On 02/01/2020 9:31 a.m., Eric Berger wrote:
Hi Gabor,
This is great, thanks. It brought the time down to about 4 seconds.
The command
do.call("merge.xts",L)
also works in this case.
Suppose that instead of the default "outer" join I wanted to use, say, a
"left" join.
Is that possible? I tried a few ways of adding the
join="left"
parameter to the do.call() command but I could not get the syntax to work
(assuming it's even possible).
This should work:
do.call("merge", c(L, join = "left"))
The second argument to do.call is a list which becomes the arguments to
the function being called. Your time series should be unnamed entries
in the list, while other arguments to merge() should be named.
Duncan Murdoch
Thanks,
Eric
On Thu, Jan 2, 2020 at 3:23 PM Gabor Grothendieck <ggrothendi...@gmail.com>
wrote:
You don't need Reduce as xts already supports mutliway merges. This
perfroms one
multiway merge rather than k-1 two way merges.
do.call("merge", L)
On Thu, Jan 2, 2020 at 6:13 AM Eric Berger <ericjber...@gmail.com> wrote:
Hi,
I have a list L of about 2,600 xts's.
Each xts has a single numeric column. About 90% of the xts's have
approximately 500 rows, and the rest have fewer than 500 rows.
I create a single xts using the command
myXts <- Reduce( merge.xts, L )
By default, merge.xts() does an outer join (which is what I want).
The command takes about 80 seconds to complete.
I have plenty of RAM on my computer.
Are there faster ways to accomplish this task?
Thanks,
Eric
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