When to define new columns in krige in R? - r

Question

I am wondering: is it ever necessary to redefine your own columns when kriging? The error below seems to indicate this:
Warning: singular model in variogram fit
> sk1 <- krige(formula=Zs~1, locations=~Xs+Ys, data=sampled, newdata=pred.grid, model=fit.sph, beta=0)
Error in `[.data.frame`(object, , -coord.numbers, drop = FALSE) :
undefined columns selected
Is there a problem that I'm not seeing? Or, do I need to define my own columns? Thanks.
The following program is completely reproducable and runnable from here down:
library(gstat)
x <- seq(0,2000,by=20)
y <- seq(0,2000,by=20)
x = sample(x,10,replace=T)
y = sample(y,10,replace=T)
z = sample(0.532:3.7,10,replace=T)
samples = data.frame(x,y,z)
# detrend the samples:
print(mean(samples$z))
#create object of class gstat
h <- gstat(formula=z~1, locations=~x+y, data=samples)
samples.vgm <- variogram(h) # create method of class "gstatVariogram"
plot(samples.vgm,main='Variogram of Samples NOT detrended') # plot method for class "gstatVariogram"
# DETREND
z = samples$z
x = samples$x
y = samples$y
trend <- lm(z~x+y)
c = trend$coefficients[[1]]
a = trend$coefficients[[2]]
b = trend$coefficients[[3]]
#z_prime = z - (a*x + b*y +c)
# SUBTRACT THE PREDICTED LINE
Xs <- c()
Ys <- c()
Zs <- c()
print('started the loop')
for (i in 1:nrow(samples)){
i = samples[i,]
x=i$x
y=i$y
z=i$z
z_prime = z - (a*x+b*y+c)
Xs <- c(Xs,x)
Ys <- c(Ys,y)
Zs <- c(Zs,z_prime)
}
sampled <- data.frame(Xs=Xs,Ys=Ys,Zs=Zs)
print(sampled)
print('the length of sampled is')
print(length(sampled[[1]]))
# "result" is the new dataset with Z's detrended
# print(levelplot(Zs~Xs+Ys,sampled))
# define the domain or kriging estimation
x <- seq(0,2000,by=20)
y <- seq(0,2000,by=20)
# make data frame with prediction locations
pred.grid <- data.frame(x=rep(x,times=length(y)),y=rep(y,each=length(x)))
#create object of class gstat
g <- gstat(formula=Zs~1, locations=~Xs+Ys, data=sampled)
sampled.vgm <- variogram(g) # create method of class "gstatVariogram"
plot(sampled.vgm,main='Variogram of Samples hopefully detrended') # plot method for class "gstatVariogram"
vg.sph <- vgm(psill=1.0,model='Sph', range = 500)
fit.sph <- fit.variogram(sampled.vgm, model = vg.sph)
sk1 <- krige(formula=Zs~1, locations=~Xs+Ys, data=sampled, newdata=pred.grid, model=fit.sph, beta=0)

Answer 1

Add library(gstat) to the top of your code, then it's reproducible.
To answer your question directly, the reason you receive an undefined columns selected error is because your newdata does not have the correct column names. The column names need to match the data column names, which are Xs and Ys in this case. Redefine the pred.grid to have columns Xs and Ys to solve your problem. I tested, and your code runs.
pred.grid <- data.frame(Xs=rep(x,times=length(y)),Ys=rep(y,each=length(x)))
As for other comments: Warning: singular model in variogram fit is a result of not being able to fit a model based on the sample semivariogram data. If you take a look at the plot of your data (below), it's very clear that no empirical function will be able to fit this. In your case, it is because you have only one point per bin (11 points total) so there's really not enough data to fit the semivariogram. Even reducing the number of bins, there still would not be enough data support to fit an empirical semivariogram.
Changing your number of samples to 500,
x = sample(x,500,replace=T)
y = sample(y,500,replace=T)
z = sample(0.532:3.7,500,replace=T)
it becomes very clear that the data you are generating are uncorrelated such that samples closer to one another in x-y space are not more similar than samples farther away (pure nugget semivariogram). Is this what you wanted?