On Thu, 24 Nov 2011, Jeff Newmiller wrote:
AFAIK all solutions to the "grow object size" problem in R involve
creation of a new object to "change" an old one. There is
considerable sophistication under the hood that allows a minimum of
intermediate objects to be created if you are careful, but actually
changing the size of an object in place is not supported.
Just for the record, it actually is (R vectors have a LENGTH and
TRUELENGTH property at C level), but this is AFAIK only used
internally.
The better question is why you care about the order of the rows of a
data frame? A good way to think of a d.f. is like a table in a DBMS:
for efficiency the cases are unordered, but you can retrieve them in
any order you want (or no order). Indeed, for efficient operations on
millions of rows an R-DBMS interface is highly recomemded -- see the
R-data manual.
That doesn't mean that you have to "copy... to a temporary and then copy back", since you
can reference chunks of an existing object without actually moving them in memory by using
indexing. But (AFAIK) you cannot escape creating at least one new copy of the data that
"becomes" the object if you use rbind to grow your object.
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Sent from my phone. Please excuse my brevity.
Sammy Zee <szee2...@gmail.com> wrote:
Is there easy way (without copying the existing rows to a temporary
location and copying back) to add a new row to a specific index
location in
an existing data frame?
Example
df = data.frame( A= c('a','b','c'), B=c(1,2,3), C=(10,20,30))
newrow = c('X', 100, 200)
I want to add the newrow as the second row to the data frame df
Please suggest a solution that is efficient for a data frame that
can have millions of rows, and I want to add a new row in any given
index location of the data frame.
Thanks,
Sammy
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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