search for: plotid

My data frame looks like: SightingID PA1 PA2 PlotID InOverlap Area1 2001 1 -99 392 Y 0.22 2002 1 -99 388 Y 0.253 2008 1 NA 104 N 0.344 2010 1 NA 71 N 0.185 2012 1 NA 61 N 0.166 2013 1 NA 61...

...l areas of different tree species in a number of research plots. Example data follow: > Trees<-data.frame(SppID=as.factor(c(rep('QUEELL',2), rep('QUEALB',3), 'CORAME', 'ACENEG', 'TILAME')), BA=c(907.9, 1104.4, 113.0, 143.1, 452.3, 638.7, 791.7, 804.3), PlotID=as.factor(c('BU3F10', rep('BU3F11',2), rep('BU3F12',5)))) > Trees SppID BA PlotID 1 QUEELL 907.9 BU3F10 2 QUEELL 1104.4 BU3F11 3 QUEALB 113.0 BU3F11 4 QUEALB 143.1 BU3F12 5 QUEALB 452.3 BU3F12 6 CORAME 638.7 BU3F12 7 ACENEG 791.7 BU3F12 8 TILAME...

...that match "plot". I've tried subset, and also the "which" command. plot <- c("plot1", "plot2", "plot3") # character vector used to select rows from data # create fake data from to try out subset v1 <- c(2,5,6,4,3) PLOTID <- c("plot1", "plot2", "plot3", "plot4", "plot5") full.data <- cbind(v1, PLOTID) full.data <- as.data.frame(full.data) # first attempt test <- which(full.data$PLOTID == plot) # second attempt test <- full.data[full.data$PLOTID ==...

...f you drop that, you'll get all points without S and P distinguished at all. If you add a groups argument, you should get them presented with different colors/symbols/etc. depending on your trellis settings (warning: untested code): par.plot(lnvol~lnden, groups = source,data=dat,sub=as.factor(plotid),col=T) Hope this helps, Matt Wiener -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of jennystadt Sent: Friday, August 25, 2006 11:28 AM To: r-help at stat.math.ethz.ch Subject: [R] Plot y ~ x under condition of variab...

...iment consists of randomized block design using fruits, vegetation and time as factors. The idea is to see if fruits, vegetation and time explain the abundance of mice. I am using tree density as a covariate. So I tried to fit the following structure to my data. > traps<-groupedData(mice~tree|plotID,outer=~time*fruit*veget|block,data=trap) but time and block are still considered to be numeric, while in fact they should be factor and ordered, respectively. > sapply(traps,data.class) mice plotID veget fruit time block tree "numeric" "ordered&quo...

...until now. #assuming independence between trees and thus ignoring the plot level. library(geepack) geeglm(formula = Y ~ Year, id = TreeID, family = binomial, corstr = "exchangeable") #using waves. But I'm wondering if this is correct. library(geepack) geeglm(formula = Y ~ Year, id = PlotID, waves = TreeID, family = binomial, corstr = "exchangeable") #using a unstructured correlation on the plot level. geeglm with unstructured correlation resulted in an out of memory error. library(Zelig) zelig(formula = Y ~ Year, model = "logit.gee", id = "PlotID", cors...

Greetings, I?ve fit the following mixed effects model using the NLME package: hd.impute.lme <- lme(I(log(HEIGHT_M - 1.37)) ~ SPECIES + SPECIES:I(1/(DBH_CM + 2.54)), random = ~ I(1/(DBH_CM + 2.54)) | PLOTID, data = trees, na.action = na.exclude) I would now like to extract a jackknife estimate of model error. I tried the following code, however, the estimate produced seems too low. ss.ok <- 0 for (i in 1:dim(trees)[1]) lme.msc <- ss.ok + sum(sd(lme(I(log(HEIGHT_M - 1.37)) ~ SPECIES +...

...ne, sometimes (case of non-sphericity), one needs to adjust for the degree of freedom (DF) used to test the significance of the "within subject factor" (Time i.e., Year in my case). My question is: How does this work with lme in R? Isn't it enough for me to specify in my model, Year × plotID as a random factor to account for the temporal autocorrelation? Or what else should I do to ensure that I have correct results from the summary function applied to my model (correct t- and p-values)? Thanks in advance. With regards, Sidzabda Djibril Dayamba, Swedish University of Agricultural Sci...

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje

Hi; I am working with the similarity matrix below and I would like to plot a two-dimensional MDS solution such as each point in the plot has a label. This is what I did: data <- read.table('c:/multivariate/mds/colour.txt',header=FALSE) similarity <- as.dist(data) distance <- 1-similarity result.nmds <- nmds(distance) plot(result.nmds) (nmds and plot.nmds as defined at

I have a text file that I have imported into R. It contains 3 columns and 316940 rows. The first column is vegetation plot ID, the second species names and the third is a cover value (numeric). I imported using the read.table function. My problem is this. I need to reformat the information as a matrix, with the first column becoming the row labels and the second the column labels and the