runsum(x, k, endrule = c("drop", "constant"), ...)

runmean(x, k, endrule = c("drop", "constant"), ...)

## S4 method for signature 'Rle'
smoothEnds(y, k = 3)

## S4 method for signature 'Rle'
runmed(x, k, endrule = c("median", "keep", "drop", "constant"), 
       algorithm = NULL, print.level = 0)

runwtsum(x, k, wt, endrule = c("drop", "constant"), ...)

runq(x, k, i, endrule = c("drop", "constant"), ...)

x <- Rle(1:10, 1:10)
  runsum(x, k = 3)
  runsum(x, k = 3, endrule = "constant")
  runmean(x, k = 3)
  runwtsum(x, k = 3, wt = c(0.25, 0.5, 0.25))
  runq(x, k = 5, i = 3, endrule = "constant")

  ## Missing and non-finite values
  x <- Rle(c(1, 2, NA, 0, 3, Inf, 4, NaN))
  runsum(x, k = 2)
  runsum(x, k = 2, na.rm = TRUE)
  runmean(x, k = 2, na.rm = TRUE)
  runwtsum(x, k = 2, wt =  c(0.25, 0.5), na.rm = TRUE)
  runq(x, k = 2, i = 2, na.rm = TRUE) ## max value in window

  ## The .naive_runsum() function demonstrates the semantics of 
  ## runsum(). This test ensures the behavior is consistent with
  ## base::sum().

  .naive_runsum <- function(x, k, na.rm=FALSE)
      sapply(0:(length(x)-k),
          function(offset) sum(x[1:k + offset], na.rm=na.rm))

  x0 <- c(1, Inf, 3, 4, 5, NA)
  x <- Rle(x0)
  target1 <- .naive_runsum(x0, 3, na.rm = TRUE)
  target2 <- .naive_runsum(x, 3, na.rm = TRUE)
  stopifnot(target1 == target2)
  current <- as.vector(runsum(x, 3, na.rm = TRUE))
  stopifnot(target1 == current)

  ## runmean() and runwtsum() :
  x <- Rle(c(2, 1, NA, 0, 1, -Inf))
  runmean(x, k = 3)
  runmean(x, k = 3, na.rm = TRUE)
  runwtsum(x, k = 3, wt = c(0.25, 0.50, 0.25))
  runwtsum(x, k = 3, wt = c(0.25, 0.50, 0.25), na.rm = TRUE)

  ## runq() :
  runq(x, k = 3, i = 1, na.rm = TRUE) ## smallest value in window
  runq(x, k = 3, i = 3, na.rm = TRUE) ## largest value in window

  ## When na.rm = TRUE, it is possible the number of non-NA 
  ## values in the window will be less than the 'i' specified.
  ## Here we request the 4th smallest value in the window,
  ## which tranlates to the value at the 4/5 (0.8) percentile.
  x <- Rle(c(1, 2, 3, 4, 5))
  runq(x, k=length(x), i=4, na.rm=TRUE)

  ## The same request on a Rle with two missing values
  ## finds the value at the 0.8 percentile of the vector 
  ## at the new length of 3 after the NA's have been removed.
  ## This translates to round((0.8) * 3).
  x <- Rle(c(1, 2, 3, NA, NA))
  runq(x, k=length(x), i=4, na.rm=TRUE)