[R] Hourly data with zoo

[steven mosher]

Gabor.. thanks.

zr <- zooreg(rnorm(24), as.chron("2011-01-01"), frequency = 24)

a couple issues:  my date data  has missing days and missing hours..
Sorry if I was not clear on
that.. I input it to a data frame and dates are of the form  20110101
and hours are in the format
0,100,200

The end goal is to create a data structure for around 200 series aligned by time

On Mon, Sep 12, 2011 at 3:48 AM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> On Mon, Sep 12, 2011 at 1:58 AM, steven mosher <moshersteven at gmail.com> wrote:
>> I have date data as a numeric and hourly data in 0 to 2300 hours in a dataframe.
>>
>> d  <-  rep(20110101,24)
>> h  <-  seq(from =  0, to  =  2300, by  = 100)
>>
>> df  <-  data.frame(LST_DATE  =  d,  LST_TIME  =  h,  data  =  rnorm(24, 0, 1))
>>
>> S  <-  chron(dates. = as.character(df$LST_DATE), times. =
>> paste(as.character(df$LST_TIME/100), ":0:0", sep  = ""),
>>           format  = c(dates  =  "Ymd",  times =  "h:m:s"))
>> X  <-  zoo(df$data, order.by = S)
>>
>> And I want to create a regular zoo series,  The above works but its
>> pretty ugly. Is there a more elegant way to do this.
>
> You probably want to create a zooreg object:
>
> library(zoo)
> library(chron)
>
> zr <- zooreg(rnorm(24), as.chron("2011-01-01"), frequency = 24)
>
> although if you really do want a zoo object that is not a zooreg
> object then you can do it like this:
>
> z <- as.zoo(zr)
>
> --
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>