can you please stop repeating this nonsense? I don't think anybody
ever claimed that vectors can be considered list.
yes, it is nonsense. yes, there is one person who repeatedly made
this
claim. please read the archives; specifically, [1]. note this
statement:
"Note that any vector can be considered to be a list."
[1] https://stat.ethz.ch/pipermail/r-help/2009-February/186932.html
Sorry, I missed that one. Unfortunately I haven't kept a copy of the
posting I was referring to and I can't find a page that allows me to
search the February archives.
when you're talking about nonsense, please attribute it to the right
person.
Well, you _are_ repeating it, aren't you? Since you are so pedantic
about the semantics of programming languages, you certainly can't
complain about that statement.
I take back everything I wrote after that on the matter. Sorry!
It's rather the other way round: lists can also be seen as vectors to
R (possibly they are implemented as such, but I don't much about the
internals of R).
not *all* lists are vectors; pairlists are not, though they are
lists.
Didn't someone just say that it would probably be best to hide
pairlists from users entirely?
but
mode(unlist(a))
# "numeric"
class(unlist(a))
# integer
Well, yes, that was the solution, wasn't it? Since a list is a
vector, you have to unlist() it explicitly to turn it into a "plain"
vector. Except that it flattens the list, which is an entirely
different operation than a mere mode change. So I don't really
understand what that comment was supposed to mean.
Honestly, I can't think of a situation where I would want to do than
in R. In a Perl script, quite likely; but this is a kind of data
manipulation that R wasn't really designed for IMHO.
irrespectively of how exotic sorting lists of vectors can be in a
system
for the manipulation of a comprehensive range of sorts of data,
having a
procedure called 'sort.list' complain about being called on a list
is a
sure source of confusion.
Only until you read ?sort, isn't it?
how much is it likely that people will want to sort complex numbers
in a
system where complex numbers are incomparable?
I can only say that I'm not likely to want that either, but obviously
someone did so it was implemented (I am sure that one factor leading
to their decision was that this method does not clash with an existing
function).
Cheers,
Stefan