[Rd] "table(droplevels(aq)$Month)" in manual page of droplevels

[Martin Maechler]

>>>>> Rui Barradas <ruipbarradas at sapo.pt>
>>>>>     on Wed, 12 Apr 2017 17:07:45 +0100 writes:

    > 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.henric at gmail.com> schrieb am
    >> 12.04.2017
>>>>> um 15:35 in
    >>> Nachricht
    >>> <b66fe849-bb8d-f00d-87e5-553f866d57e0 at 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,

Indeed, we hope..  Thanks to the helpers!

Ulrich, please note that in the end this was all  because you're
still learning to understand R (e.g., data frames !) better.

As such this was completely inappropriate for R-devel and should
have gotten to the R help list  R-help.

With regards,
Martin Maechler, ETH Zurich

    > 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 at r-project.org mailing list
    >>>>> https://stat.ethz.ch/mailman/listinfo/r-devel
    >>>>> 
    >>> 
    >>> 
    >>> 
    >>> 
    >>> 
    >> 
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