[R] outer join of xts's

Enrico Schumann e@ @end|ng |rom enr|co@chum@nn@net
Thu Jan 2 12:49:26 CET 2020

Quoting Eric Berger <ericjberger using gmail.com>:

> 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

Since you already know the number of series and all possible timestamps,
you could preallocate a matrix (number of timestamps times number of series).
You could use the fastmatch package to match the timestamps against the rows.
This what 'pricetable' in the PMwR package does.  Calling

     do.call(pricetable, L)

should give you matrix of the merged series, with an attribute 'timestamp',
from which you could create an xts object again.

I am not sure if it is the fastest way, but it's probably faster than calling
merge repeatedly.

kind regards
     Enrico  (the maintainer of PMwR)

Enrico Schumann
Lucerne, Switzerland

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