will update it so that this requirement can be relaxed.

For row 3, it is a different question, rather conceptional. because, for this scenario, you have two chunks of wear periods and they should be separated. If you connect them as you did in row 3, you are likely increase duration of active or sedentary bouts (and reduce the number of bouts), which will change the value. For example, consider the following example (A is active, and S is sedentary)

A A A NA NA NA A A A S S

That is 3 minutes of A, 3 minutes of NA, and 3 minutes of A again and 2 minutes of S. The average A duration is (3+3)/2 = 3
if you simply remove the NA like this. the average duration of A changes from 3 to 6.