Commenting on this, is there a strong argument against modifying
relevel() to reorder more than one level at a time?
I started a topic a while back ("recursive relevel",
https://stat.ethz.ch/pipermail/r-help/2009-January/184397.html) and I've
happily used the proposed change since then by overloading
stats:::relevel.factor.
I'm sure there should be additional checking (for which I'm not versed
enough in programming to propose a real patch) but the enhanced
functionality seems very desirable (especially for plotting).
Best regards,
baptiste
Peter Dalgaard wrote:
Mark Difford wrote:
Hi Mao,
I am confused. And, I want to know how to assign a wanted order to factor
levels, intentionally?
You want ?relevel. Although the documentation leads one to think that it can
only be used to set a reference level, with the other levels being moved
down, presently it can in fact be used to set any order you wish. For a
factor with just a few levels you could simply use an index into the default
order.
##
new_d <- d
c(5,1,6:10,2:4)
new_d$population <- relevel(d$population,
levels(d$population)[c(5,1,6:10,2:4)])
Ignore the warning. Note that relevel can also be used "on-the-fly," so
without permanently changing level-order.
Now that's a dangerous strategy! You're relying on undocumented
behaviour and ignoring a warning message to boot. If someone implements
a check that ref is a scalar as assumed, you're shot.
Better to have a look at why stats:::relevel.factor currently works and
use the same mechanism:
lev <- levels(x)
if (is.character(ref))
ref <- match(ref, lev)
if (is.na(ref))
stop("'ref' must be an existing level")
nlev <- length(lev)
if (ref < 1 || ref > nlev)
stop(gettextf("ref = %d must be in 1:%d", ref, nlev),
domain = NA)
factor(x, levels = lev[c(ref, seq_along(lev)[-ref])])
and if you assume an integer reordering in ref, this reduces to
lev <- levels(x)
factor(x, levels = lev[ref])
and if ref is a character vector, plain
factor(x, levels=ref)
should do.
(Or, you can go "full monty" and retain all the checks an balances, just
cure the warning using
if (any(is.na(ref))
stop("'ref' must contain existing levels")
...
if (any(ref < 1 | ref > nlev))
Maybe also check !any(duplicated(ref)) for good measure
)
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