How can I query an xts time range using indexes rather than string

Question

I can query an xts time range by using two time strings separated by "/":

library(xts)
set.seed(1234)
a = xts(1:10, as.POSIXlt(1366039619, tz="", origin="1970-01-01") + rnorm(10, 0, 3))
                           [,1]
2013-04-15 11:26:51.962906    4
2013-04-15 11:26:55.378802    1
2013-04-15 11:26:56.329886   10
2013-04-15 11:26:57.275780    7
2013-04-15 11:26:57.306643    9
2013-04-15 11:26:57.360104    8
2013-04-15 11:26:59.832287    2
2013-04-15 11:27:00.287374    5
2013-04-15 11:27:00.518167    6
2013-04-15 11:27:02.253323    3

> a['2013-04-15 11:26:57/2013-04-15 11:26:58']
                           [,1]
2013-04-15 11:26:57.275780    7
2013-04-15 11:26:57.306643    9
2013-04-15 11:26:57.360104    8

How can I run the same range query on a different xts object using the POSIXlt objects index(a[4]) and index(a[7])? Do I have to convert the indexes to strings or there is a faster way using integer values, like the number of secs since the epoch embedded in POSIXlt?

@geektrader I used 4,7 as an example of what I want to achieve, but my input must be two POSIXlt objects. I'm trying to understand if I have to use strings (costing more time) or I can go with the indexes directly. — Robert Kubrick, Apr 26 '13 at 15:41
I think you can subset with indexes directly, no penalty there. — CHP, Apr 26 '13 at 15:44
@GSee Yes, it's the same question. Couldn't find it in my search. — Robert Kubrick, Apr 26 '13 at 17:33

score 0 · Accepted Answer · answered Apr 26 '13 at 15:58

0

ISO-style character subsetting by range (i.e. a length 1 vector containing : or /) is very fast. So you should use something like this :

a[paste(index(a)[4],index(a)[7],sep='::')]
                    [,1]
2013-04-15 17:26:57    7
2013-04-15 17:26:57    9
2013-04-15 17:26:57    8
2013-04-15 17:26:59    2

answered Apr 26 '13 at 15:58

agstudy

119,832
17
199
261

1

Right. So basically build a string. – Robert Kubrick Apr 26 '13 at 16:02
That's what I was trying to avoid to save some time (building the string), but if it's the only solution... – Robert Kubrick Apr 26 '13 at 16:05
@RobertKubrick it is not the only solution but it is the fastest way to subset an xts object. – agstudy Apr 26 '13 at 16:25
I'm not worry about the subsetting, I'm worried about building query strings millions of times in my program when I already have the indexes. – Robert Kubrick Apr 26 '13 at 16:34
@RobertKubrick if you have indxes you can do `a[4:7]`. – agstudy Apr 26 '13 at 16:38
Again the goal is to use POSIXlt indexes, not object offsets. I have to reference the dates range in a different xts object. It sounds like the only way is to build a string in this case. – Robert Kubrick Apr 26 '13 at 17:01
@RobertKubrick, if you're worried about performance, use POSIXct as the indexClass instead of POSIXlt – GSee Apr 26 '13 at 17:03
@GSee I still would have to build the string to query a different object. I'm looking to use the seconds since the epoch to query a different xts object range. – Robert Kubrick Apr 26 '13 at 17:04

score 0 · Answer 2 · answered Apr 26 '13 at 16:52

0

a[1:10] would extract the first 10 rows, and you don't even have to tell it which columns you want a[30:32] would give you the 30th through the 32 rows.

answered Apr 26 '13 at 16:52

user2004820

321
2
11

How can I query an xts time range using indexes rather than string

2 Answers2