ls() pattern question
Thanks Andrew. it works well. --- Kai
On Wednesday, July 14, 2021, 05:22:01 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote:
Actually fun( param != something..) is syntactically incorrect in the first place for any function! ls sees "pat != whatever"? as the "name" argument of ls() and can't make any sense of it, of course. Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Jul 14, 2021 at 5:01 PM Andrew Simmons <akwsimmo at gmail.com> wrote:
Hello,
First, `ls` does not support `!=` for pattern, but it's actually throwing a
different error. For `rm`, the objects provided into `...` are substituted
(not evaluated), so you should really do something like
rm(list = ls(pattern = ...))
As for all except "con", "DB2", and "ora", I would try something like
setdiff(ls(), c("con", "DB2", "ora"))
and then add `rm` to that like
rm(list = setdiff(ls(), c("con", "DB2", "ora")))
On Wed, Jul 14, 2021 at 7:41 PM Kai Yang via R-help <r-help at r-project.org>
wrote:
Hello List,
I have many data frames in environment.? I need to keep 3 data frames
only, con DB2 and ora.
I write the script to do this.
rm(ls(pattern != c("(con|DB2|ora)")))
but it give me an error message:
Error in rm(ls(pattern != c("(con|DB2|ora)"))) :
? ?... must contain names or character strings
I think the pattern option doesn't support != ? and is it possible to fix
this?
Thank you,
Kai
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