similar to: geoR, plot of variog4 lines incomplete

Mac OS 10.3.5, R 2.0.0 latest version of geoR I have an incomplete 5 x 20 spatial array of samples (60 out of 100 possible locations) for which I would like to calculate directional variograms using variog4(). Unfortunately, I can't get it to return all 4 directions. It returns variograms for 45, 90, and 135 degrees, omitting 0 degrees (pi/4, pi/2, 3pi/4, omitting 0). If I specify 0

Hello, I am attempting to get the standard deviation in multiple distance bins in my spatial data. It appears as though the 'variog' command in the geoR package will do the trick, as one of the outputs from 'variog' is 'variog$sd', which, according to the manual, is the "standard deviation of the values in each bin". However, when I run this command, the

Hi, I'm a new R user. My goal is to do a variogram using geoR. I started by trying to do the example in the geoR Illustrative Session using my own data. I am able to read in my Ascii data using: D <- matrix(scan("file.dat", n=530*3), 530,3, byrow=TRUE). Then I use: as.geodata(D, coords.col=1:2, data.col=3) to make the object D geodata. I check the descriptive statistics,

Hi, I'm writing some R code that requires a massive amount of looping and would ideally like to write it so that it avoid the use of "for" loop ... however I'm having some trouble. Very briefly, the basic idea is to implement a binary partitioning algorithm to determine the optimal cutpoint based on deviance measures obtained from likelihood estiamtes. This is in the

Please see the code below. When I try to run the variogram - vg.deft<-variog(rd,uvec=10) I keep getting this error- variog: computing omnidirectional variogram Error in vector("double", length) : vector size specified is too large Also, when I try to define distance-based neighborhood - nb.tr=dist.neighbors(tr.locs,2) I get this error - Error in vector("double", length) :

Dear GeoR-er, If I use the variog function in the latest release of geoR, the first lag is always ignored. For instance, if you read in geodata, calculates the variogram using the variog function and give in a uvec like uvec=seq(0,max,by=2.44), it only starts giving results from distance=4.88 and ignores 2.44! This wasn't the case in former versions of geoR. Is this done on purpose? Why?

ComRades: I am geeting the error message Error in ldots[[which(MET)]] : attempt to select less than one element when I try to fit the geostatistical model with the likfit() function of geoR. I have tried with old data for which likfit() successfully maximised the likelihood in previous versions of geoR, and yet the current version fails. I have tried in Windows Vista and Windows XP (I haven't

Hi all, I use kriging to interpolate the precipitation from stations, but the map of this results show lots of stripes. (please see the attachment)I think there's something wrong with the setting of the dimension of this matrix, however, I have no idea how to know or test to see if this setting is correct or not.I've tried to switch the latitude and longitude, but still got the same

Hello All! I calculate a variogram using the function variog (package geoR) afterwards I use variofit to fit a spherical model (see code below). Now I just changed the units of the variable (in this case MPa to kPa just a factor of 1000). If I do so, I get a different fit and therefore different ranges etc. Why? The semi-variance is of course 6 orders of magnitude higher but the values

I wonder what is the default value for the argument 'cutoff' when not specified in the variogram.formula function of gstat. Computing variogram envelops within gstat, I am comparing the results obtained with variog in geoR and variogram in gstat, and it took me a while before understanding that the cutoff default value is not the maximum distance. Can Edzer tell us about it? All the

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Hi R-masters! I trying model Heart disease mortality in my country with a lme model like this: m1.lme<-lme(log(rdeath)~age*year,random=~age|year,data=dados) where: rdeath is rate of mortality per 100000 person per age and year age: age of death (22 27 32 37 42 47 52 57 62 67 72 77 82) year: year of death (1980:2002) I don?t have problem to fit the model, but in

Hello, my question is about the data handling. I have a data set that is lined as: 4 1 17 1 1 -5.1536 -0.1668 -2.3412 -0.5062 0.9621 0.3640 0.3678 -0.5081 -0.2227 0.8142 -0.0389 -0.0445 -0.0578 -0.1175 -0.1232 0.8673 -0.1033 -0.0796 -0.0341 -0.1716 -0.1801 -0.7014 0.6578 0.5611 4 1 17 2 1 -5.1536 -0.1668 -2.3412 -0.5062 0.9621 0.3640 0.3678 -0.5081 -0.2227 0.8142 -0.0389 -0.0445

Hello everybody! I have an ARIMA model for a time series. This model was obtained through an auto.arima function. The resulting model is a ARIMA(2,1,4)(2,0,1)[12] with drift (my time series has monthly data). Then I perform a 12-step ahead forecast to the cited model... so far so good... but when I look the plot of my forecast I see that the result is really far from the behavior of my time

Dear Aaron, I am really a tool user and not a tool maker (actually an ecologist doing some biostatistics)... so, I take the liberty of sending a copy of this e-mail to the r-help list where capable computer persons and true statisticians may provide more relevant information and also to Paulo Ribeiro and Peter Diggle, the authors of geoR.. I really feel that your huge matrix cannot be

I wonder if someone could explain why, when I perform confirmatory factor-analysis model using polychoric correlations why I do not get an estimated confidence interval for the RMSEA. My experience with these type models is that I would obtain a confidence interval estimate. I did not get any warning messages with the output. RESULTS: Model Chisquare = 1374 Df = 185 Pr(>Chisq) = 0

TOPIC My question regards the philosophy behind how R implements corrections to chi-square statistical tests. At least in recent versions (I'm using 2.13.1 (2011-07-08) on OSX 10.6.8.), the chisq.test function applies the Yates continuity correction for 2 by 2 contingency tables. But when used as a goodness of fit test (GoF, aka likelihood ratio test), chisq.test does not appear to implement

I have what R seems to consider a very large dataset, a 12MB text file of lat,long,and height values, 130,000 rows to be exact. Here's what I get: Thomas Colson North Carolina State University Department of Forestry and Environmental Resources (919) 673 8023 tom_colson at ncsu.edu Calendar: www4.ncsu.edu/~tpcolson

dear list I am student M.S. statistics in department statistics . I am working in the function "nls" in the [R 2.3.1] with 246 data and want to fit the "exp" model to vectors( v and u ) but I have a problem to use it u 5.000000e-13 2.179057e+03 6.537171e+03 1.089529e+04 1.525340e+04 1.961151e+04 2.396963e+04 2.832774e+04 3.268586e+04 3.704397e+04 4.140209e+04

Um texto embutido e sem conjunto de caracteres especificado associado... Nome: n?o dispon?vel Url: https://stat.ethz.ch/pipermail/r-help/attachments/20060628/2ff7714d/attachment.pl