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....@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@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/ ______________________________________________ R-help@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.
