predict.Krig: Evaluation of Krig spatial process estimate.

Description

Provides predictions from the Krig spatial process estimate at arbitrary points, new data (Y) or other values of the smoothing parameter (lambda) including a GCV estimate.

Usage

## S3 method for class 'Krig':
predict(
object, x = NULL, Z = NULL, drop.Z = FALSE, just.fixed
                 = FALSE, lambda = NA, df = NA, model = NA,
                 eval.correlation.model = TRUE, y = NULL, yM = NULL,
                 verbose = FALSE, ...)
## S3 method for class 'Krig':
predict.derivative(object, x = NULL,  verbose = FALSE,...)

Arguments

Value

Vector of predicted responses or a matrix of the partial derivatives.

Details

The main goal in this function is to reuse the Krig object to rapidly evaluate different estimates. Thus there is flexibility in changing the value of lambda and also the independent data without having to recompute the matrices associated with the Krig object. The reason this is possible is that most on the calculations depend on the observed locations not on lambda or the observed data. Note the version for evaluating partial derivatives does not provide the same flexibility as predict.Krig and makes some assumptions about the null model (as a low order polynomial) and can not handle the correlation model form.

Examples

Run this code

Krig(ozone$x,ozone$y, theta=50) ->fit
  predict( fit) # gives predicted values at data points should agree with fitted.values
                #  in fit object 

# predict at the coordinate (-5,10)
  x0<- cbind( -5,10) # has to be a  1X2 matrix
  predict( fit,x= x0)

# redoing predictions at data locations:
   predict( fit, x=ozone$x)

# only the fixed part of the model
  predict( fit, just.fixed=TRUE) 

# evaluating estimate at a grid of points 
  grid<- make.surface.grid( list( seq( -40,40,,15), seq( -40,40,,15)))
  look<- predict(fit,grid) # evaluate on a grid of points

# some useful graphing functions for these gridded predicted values
  out.p<- as.surface( grid, look) # reformat into $x $y $z image-type object
  contour( out.p) 

# see also the functions predict.surface and surface 
# for functions that combine these steps 
   

# refit with 10 degrees of freedom in surface
  look<- predict(fit,grid, df=15)
# refit with random data 
  look<- predict( fit, grid, y= rnorm( 20))


# finding partial derivatives of the estimate
#
# find the partial derivatives at observation locations
# returned object is a two column matrix. 
# this does not make sense for the exponential covariance
# but can illustrate this with a thin plate spline with
# a high enough order ( i.e. need m=3 or greater)
# 
  data(ozone2)
# the 16th day of this ozone spatial dataset
  fit0<- Tps( ozone2$lon.lat, ozone2$y[16,], m=3)
  look1<- predict.derivative.Krig( fit0)
# for extra credit compare this to
  look2<- predict.derivative.Krig( fit0, x=ozone2$lon.lat)  
# (why are there more values in look2)