[R] Hourly data with zoo
moshersteven at gmail.com
Mon Sep 12 17:57:19 CEST 2011
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
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:
> 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)
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
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