Ok thanks. If the gc( works then I can first do the assignment to the new
variable and then remove the old variable....

But I found gc() seems not always working...

Any more thoughts?
On Jan 21, 2012 8:04 PM, "R. Michael Weylandt" <michael.weylandt@gmail.com>
wrote:

> Short answer: I'm pretty sure it's impossible to remove elements "in
> place".
>
> Medium answer: If you are worried about memory usage, I might suggest
> using the data.table() package. It's pretty efficient for most things,
> though you'll loose xts specific time-functionality. You might be able
> to wedge time stamps in though smart use of the keys. (I'm not
> familiar enough with the package to say for sure.) Pondering aloud,
> merging data.table and xts functionalities sounds like an
> exceptionally fun project and a new package idea. I might do that when
> life gets 80% less busy....
>
> Long answer: it *may* be possible using the .Call() interface and
> playing very deeply with the internals (I haven't looked at them
> recently enough to say one way or the other) but it's exceptionally
> dangerous since R uses copy-on-change semantics. If you are determined
> enough, I think I can see a way that it's doable, but it's such a
> non-R thing to do that I don't think you'll get much help.
>
> Michael
>
> On Sat, Jan 21, 2012 at 7:13 PM, Michael <comtech.usa@gmail.com> wrote:
> > I hope I can do:
> >
> > x[ !Conditions, ] = NULL
> >
> > or something like this...
> >
> > On Sat, Jan 21, 2012 at 6:13 PM, Michael <comtech.usa@gmail.com> wrote:
> >
> >> Hi all,
> >>
> >> How do I remove a row in an xts object and the operation has to be in
> >> place?
> >>
> >> I knew I can do
> >>
> >> y=x[ Conditions, ]
> >>
> >> where x is the original time series xts object...
> >>
> >> But if x and y both are huge,
> >>
> >> this operation can consume a lot of memory...
> >>
> >> Any thoughts?
> >>
> >> Thanks a lot!
> >>
> >>
> >>
> >
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> >
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