R help - Feb 2011 - summing 15 minute precip data to daily

Hi all,
i'm sure there is an easy way to do this, but i'm stumped, so any help
would
be appreciated.

i have a single column of data for precipitation every 15 minutes over a
year.  i want to sum the precip to daily data.
so the first 96 records = the first day, the second 96 records = the second
day, and so on....
is there a way to write a for loop that would sum the data to daily, and
write each value to a second object so i end up with a file of daily precip?

thanx,
Janet

	[[alternative HTML version deleted]]

Hi Janet,

One relatively simple way would be to transofrm the data into a 96 x
Ndays matrix and use colSums().  Of course, lets say on one day, the
measurement tool had technical difficulties and missed two
observations, then you only have 94 observations for that day, you
will need a fancier solution that deals with time not number of
observations.  Below is an example.

Cheers,

Josh


## Imaginary precipitation data for 7 days
set.seed(10)
x <- rnorm(96*7, 1, .1)
## An alternate way you may have the data stored
xalt <- data.frame(precip = x)

## Assuming _no_ missing observations
colSums(matrix(x, nrow = 96))

## alternate version
colSums(matrix(xalt$precip, nrow = 96))


On Thu, Feb 17, 2011 at 11:56 AM, Janet Choate <jsc.eco at gmail.com>
wrote:
> Hi all,
> i'm sure there is an easy way to do this, but i'm stumped, so any
help would
> be appreciated.
>
> i have a single column of data for precipitation every 15 minutes over a
> year. ?i want to sum the precip to daily data.
> so the first 96 records = the first day, the second 96 records = the second
> day, and so on....
> is there a way to write a for loop that would sum the data to daily, and
> write each value to a second object so i end up with a file of daily
precip?
>
> thanx,
> Janet
>
> ? ? ? ?[[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
>


-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

Janet:

The zoo package and aggregate.zoo should do the trick.  I have done this
many times with these tools.
HTH,

Stephen

On Thu, 2011-02-17 at 11:56 -0800, Janet Choate wrote:
> Hi all,
> i'm sure there is an easy way to do this, but i'm stumped, so any
help would
> be appreciated.
> 
> i have a single column of data for precipitation every 15 minutes over a
> year.  i want to sum the precip to daily data.
> so the first 96 records = the first day, the second 96 records = the second
> day, and so on....
> is there a way to write a for loop that would sum the data to daily, and
> write each value to a second object so i end up with a file of daily
precip?
> 
> thanx,
> Janet
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> 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.

On Thu, Feb 17, 2011 at 3:39 PM, Stephen Sefick <ssefick at gmail.com>
wrote:
> Janet:
>
> The zoo package and aggregate.zoo should do the trick. ?I have done this
> many times with these tools.
Here is some code. Suppose precip is the data.  We convert it to a zoo
object with date/times assuming the first one is at the start of
2000-01-01.  Then we aggregate them by day:
> library(zoo)
> library(chron)
> precip <- 1:200 # sample data
> z <- zooreg(precip, start = as.chron("2000-01-01 00:00:00"),
freq = 96)
> z.ag <- aggregate(z, as.Date)
> z.ag
2000-01-01 2000-01-02 2000-01-03
      4656      13872       1572


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