library(sp)
data(meuse)
coordinates(meuse)<-~x+y
v<-variogram(log(zinc)~1,meuse)
v
np dist gamma dir.hor dir.ver id
1 57 79.29244 0.1234479 0 0 var1
2 299 163.97367 0.2162185 0 0 var1
3 419 267.36483 0.3027859 0 0 var1
4 457 372.73542 0.4121448 0 0 var1
.
.
v1<-variogram(log(zinc)~x+y,meuse)
v1
np dist gamma dir.hor dir.ver id
1 57 79.29244 0.1060834 0 0 var1
2 299 163.97367 0.1829983 0 0 var1
3 419 267.36483 0.2264256 0 0 var1
4 457 372.73542 0.2847192 0 0 var1
.
.
From above following code and output I can found that log(zinc)~1 and
log(zinc)~x+y returns the different gamma value for v and v1. What's the basic difference between these two operation?
help(variogram) told me that "formula defining the response vector and (possible) regressors, in case of absence of regressors, use e.g. z~1;".But,I didn't understand clearly these sentence! could anyone please tell me in details that when should I use z~1 or when should I use z~LON+LAT?
The first argument to variogram defines the model for the mean structure. in case of a constant mean, the model contains only an intercept, hence log(zinc)~1. If the mean is modelled as a linear regression model in x and y, then use log(zinc)~x+y. In that case, ordinary least squares residuals are computed as the basis for the semivariogram values, instead of measured log(zinc) values.