Den 2015-10-09 kl. 12:14, skrev Martin Maechler:
I ran into a problem where I actually need rank(, ties.method="last"). It would
be great to have this feature in base and it's also simple to get (see below).
Thanks & cheers,
Marius
rank2 <- function (x, na.last = TRUE, ties.method = c("average",
"first", "last", # new "last"
"random", "max", "min"))
{
nas <- is.na(x)
nm <- names(x)
ties.method <- match.arg(ties.method)
if (is.factor(x))
x <- as.integer(x)
x <- x[!nas]
y <- switch(ties.method, average = , min = , max = .Internal(rank(x,
length(x), ties.method)), first = sort.list(sort.list(x)),
last = sort.list(sort.list(x, decreasing=TRUE), decreasing=TRUE), #
change
random = sort.list(order(x, stats::runif(sum(!nas)))))
if (!is.na(na.last) && any(nas)) {
yy <- NA
NAkeep <- (na.last == "keep")
if (NAkeep || na.last) {
yy[!nas] <- y
if (!NAkeep)
yy[nas] <- (length(y) + 1L):length(yy)
}
else {
len <- sum(nas)
yy[!nas] <- y + len
yy[nas] <- seq_len(len)
}
y <- yy
names(y) <- nm
}
else names(y) <- nm[!nas]
y
}
## MWE
x <- c(10, 11, 11, 12, 12, 13)
rank(x, ties.method="first")
rank2(x, ties.method="last")
Indeed, this makes sense to me, and is easy enough to document
and maintain, and preferable to asking useRs to use rev(.) and
similar "easy" (but somewhat costly for large data!)
transformations to get the same....
Or have (Marius Hofert and I) overlooked something obvious ?
I think so: the code above doesn't seem to do the right thing. Consider
the following example:
> x <- c(1, 1, 2, 3)
> rank2(x, ties.method = "last")
[1] 1 2 4 3
That doesn't look right to me -- I had expected
> rev(sort.list(x, decreasing = TRUE))
[1] 2 1 3 4
Henric Winell
Martin Maechler,
ETH Zurich
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