similar to: Nested model and variance partitioning

>From the following: basin.map <- readAsciiGrid("c:/temp/area.asc", colname="area") I have a SpatialGridDataFrame which has the x and y cordinate of a cell, and the drainage area of that cell. There are many cells with a low drainage area (in my case, 33000 with an area of 37.16) and one cell with the highest drainage area (again, in my case, a drainage area of of

Hi I am having difficulty interpretive the multiple regression output. I would like to know what it means when one of the factors is assigned as the intercept? In my data I am looking at the relationship between environmental parameters and biological production. One of my variables in the analysis is substratum type and gravel is identified as the intercept and the P-value is significant,...

I'm trying to calculate a cumulative area distribution (graph) of drainage areas. This is defined as P(A > A*). Simple in principle. I can do this in excel, with "COUNTIF", which will count the number of cells in the row "area" that have area A, then determine, for each cell in the row "area, how many cells exceede that area, then dividing that number by the total

I'm trying to develop a linear model for crop productivity based on variables published as part of the SSURGO database released by the USDA. My default is to just run lm() with continuous predictor variables as numeric, and discrete predictor variables as factors, but some of the discrete variables are ordinal (e.g. drainage class, which ranges from excessively drained to excessively poorly

Hi useRs, I'm trying for the first time to use a sem. The model finally runs, but gives a warning saying : "In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : Could not compute QR decomposition of Hessian. Optimization probably did not converge. " I found in R-help some posts on this warning, but my attemps to modify the code didn't change

I need to make a plot illustrating main characterisitig of river drainage data. For this I have 2 questions: how can I rotate a histogram -90? (or 270?) (like the horizontal=TRUE with plot)? how can I use split.screen to produce 3 plot with uneuqal size (1/5, 2/5, 2/5 of the screen width)? thank you very much in advance for your help fabian

Dear R users, I am sorry to ask you such a pathetic newbie question, but how does one append data at the end of a data frame? I am working with GRASS/R library, but the question is about R. I have a data.frame containing the following variables basinID, distoutlet, drainage_area, slope These variables are stored for all pixels of Grass Raster objects. For each drainage basin (basinID), I'd

I would consider this is a question for a statistics forum such as stats.stackexchange.com, not R-help, which is about R programming. They do sometimes intersect, as here, but I think you need to *understand what you're doing* before you write the R code to do it. Obviously, IMO. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and

I'm trying to read some web tables directly into R. These are both genome sequencing projects (eukaryotes and metagenomes) from NCBI and look very similar; however, only the first one works. http://www.ncbi.nlm.nih.gov/genomes/leuks.cgi http://www.ncbi.nlm.nih.gov/genomes/lenvs.cgi I added ?dump=selected to the end of the url string to get a tab- delimited file (which is what happens

All, has anyone any experience in using Wifi smartphones as SIP clients? Does this work properly? What models/brands are optimal for this (in terms of ease of use, battery life etc)? Thx!! B.

Hello, Despite my search, I didn't find a post-hoc test for an ANCOVA. I used the functions aov() and lm() to run the ANCOVA then I tried TukeyHSD() but it didn't work (because of the covariable is a continuous variable?). Furthermore, I would like to plot the adjusted values (i.e. the values of the tested variable taking into account the covariable). Thanks for your help! N. Poulet

Hi, Just something I don't understand: data <- data.frame(V1=c(1:12),F1=c(rep("a",4),rep("b",4),rep("c",4))) data_ac <- data[which(data$F1 !="b"), ] levels(data_ac$F1) Why the level "b" is always present ? thanks Tristan, R 2.0.1 for Linux Fedora 3 -- ------------------------------------------------------------ Tristan

One possibility ist boxplot(sapply(ListOfNames, get, env = .GlobalEnv)) Hope that this helps Thomas -----Urspr??ngliche Nachricht----- Von: Lefebure Tristan [mailto:Tristan.Lefebure at univ-lyon1.fr] Gesendet: Mittwoch, 7. Juli 2004 10:34 An: r-help at stat.math.ethz.ch Betreff: [R] boxplot a list of objects Hi list, #Imagine we have vectors of different length (in practice 100 vectors):

Hi I use RSvgDevice to output plot, and modify them using svg editor (inkscape or sodipodi on Linux). Some month ago, results were perfect. I did exactly the same analysis today on the same data, and unfortunatly the results are different. While looking to the svg file, it seems that all information concerning "fill" and "stroke" of objects are lost. The consequence is that

sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. > Hello > I have a data array with soil variables (caperf), in which the variable "clay" is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered > >

Hi all, I am wondering whether there is a small bug in the binom.test function of the ctest library (I'm using R 1.6.0 on windows 2000, but Splus 2000 seems to have the same behaviour). Or perhaps I've misunderstood something. the command binom.test(11,100,p=0.1) and binom.test(9,100,p=0.1) give different p-values (see below). As 9 and 11 are equidistant from 10, the mean of the

Colin, If your propose is to create a ternary plot with points and vectors, I think easier do this with graphics based plots instead of trellis based plots. Although, with little work you can do with trellis too. I gave you a reproducible code to put an arrow in a ternary plot. I use the function locator() to extract coordinates. The code is the following

Hello, I have a population of 2000+ zoo time series (but my environment also contains objects that are not zoo time series). I'm trying to calculate the latest 90 days Z-Score of all zoo time series, using the following code: LZS<-function(ser) { temp<-window(ser,start=Sys.Date()-90) last((temp-mean(temp))/sd(temp)) } sapply(ls(), LZS ) The LZS function works on individual zoo time

Hi list, #Imagine we have vectors of different length (in practice 100 vectors): a<-c(1:10) b<-c(1:20) c<-c(1:30) #then we got a list of the names of those objects: list<-c("a","b","c") #I don't find how to boxplot them using a less stupid way than : boxplot(get(list[1]),get(list[2]),get(list[3])) Thanks for any advice ! --

Dear R users, I'd like to fit different longevities for males and females (µm vs µf below) with a Weibull interval-censored survival model. Unfortunately more than 50% of the individuals in the experiment died before their sex could be assigned. Hence I was wondering about fitting the proportion of females in the model ( "prop" below) with a likelihood function such as: