Hello,
Inline.
Em 12-04-2017 16:40, Henric Winell escreveu:
(Let's keep the discussion on-list -- I've added back R-devel.)
On 2017-04-12 16:39, Ulrich Windl wrote:
Henric Winell <nilsson.hen...@gmail.com> schrieb am 12.04.2017
um 15:35 in
Nachricht <b66fe849-bb8d-f00d-87e5-553f866d5...@gmail.com>:
On 2017-04-12 14:40, Ulrich Windl wrote:
The last line of the example in droplevels' manual page seems to
be incorrect to me. I think it should read:
"table(droplevels(aq$Month))". Amazingly (I don't understand)
both variants seem to produce the same result (R 3.3.3): ---
The manual says that "The function 'droplevels' is used to drop
unused levels from a 'factor' or, more commonly, from factors in a
data frame." and, as documented, the 'droplevels' generic has
methods for objects of class "data.frame" and "factor". So, your
being amazed is a bit surprising given that 'aq' is a data frame.
The "surprising" thing is the syntax: I was unaware that '$' is a
generic operator that can be applied to the result of a function
(i.e.: droplevels); I thought it's kind of a special variable syntax.
Then your surprise is unrelated to the use of 'droplevels'.
Since the 'droplevels' method for objects of class "data.frame" returns
a data frame, the extraction operator '$' works directly on the
resulting object. So, 'droplevels(aq)$Month' is essentially the same as
aq <- droplevels(aq)
aq$Month
> Isn't there also the syntax ``droplevels(aq)["Month"]''?
Sure, and there are even more ways to do subsetting. But this is basic
stuff and therefore off-topic for R-devel. Please see the manual
(?Extract) or, e.g., Chapter 3 of Hadley Wickham's "Advanced R".
But note that droplevels(aq)["Month"] and droplevels(aq)$Month are _not_
the same. The first returns a data.frame (with just one vector), the
latter returns a vector. To return just a vector you could also use
droplevels(aq)[["Month"]]
which is preferable for programming, by the way. The '$' operator should
be reserved for interactive use only.
Hope this helps,
Rui Barradas
Henric Winell
Regards, Ulrich
Henric Winell
aq <- transform(airquality, Month = factor(Month, labels =
month.abb[5:9])) aq <- subset(aq, Month != "Jul")
table(aq$Month)
May Jun Jul Aug Sep 31 30 0 31 30
table(droplevels(aq)$Month)
May Jun Aug Sep 31 30 31 30
table(droplevels(aq$Month))
May Jun Aug Sep 31 30 31 30
--- For the sake of learners, try to keep the examples simple
and useful, even though you experts want to impress the
newbees...
Ulrich
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
