xts() speed on data with date index

An example to clarify/confuse ;)

system.time(px<-xts(pr,t))

user  system elapsed
  0.001   0.001   0.002

system.time(dx<-xts(pr,d))

user  system elapsed
  0.164   0.004   0.168

indexClass(dx)

[1] "Date"

indexClass(px)

[1] "POSIXt"  "POSIXct"

##  in POSIXct time as expected

head(px)

[,1]       [,2]       [,3]       [,4]
2009-07-26 09:39:32  0.6739000  0.6124116 -0.5820407  0.1022264
2009-07-27 09:39:32 -0.7806308 -0.4988508 -0.5448239 -0.5649892
2009-07-28 09:39:32  1.0275375 -0.8538655 -0.6098992  1.9661175
2009-07-29 09:39:32  0.6268532 -0.3144514  1.0142005 -0.2303953
2009-07-30 09:39:32 -3.8062200 -0.3875341 -0.9251609 -0.4677076
2009-07-31 09:39:32 -1.1192617 -0.6575085 -0.9918533 -0.8743504

head(dx)

[,1]       [,2]       [,3]       [,4]
2009-07-26  0.6739000  0.6124116 -0.5820407  0.1022264
2009-07-27 -0.7806308 -0.4988508 -0.5448239 -0.5649892
2009-07-28  1.0275375 -0.8538655 -0.6098992  1.9661175
2009-07-29  0.6268532 -0.3144514  1.0142005 -0.2303953
2009-07-30 -3.8062200 -0.3875341 -0.9251609 -0.4677076
2009-07-31 -1.1192617 -0.6575085 -0.9918533 -0.8743504

indexClass(px) <- "Date"

### simply convert the indexClass -- this is essentially costless,
### and now you have an xts object 'indexed' by Date

head(px)

[,1]       [,2]       [,3]       [,4]
2009-07-26  0.6739000  0.6124116 -0.5820407  0.1022264
2009-07-27 -0.7806308 -0.4988508 -0.5448239 -0.5649892
2009-07-28  1.0275375 -0.8538655 -0.6098992  1.9661175
2009-07-29  0.6268532 -0.3144514  1.0142005 -0.2303953
2009-07-30 -3.8062200 -0.3875341 -0.9251609 -0.4677076
2009-07-31 -1.1192617 -0.6575085 -0.9918533 -0.8743504

system.time(indexClass(px) <- "Date")

user  system elapsed
  0.001   0.001   0.001

Now convert back, and see if something like merge cares about the
index class... (hint: it doesn't)

system.time(indexClass(px) <- c("POSIXt","POSIXct"))

user  system elapsed
  0.001   0.000   0.000

indexClass(px)

[1] "POSIXt"  "POSIXct"

head(px)

[,1]       [,2]       [,3]       [,4]
2009-07-26 09:39:32  0.6739000  0.6124116 -0.5820407  0.1022264
2009-07-27 09:39:32 -0.7806308 -0.4988508 -0.5448239 -0.5649892
2009-07-28 09:39:32  1.0275375 -0.8538655 -0.6098992  1.9661175
2009-07-29 09:39:32  0.6268532 -0.3144514  1.0142005 -0.2303953
2009-07-30 09:39:32 -3.8062200 -0.3875341 -0.9251609 -0.4677076
2009-07-31 09:39:32 -1.1192617 -0.6575085 -0.9918533 -0.8743504

head(dx)

[,1]       [,2]       [,3]       [,4]
2009-07-26  0.6739000  0.6124116 -0.5820407  0.1022264
2009-07-27 -0.7806308 -0.4988508 -0.5448239 -0.5649892
2009-07-28  1.0275375 -0.8538655 -0.6098992  1.9661175
2009-07-29  0.6268532 -0.3144514  1.0142005 -0.2303953
2009-07-30 -3.8062200 -0.3875341 -0.9251609 -0.4677076
2009-07-31 -1.1192617 -0.6575085 -0.9918533 -0.8743504

system.time(merge(dx,dx))

user  system elapsed
  0.001   0.000   0.002

system.time(merge(px,px))

user  system elapsed
  0.001   0.000   0.002

### Caveat: converting back and forth is a bit silly, and possibly
perilous, but it is
### illustrative of what you can do with xts

HTHsomewhat

Jeff

I like xts because ?subsetting is easier. ?Something like x["2009-08"].
But it seems that xts() is a little slower than zoo() when converting
data with date index instead of time index.

pr<-matrix(rnorm(20000),nrow=5000,ncol=4)
t<-Sys.time()+(1:5000)*24*3600
d<-as.Date(t)
system.time(x<-zoo(pr,t))

? user ?system elapsed
? ? ?0 ? ? ? 0 ? ? ? 0

system.time(x<-zoo(pr,d))

? user ?system elapsed
? ? ?0 ? ? ? 0 ? ? ? 0

system.time(x<-xts(pr,t))

? user ?system elapsed
? ? ?0 ? ? ? 0 ? ? ? 0

system.time(x<-xts(pr,d))

? user ?system elapsed
? 0.34 ? ?0.00 ? ?0.35

Regards,
Michael

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Jeffrey Ryan
jeffrey.ryan at insightalgo.com

ia: insight algorithmics
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