6

Some has already been said by Spacedman in the comments. This warning may not pose a problem, if the variogram looks good. A good option might be to initialize some of the variogram parameters in vgm("Sph"). I usually take these values as default: vario <- variogram(Temperature ~1, WU_data_spatial) vario.fit <- fit.variogram(vario, vgm(psill=max(...


4

If you make your variogram go a bit further you can start to see some structure: > pr.v <- variogram(prodp ~1, tzprice1_v, cutoff=1500, width=1500/8) > plot(pr.v) You can then fit a variogram if you use a different fit weighting method: > fv = fit.variogram(pr.v, vgm(psill=.05,"Exp",range=80,nugget=.15),fit.method=1) > plot(pr.v, fv) Or ...


4

I am wondering if someone can explain why the nugget seems so high? Does this mean that even at similar locations, there is still a relatively high degree of difference? Yes, a high nugget effect (high semivariance at origin) tells there is a weak (or none) spatial dependence (autocorrelation) among sample data at small distances. It could be the case the ...


4

Maybe I should begin by stating that "the summit issue is reflecting how IDW models the surface, rather than weighting". IDW sees the predicted surface as an averaging model, while Spline tries to minimize abrupt change to make 'smooth rubber sheet' and Kriging tries to minimize errors. (I hope this makes my point clear). Let me focus on the difference ...


3

The text size of bottom-right annotations is hard-coded in autokrige.vgm.panel.r, which runs under the hood of automap's plot() method. The latter function is also responsible for creating point labels that are passed to gstat::vgm.panel.xyplot(), which is hard-coded as well. Now, you might get rid of these point labels through something like ## sample data ...


3

This example comes from the book Statistics and Data Analysis in Geology (3rd ed. 2002, by John C. Davis) pp. 422, Figure 5-95: and it is specified that: We will assume that a prior structural analysis bas produced the experimental semivartogram and model shown in Figure 5-96; the model is linear with a slope of 4.0 m2 /km within a neighborhood of 20 km. ...


3

Yes, this is correct. When you print the model by typing model.vari you'll see sill values, split up in a nugget component (the offset) and the exponential component. The sum of these two is usually indicated by "the sill value" (i.e., around 25).


3

Finally, I found this template in SAGA GUI (2.0.1), in the interactive mode of Ordinary Kriging (global). So a Spherical model can be like: And I got this; [Note] If anybody kindly volunteer to test this workflow, this sample point data and parameters are reproduced by R commands: library(sp) data(meuse, package= "sp") coordinates(meuse)= ~x + y ...


2

You would calculate it exactly as you would any variogram: by estimating the values of (Z(i) - Z(j))^2/2 as a function of the distance between data locations i and j. You have many choices of distance, but the natural ones would be either distance along the routes or travel time along the routes. If the routes are one-way, additional techniques borrowed ...


2

Yes, this is true. Your professor is right. Look for function fn_exponential in the source code if you want to be sure.


2

It depends on the fit.method chosen in the function fit.variogram. If you use: fit.method = 0, no fit fit.method = 1, weights are equal to Nj fit.method = 2, weights are equal to Nj / {γ(hj)}2 fit.method = 5, REML (restricted maximum likelyhood) fit.method = 6, no weights (OLS = Ordinary Least Squares) fit.method = 7, Nj / hj2 Please see this: Table 4.2: ...


2

Spatial process can be introduced by the underlying data that went into a probabilistic estimate, the model itself and the grain of the raster. You also have the addition of potentially of introduced nonstationarity in the random field. How do you plan on teasing out these confounding factors in understanding the anisotropy of the data? What exactly is the ...


2

You should remove your NA in your data frame first using the function na.omit (see this: remove-rows-with-nas-in-data-frame) and then apply the variogram function. # Load libraries library('gstat') # Example data newdf <- data.frame("id" = 1:10, "yield_cleaned" = runif(n = 10, min = 0, max = 1), "x" = sample(x = ...


2

There's massive literature for sampling schemes for spatial interpolation. Here's some thoughts: On 1: yes they will generate different empirical variograms and different kriged maps. But if the underlying data is independent of your sampling scheme then the outputs should be in agreement within standard errors. But if your data are somehow related to your ...


1

You will find all you need in the excellent (and didactic) technical note from Rossiter (2012)*: Technical Note: Co-kriging with the gstat package of the R environment for statistical computing. Co-kriging will use different functions from those with univariate kriging (for example, ordinary kriging). The datasets (target and co-variables) should ...


1

I would go for the automap package: library(automap) fit.vgm = autofitVariogram(prodp ~1, tzprice1_v, model = "Exp") plot(pr.v, fit.vgm$var_model) On the other hand though, I'm not a statistician - I work with variograms more empirically which I guess I shouldn't recommend. I am not able to answer your question about possible ways of estimating variogram ...


1

Is it true that we're just using a different set of weight values than the 1/distance in IDW? Yes, both IDW interpolation and Ordinary Kriging (OK) will calculate weights based on distance, but with different criteria. In both methods, weights do not depend on sample values. The answer from Dahn Jahn in Ordinary kriging example step by step? is very ...


1

You can find the equation of the Spherical model for example on Wikipedia: I think the last option from the drop-down menu is what you're looking for.


1

Interesting method to use; have you considered running the model with equal numbers of pseudo absences and then increasing or decreasing the proportion of known absences to pseudo absences to see if a trend appears in how it effects the model output? This approach may also give you insight into the uncertainty of the model as well.


1

I found a very simple solution for obtaing the fitted gamma values within gstat package. It's the variogramLine() function. A simple code is attached here. # Empirical variogram ev = variogram("pH", data=data,....) fv = fit.variogram(q, "Sph", ....) fitted=variogramLine(v.fit, maxdist=max(q$dist), dist_vector=q$dist) fitted # see what are the values of ...


1

I couldn't answer your question using gstat package. However, you can also use geoR package to fit a variogram model to an empirical variogram and analyse fitted and residuals values. I give you a reproducible example below: # Load libraries ---------------------------------------------------------- library("sp") library("geoR") # Load data --------------...


1

From the documentation: fit.variogram.reml package:gstat R Documentation REML Fit Direct Variogram Partial Sills to Data Description: Fit Variogram Sills to Data, using REML (only for direct variograms; not for cross variograms) You've set range to NA, and this only fits sills, not ranges. Use fit.variogram or fit....


1

I think the goodness of fit measure you are looking for is a chi-squared. For which (I believe) your 'minimum weighted sum of squares' IS your goodness of fit statistic. Assuming this or this is what you are doing. Reading about the packages on CRAN can sometimes be helpful. This thread will also be relevant. Or you could google: "calculate R2 weighted sum ...


1

The semivariogram is a function of difference in DEM values over distance. The x-axis is distance, the y-axis is something like average height difference for two random points that distance apart. Close together points have a small difference on average, far apart points have a large difference on average. So the semivariogram doesn't level off at a value ...


Only top voted, non community-wiki answers of a minimum length are eligible