Tom, in your example below, x contains the kriging variance; points with
zero kriging variance must be observation locations, with the predicted
value equal to the observation.
In case the nugget in m would have been replaced by a measurement error
component (Err in m and m1 below), you would not have this effect, and
also have no discontinuity in the interpolated surface at observation
locations:
m = vgm(1, "Sph", 900, Err = 1)
v = variogram(log(zinc)~1, meuse)
m1 = fit.variogram(v, m)
m1
model psill range
1 Err 0.05066243 0.0000
2 Sph 0.59060780 897.0209
m2 = fit.variogram(v, vgm(1, "Sph", 900, 1))
m2
model psill range
1 Nug 0.05066243 0.0000
2 Sph 0.59060780 897.0209
as(krige(log(zinc)~1, meuse, meuse[1,], m2), "data.frame")
[using ordinary kriging]
x y var1.pred var1.var
1 181072 333611 6.929517 1.110223e-16
as(krige(log(zinc)~1, meuse, meuse[1,], m1), "data.frame")
[using ordinary kriging]
x y var1.pred var1.var
1 181072 333611 6.884401 0.03648868
# note identical predictions but different variances
# for other locations:
as(krige(log(zinc)~1, meuse, meuse.grid[1,], m2), "data.frame")
[using ordinary kriging]
x y var1.pred var1.var
1 181180 333740 6.499624 0.3198084
as(krige(log(zinc)~1, meuse, meuse.grid[1,], m1), "data.frame")
[using ordinary kriging]
x y var1.pred var1.var
1 181180 333740 6.499624 0.269146
On 11/27/2014 02:35 PM, Tomislav Hengl wrote:
Dear list,
I have a question about licensing the data that is produced by spatial
prediction from point data. My ideas is that a map produced by using
e.g. geostatistics from point data is a new data product and as such
does not falls under the regulations of the original license used for
the point data (so if the license for the point data is restrictive, the
license for the output maps does not have to respect this). Consider for
example:
R> library(gstat)
R> library(sp)
R> demo(meuse, echo=FALSE)
R> m <- vgm(.59, "Sph", 874, .04)
R> x <- krige(log(zinc)~1, meuse, meuse.grid, model = m)
The produced map "x" can be considered a new data product. There is
absolutely no way that one could reproduce the original input points
from this map, hence it should be considered "a non-derivative work".
Only if we would derive a map using interpolation technique that allows
re-constructions of points (e.g. Thiessen polygons) the license would
need to be respected.
Or am I mistaken? (I know this is a type of a question for lawyers in
fact, but any experience / opinion you have is welcome)
http://www.publicdomainsherpa.com/derivative-work.html
"To qualify as a derivative work, the derivative must use a substantial
amount of the prior work?s expression. How much? Enough so that the
average person would conclude that it had been based on or adapted from
the prior work"
thank you,