Re: R-beta: 'all.names' function -- failing as.list( _function_ )

Ross Ihaka (ihaka@stat.auckland.ac.nz)
Tue, 29 Apr 1997 10:25:07 +1200 (NZST)


Date: Tue, 29 Apr 1997 10:25:07 +1200 (NZST)
From: Ross Ihaka <ihaka@stat.auckland.ac.nz>
Message-Id: <199704282225.KAA25184@stat1.stat.auckland.ac.nz>
To: r-help@stat.math.ethz.ch
Subject: Re: R-beta: 'all.names' function -- failing   as.list( _function_ )
In-Reply-To: <6r67x7xapp.fsf@franz.stat.wisc.edu>
	<6r67x7xapp.fsf@franz.stat.wisc.edu>

Douglas Bates writes:
 > Some of the "eternal truths" about the S language are:
 >  - every object has a mode obtainable by mode(object)
 >  - every object has a length obtainable by length(object)
 >  - every object can be coerced to a list of the same length 
 > One can imagine that code that messes around with functions and other
 > expressions in R will break fairly quickly when these conditions do
 > not hold.  I don't know how much work would be involved in patching
 > over these differences between R and S but I suspect it would not be a
 > trivial undertaking.

Our basic philosopy is one of bringing the mountain to the prophet.
When someone finds an incompatibility important enough to bring to our
attention we will try to find a way to eliminate it or to cover it up.
[ This is not an open invitation, only ask if you need it please :-) ].

Many of the incompatibilities result from us not being familiar with
some of the inner mysteries of S - these are generally pretty easy to
fix.  Some incompatibilities however result from the fact that R
started life as a kind of Lisp interpreter.  These can be quite a bit
harder to fix.

The biggest of these changes is to make it so that users see "generic
vectors" rather than lists composed of "cons-cells".  Doing this will
enable us to eliminate vast tracts of code which do subset extraction
/ mutation for lists.  A consequence of this will be clean up of the
data frame code.

This is a fundamental change and will break absolutely everything.  It
has been delayed to next (Southern) summer.
	Ross
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