similar to: Duda interpolación (package ' gstat ')

Sale plano sí. Ya se que sin tener los datos y el código es un poco difícil, pero es que mis datos ocupan mucho, es imposible. Seguiré mirando por internet. Muchas gracias Rubén. Un saludo, > To: r-help-es en r-project.org > From: rubenfcasal en gmail.com > Date: Thu, 6 Aug 2015 14:21:47 +0200 > Subject: Re: [R-es] Duda interpolación (package ' gstat ') > > Hola

Hi All, I am a PhD student in forestry science and I am working with LiDAR data set (huge data set). I am a brand-new in R and geostatistic (SORRY, my background it?s in forestry) but I wish improve my skill for improve myself. I wish to develop a methodology to processing a large data-set of points (typical in LiDAR) but there is a problem with memory. I had created a subsample data-base but

Hola, buen día a todos. He estado mirando cómo interpolar unos datos anuales con el método IDW, pero no consigo aún realizarlo. idw_models <- list() for (i in 1:ncol(data_y_mt)) { #Extract the current variable current_var <- names(data_y_mt)[i] #Create an IDW model for the current variable idw_model <- idw(current_var, sta_mt, grid, p = 1) idw_models[[i]] <- idw_model }

Hola, cordial saludo, estoy trabajando con geoestadística y tengo la siguiente inquietud ¿Existe algún criterio óptimo para determinar el número de bins en un semivariograma? sé que el paquete geoR tiende a tomar por defecto 13 bins y gstat 15 ¿pero existe algún criterio detrás de dicha elección? Gracias de antemano a quien pueda resolver mi duda. -- *Carlos A. Taimal Yepes* *Estudiante de

Hi all; I've already send a similar e-mail to the list and Prof. Brian Ripley answered me but my doubts remain unresolved. Thanks for the clarification, but perhaps I wasn't clear enough in posting my questions. I've got a postgres database which I read into R. The first column is Timestamp with timezone, and my data are already in UTC format. An 'printed' extract of R

Hi all, I've got some problems with irts objects, one of which could be a bug: 1) Read a table with several columns from Postgres and the first column is Timestamp with timezone (this is OK). An extract is: raincida$ts: [2039] "25/03/2000 22:00:00 UTC" "25/03/2000 23:00:00 UTC" [2041] "26/03/2000 00:00:00 UTC" "26/03/2000 01:00:00 UTC" [2043]

Emilio, muchas gracias. Tuve que rehacer el código para entender mejor todo, pero me sale el siguiente error: *Error in as.data.frame.default(x[[i]], optional = TRUE) : cannot coerce class ?structure("RasterLayer", package = "raster")? to a data.frame* El código es el siguiente: *for (i in 1:nrows_mt) { data_it_mt <-

Hi Gabor and everybody; Thanks Gabor, with the alternative step you've told me the problem is resolved. Comparing the two procedures: Extract from the source 'character' data: > rain$ts[2039:2046] [1] "25/03/2000 22:00:00 UTC" "25/03/2000 23:00:00 UTC" [3] "26/03/2000 00:00:00 UTC" "26/03/2000 01:00:00 UTC" [5] "26/03/2000 02:00:00

Dear R-helpers, I'm for sure not familiar with R, but it seem like a nice sofware tool, so I've decided to try using it. Here is my problem I just can't figure out: I'd like to do least square fit of a straight horizontal (a = 0) line y = ax + b through some data points x = (3,4,5,6,7,8) y = (0.62, 0.99, 0.83, 0.69, 0.76, 0.82) How would i find b????? All the best, Ked

Dear All, Is there anybody has the experience in using variogram(gstat) ? Please kindly give me some hints about the results. I used variogram() to build a semivariogram plot as: tr.var=variogram(Incr~1,loc=~X+Y,data=TRI2TU,width=5) then fir the variogram to get the parameters as: v.fit = fit.variogram(tr.var,vgm(0.5,"Exp",300,1)) v.fit model psill range 1 Nug 1.484879

All, This is a fresh, un-f#$ked-with OpenSUSE install after adding the Tumbleweed repository and doing a dup. I installed the Hypervisor with Tools for which it prompted me for Xen or QEMU and I chose Xen. No bridge was made. When I go to create a fully virtualised machine, it nao has the error: Error: Device 0 (vif) could not be connected. Could not find the bridge, and none was specified.

I have been trying to get wine up and running for some time. I work with the Debian Linux system with a KDE Desktop. Debian is supplying version 1.0.1 . I just installed the .deb file from the Wine website which is version 1.1.24. That is the highest version available for the Debian system. All of the versions that I have tried have the same thing wrong with the graphic interface. I will use

I can not answer this (message only to me). Certainly, some of you do.. Martin ------- Date: Wed, 18 Nov 1998 13:22:47 -0500 (EST) From: "J. Andy Royle" <jar@oriole.er.usgs.gov> To: Martin Maechler <maechler@stat.math.ethz.ch> Subject: Re: [R] loading fortran with Redhat 5.1 In-Reply-To: <199811181812.TAA26068@sophie.ethz.ch> Hi Martin, The reason I was

Hi, Is there any way that I can supress error messages so that they don't stop for loops running? I'm using the gstat package and have created a variogram model for the double exponential model: "dexp <- vgm( .5, "Exp", 10,.1, add.to = vgm(.5, "Exp", 100))" and fitted it using "dexp.fit <- fit.variogram(temp2, dexp, fit.method=6,

Dear Members, I am trying to fit a variogram model using fit.variogram function from the gstat package. The figure showing my experimental variogram can be seen here: https://i.stack.imgur.com/UZXw4.png My code line for this operation is: > c2.vgm.fit<-fit.variogram(c2.vgm.exp,vgm(nugget=0, psill=400,model="Exp",range =40000),fit.method = 7) The system throws following error

Dear all, I am trying to draw a weighted cumulative distribution (as defined here http://rss.acs.unt.edu/Rdoc/library/spatstat/html/ewcdf.html) with ggplot2 however the syntax temp<-qplot(X,weight=weight,data=data,stat = "ecdf", geom = "step",colour=factor(year)) seems not to produce exactly the right figure (the values seems higher at some points)... I am wrong in the

Dear All, I run following code to estimate the blocks using cokriging ( my data set has more than 50,000 data points). All the things run finely but Once I run the predict.gstat function it gave the error message - "memory.c", line 58: can't allocate memory in function m_get(). I run this code on LINUX sever but result is same. Would any one please be able to give a solution for

Hi! First of all: I'm a newbie to both statistics and R, so please be patient with me... I do however, like R because I've been programming (pascal, IDL, perl, C etc) and designing models since -92, but never related to statistics. Ok, here we go: I've got a set of 15 people, all of them observed over 10 weeks (10 analysed blood samples) with - let us kall it the A-value - influenced

Hi all! I've got a datstructure like this: subject week value 1 1 4 2 1 8 3 1 3 4 1 5 1 2 5 2 2 6 3 2 2 4 2 6 1 3 3 2 3 7 3 3 3 4 3 7 I'd like to plot the mean of 'value' against week. Is there a direct way of doing this or do I have to make a new structure with the calculated values and then plot it? All the best! -- ########################################### Kare

Hi all. If I run the simple regression when x is a categorical variable ( x <- factor(x) ): > MyFit <-coxph( Surv(start, stop, event) ~ x ) How can I get the overall p-value on x other than for each dummy variable? > anova(MyFit) does NOT provide that information as previously suggested on the list. All the best, Kare [[alternative HTML version deleted]]