Detect and replace omitted data
Prompted by David's xtabs() suggestion, one way to do what I think the
OP wants is to
* define day and unit as factors whose levels comprise the full range
of desired values;
* use xtabs();
* return the result as a data frame.
Something like
x <- data.frame( day = factor(rep(c(4, 6), each = 8), levels = 4:6),
unit = factor(c(1:8, seq(2,16,2)), levels = 1:16),
value = floor(rnorm(16,25,10)) )
as.data.frame(with(x, xtabs(value ~ unit + day)))
HTH,
Dennis
On Tue, Oct 18, 2011 at 11:33 AM, David Winsemius
<dwinsemius at comcast.net> wrote:
On Oct 18, 2011, at 2:24 PM, Sarah Goslee wrote:
Hi Jonny, On Tue, Oct 18, 2011 at 1:02 PM, Jonny Armstrong <jonny5armstrong at gmail.com> wrote:
I am analyzing the spatial distribution of fish in a stream. The stream is divided into equally sized units, and the number of fish in each unit is counted. My problem is that my dataset is missing rows where the count in a unit equals zero. I need to create zero data for the missing units. For example: day<-(c(rep(4,8),rep(6,8))) unit<-c(seq(1,8,1),seq(2,16,2)) value<-floor(rnorm(16,25,10)) x<-cbind(day,unit,value)
Thanks for the actual reproducible example.
x ? ? day unit value ?[1,] ? 4 ? ?1 ? ?19 ?[2,] ? 4 ? ?2 ? ?15 ?[3,] ? 4 ? ?3 ? ?16 ?[4,] ? 4 ? ?4 ? ?20 ?[5,] ? 4 ? ?5 ? ?17 ?[6,] ? 4 ? ?6 ? ?15 ?[7,] ? 4 ? ?7 ? ?14 ?[8,] ? 4 ? ?8 ? ?29 ?[9,] ? 6 ? ?2 ? ?18 [10,] ? 6 ? ?4 ? ?22 [11,] ? 6 ? ?6 ? ?27 [12,] ? 6 ? ?8 ? ?16 [13,] ? 6 ? 10 ? ?45 [14,] ? 6 ? 12 ? ?36 [15,] ? 6 ? 14 ? ?34 [16,] ? 6 ? 16 ? ?13 Lets say the stream has 16 units. For each day, I want to fill in rows for any missing units (e.g., units 9-16 for day 4, the odd numbered units on day 6) with values of zero.
I could not figure out what you wanted precisely. If "day" is the row designator, and you want values by 'unit' and 'day' with zeros for the missing, then that is exactly what `xtab` delivers:
xtabs(value ~ day+unit, data=x)
? unit day ?1 ?2 ?3 ?4 ?5 ?6 ?7 ?8 10 12 14 16 ?4 25 34 ?3 25 38 18 19 33 ?0 ?0 ?0 ?0 ?6 ?0 22 ?0 42 ?0 37 ?0 ?4 12 31 17 28 You cannot get much more concise than that. -- david.
Here's one option, though it may not be terribly concise: all.samples <- expand.grid(day=unique(x[,"day"]), unit=1:16) all.samples <- all.samples[order(all.samples[,"day"], all.samples[,"unit"]),] x.final <- merge(x, all.samples, all.y=TRUE) x.final[is.na(x.final[,"value"]), "value"] <- 0 Sarah
Does anyone know a relatively concise way to do this? Thank you. ? ? ? [[alternative HTML version deleted]]
-- Sarah Goslee http://www.functionaldiversity.org
______________________________________________ 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.
David Winsemius, MD West Hartford, CT
______________________________________________ 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.