On Nov 11, 2009, at 4:45 PM, Carl Witthoft wrote:
> By Bill Dunlap:
quote:
> x <- c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10)
> rev(cumsum(rev(is.na(x))))
[1] 1 1 1 1 0 0 0 0 0 0
A more natural way to do this is
> cumsum(is.na(c(NA,x[-length(x)])))
[1] 1 1 1 1 2 2 2 2 2 2
endquote
Both of which suggest the original problem could also be dealt with
by using rle(). Something like
xna<-is.na(x)
rle(xna)
and then apply cumsum to sections of x based onthe lengths returned
by rle
Carl
Which would be something along the lines of the following:
rle.x <- rle(!is.na(x))$lengths
> rle.x
[1] 3 1 6
> as.numeric(unlist(sapply(split(x, rep(seq(along = rle.x), rle.x)),
cumsum)))
[1] 1 3 6 NA 5 11 18 26 35 45
Which is what I had been working on until I saw Bill's more elegant
"one-liner" solution... :-)
The interim use of split() gets you 'x' split by where the NA's occur:
> rep(seq(along = rle.x), rle.x)
[1] 1 1 1 2 3 3 3 3 3 3
> split(x, rep(seq(along = rle.x), rle.x))
$`1`
[1] 1 2 3
$`2`
[1] NA
$`3`
[1] 5 6 7 8 9 10
and then you use cumsum() in sapply() (or lapply()) on each list
element and coerce the result back to an un-named numeric vector.
Regards,
Marc Schwartz
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