[Rd] bug in na.contiguous? Doesn't give the first tied stretch if it is at the start

[Georgi Boshnakov]

Hi.

The description of na.contiguous says:

"Find the longest consecutive stretch of non-missing values in a
     time series object.  (In the event of a tie, the first such
     stretch.)"

But this seems not to be the case if one of the tied longest stretches is at the start of the sequence/series. In the following example, there are three stretches of length 3, so I expect the result to be [1 2 3]. But:

> x <- c(1:3, NA, NA, 6:8, NA, 10:12)
> x
[1]  1  2  3 NA NA  6  7  8 NA 10 11 12
> na.contiguous(x)
[1] 6 7 8
## expected: [1] 1 2 3

(I have stripped attributes from the output for clarity.)

Below is the beginning of stats:::na.contiguous.default.
The source of the issue appears to be  the line containing the assignment to 'seg' (marked with exclamation marks).
The calculation leading to it does

cumsum(!good)

where !good in this case is

[1] FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE

And its cumsum is:

[1] 0 0 0 1 2 2 2 2 3 3 3 3


Then the assignment to 'seg' below picks the first longest stretch and subtracts 1, since the cumsum at indices corresponding to FALSE stays constant but the length of the constant stretch is one more then the number of FALSEs, ... except for the stretch at the start of the series which is not preceded by TRUE! So it is missed.

One way to patch this could be by the two commented assignments added by me to the code below to prepend a 0 to tt and then drop the first element of 'keep' to allow correct indexing later.

Georgi Boshnakov


> stats:::na.contiguous.default
function (object, ...)
{
    tm <- time(object)
    xfreq <- frequency(object)
    if (is.matrix(object))
        good <- apply(!is.na(object), 1L, all)
    else good <- !is.na(object)
    if (!sum(good))
        stop("all times contain an NA")
    tt <- cumsum(!good)
## tt <- c(0, tt)
    ln <- sapply(0:max(tt), function(i) sum(tt == i))
    seg <- (seq_along(ln)[ln == max(ln)])[1L] - 1          ## !!!
    keep <- (tt == seg)
## keep <- keep[-1]
    st <- min(which(keep))

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