Extrapolation beyond a grid of data

Question

I have a grid of (x, y, z) values and I want a function that can approximate z-values when it is given (x,y) points that lie beyond the grid.

I have tried to solve the question using the Akima package (code block 3), but I can't seem to get the interp function to work with the linear=FALSE option that is required to extrapolate beyond the grid.

Data:

# Grid data
x <- seq(0,1,length.out=6)
y <- seq(0,1,length.out=6)
z <- outer(x,y,function(x,y){sqrt(x^2+y^3)})

Visualize data (not essential for question):

## Visualize the data - Not important for question ##
  jet.colors <- colorRampPalette( c("Royal Blue", "Lime Green") )
  nbcol <- 100
  color <- jet.colors(nbcol)
  nrz <- nrow(z)
  ncz <- ncol(z)
  zfacet <- z[-1, -1] + z[-1, -ncz] + z[-nrz, -1] + z[-nrz, -ncz]
  facetcol <- cut(zfacet, nbcol)
  pmat <- persp(x, y, z, d = 1,r = 4,
                ticktype="detailed", 
                col = color[facetcol], 
                main = "Title",
                xlab="x value", 
                ylab = "y value", 
                zlab= "z value",
                scale=FALSE,
                expand=0.6, 
                theta=-40, 
                phi=25)
## End visualization ##

My attempt at solving the question using Akima package

library(akima)

# Vectorize the grid:
zz <- as.vector(z)
# create all combinations of x and y
xy <- expand.grid(x,y)

# What we want:
sqrt(0.7^2 + 0.7^3) # c(0.7, 0.7) = 0.9126883
sqrt(0.7^2 + 1.2^3) # c(0.7, 1.2) = 1.489295

# We get a result for the first point inside the grid,
# but not for the second one outside the grid.
# This is expected behaviour when linear=TRUE:
interp(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), linear=TRUE)
# = (0.929506, NA)

# When LINEAR = FALSE we get z= 0, 0!!
interp(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), linear=FALSE, extrap = TRUE)
# = (0, 0)

# Dropping extrap=TRUE we see that both are actually NA in this case
interp(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), linear=FALSE)
# = (NA, NA)
# What is going on?

Using R 3.1.3 and akima_0.5-11.

Answer 1

Got a result by using interp.old() directly:

interp.old(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), ncp=2, extrap=TRUE)
# = (0.9211469, 1.472434)

I am not sure if this is a bug in the interp() function. interp() is a wrapper that calls interp.old() if linear=TRUE and interp.new() if linear=FALSE. The interp.new() function seems to fail.

Note that the use of the ncp parameter is deprecated. From the manual:

ncp - deprecated, use parameter linear. Now only used by interp.old(). meaning was: number of additional points to be used in computing partial derivatives at each data point. ncp must be either 0 (partial derivatives are not used), or at least 2 but smaller than the number of data points (and smaller than 25)