Dear all, I would like to ask two questions on handling "zoo" object
1. Suppose I have following matrix :
mat1 <- matrix(rnorm(60), nrow=12)
Now I want to create a zoo object with index should represent the month
names for each row. For example first row has index "jan", 2nd row is
"feb"
and so on.
2. Again I have a daily time series (zoo object) like
daily.dat <- zooreg(rnorm(500), start=as.Date("2000-01-01"),
frequency=1)
Now I want to aggregate the data with standard deviation for each month.
What I mean is that I want to create a zoo time series vector with length
12, wherein 1st element would represent the SD for "ALL" daily data
corresponding Jan, 2nd is for SD from all data corresponding Feb, and so on
upto Dec.
I anyone please help me how to tackle them, I would be grateful.
Thanks
--
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come with zoo:
vignette("zoo")
vignette("zoo-quickref")
vignette("zoo-faq")
On Wed, Feb 24, 2010 at 4:52 AM, Bogaso <bogaso.christofer at gmail.com>
wrote:>
> Dear all, I would like to ask two questions on handling "zoo"
object
>
> 1. Suppose I have following matrix :
>
> ?mat1 <- matrix(rnorm(60), nrow=12)
>
> Now I want to create a zoo object with index should represent the month
> names for each row. For example first row has index "jan", 2nd
row is "feb"
> and so on.
> mat <- matrix(1:60, 12)
> library(zoo)
> z <- zoo(mat, factor(month.abb, levels = month.abb)); z
Jan 1 13 25 37 49
Feb 2 14 26 38 50
Mar 3 15 27 39 51
Apr 4 16 28 40 52
May 5 17 29 41 53
Jun 6 18 30 42 54
Jul 7 19 31 43 55
Aug 8 20 32 44 56
Sep 9 21 33 45 57
Oct 10 22 34 46 58
Nov 11 23 35 47 59
Dec 12 24 36 48 60
> # or perhaps using yearmon class
> zym <- zooreg(mat, start = as.yearmon("2010-01"), freq = 12);
zym
Jan 2010 1 13 25 37 49
Feb 2010 2 14 26 38 50
Mar 2010 3 15 27 39 51
Apr 2010 4 16 28 40 52
May 2010 5 17 29 41 53
Jun 2010 6 18 30 42 54
Jul 2010 7 19 31 43 55
Aug 2010 8 20 32 44 56
Sep 2010 9 21 33 45 57
Oct 2010 10 22 34 46 58
Nov 2010 11 23 35 47 59
Dec 2010 12 24 36 48 60
>
>
> 2. Again I have a daily time series (zoo object) like
>
> daily.dat <- zooreg(rnorm(500), start=as.Date("2000-01-01"),
frequency=1)
>
> Now I want to aggregate the data with standard deviation for each month.
> What I mean is that I want to create a zoo time series vector with length
> 12, wherein 1st element would represent the SD for "ALL" daily
data
> corresponding Jan, 2nd is for SD from all data corresponding Feb, and so on
> upto Dec.
>
> I anyone please help me how to tackle them, I would be grateful.
> set.seed(1)
> daily.dat <- zooreg(rnorm(500), start=as.Date("2000-01-01"),
frequency=1)
> aggregate(daily.dat, as.yearmon, sd)
Jan 2000 Feb 2000 Mar 2000 Apr 2000 May 2000 Jun 2000 Jul 2000 Aug 2000
0.9370553 0.7743405 0.9514569 0.8813805 0.9188229 1.0969453 1.0327717 1.0392682
Sep 2000 Oct 2000 Nov 2000 Dec 2000 Jan 2001 Feb 2001 Mar 2001 Apr 2001
1.0276672 1.0173960 0.8953093 1.0964024 0.9696928 0.9507952 1.1275021 1.1410237
May 2001
1.5286202
Thanks Gabor for this reply. Your first reply answers my question, however end is not. Here you said :> set.seed(1) > daily.dat <- zooreg(rnorm(500), start=as.Date("2000-01-01"), frequency=1) > aggregate(daily.dat, as.yearmon, sd)Jan 2000 Feb 2000 Mar 2000 Apr 2000 May 2000 Jun 2000 Jul 2000 Aug 2000 0.9370553 0.7743405 0.9514569 0.8813805 0.9188229 1.0969453 1.0327717 1.0392682 Sep 2000 Oct 2000 Nov 2000 Dec 2000 Jan 2001 Feb 2001 Mar 2001 Apr 2001 1.0276672 1.0173960 0.8953093 1.0964024 0.9696928 0.9507952 1.1275021 1.1410237 May 2001 1.5286202 But what I want is, to aggregate "all" data on a particular month for all years and apply statistics on that aggregated data. Therefore I should get only 12 values (for 12 months) not 17 values. Is there any way to do that? Thanks, -- View this message in context: http://n4.nabble.com/Question-on-Zoo-object-tp1567207p1567464.html Sent from the R help mailing list archive at Nabble.com.
Try this:> aggregate(daily.dat, format(time(daily.dat), "%m"), sd)01 02 03 04 05 06 07 08 0.9473855 0.8577099 1.0614892 1.0120308 1.1301652 1.0969453 1.0327717 1.0392682 09 10 11 12 1.0276672 1.0173960 0.8953093 1.0964024 On Wed, Feb 24, 2010 at 8:31 AM, Bogaso <bogaso.christofer at gmail.com> wrote:> > Thanks Gabor for this reply. Your first reply answers my question, however > end is not. Here you said : > >> set.seed(1) >> daily.dat <- zooreg(rnorm(500), start=as.Date("2000-01-01"), frequency=1) >> aggregate(daily.dat, as.yearmon, sd) > ?Jan 2000 ?Feb 2000 ?Mar 2000 ?Apr 2000 ?May 2000 ?Jun 2000 ?Jul 2000 ?Aug > 2000 > 0.9370553 0.7743405 0.9514569 0.8813805 0.9188229 1.0969453 1.0327717 > 1.0392682 > ?Sep 2000 ?Oct 2000 ?Nov 2000 ?Dec 2000 ?Jan 2001 ?Feb 2001 ?Mar 2001 ?Apr > 2001 > 1.0276672 1.0173960 0.8953093 1.0964024 0.9696928 0.9507952 1.1275021 > 1.1410237 > ?May 2001 > 1.5286202 > > But what I want is, to aggregate "all" data on a particular month for all > years and apply statistics on that aggregated data. Therefore I should get > only 12 values (for 12 months) not 17 values. Is there any way to do that? > > Thanks, > -- > View this message in context: http://n4.nabble.com/Question-on-Zoo-object-tp1567207p1567464.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >