R: Bivariate Linear Interpolation

linearInterp {fBasics}

R Documentation

Bivariate Linear Interpolation

Description

Bivariate Linear Interpolation. Two options are available gridded and pointwise interpolation.

Usage

 
linearInterp(x, y = NULL, z = NULL, gridPoints = 21,
    xo = seq(min(x), max(x), length = gridPoints),
    yo = seq(min(y), max(y), length = gridPoints))
    
linearInterpp(x, y = NULL, z = NULL, xo, yo)

Arguments

`x, y, z`	for `linearInterp` the arguments `x` and `y` are two numeric vectors of grid pounts, and `z` is a numeric matrix or any other rectangular object which can be transformed by the function `as.matrix` into a matrix object. For `linearInterpp` we consider either three numeric vectors of equal length or if `y` and `z` are NULL, a list with entries `x`, `y`, `z`, or named data frame with `x` in the first, `y` in the second, and `z` in the third column.
`gridPoints`	an integer value specifying the number of grid points in `x` and `y` direction.
`xo, yo`	for `linearInterp` two numeric vectors of data points spanning the grid, and for `linearInterpp` two numeric vectors of data points building pairs for pointwise interpolation.

Value

linearInterp

returns a list with at least three entries, x, y and z. Note, that the returned values, can be directly used by the persp and contour 3D plotting methods.

linearInterpp

returns a data.frame with columns "x", "y", and "z".

Examples

 
## linearInterp -
   # Linear Interpolation:    
   if (require(akima)) {
     set.seed(1953)
     x = runif(999) - 0.5
     y = runif(999) - 0.5
     z = cos(2*pi*(x^2+y^2))
     ans = linearInterp(x, y, z, gridPoints = 41)
     persp(ans, theta = -40, phi = 30, col = "steelblue",
        xlab = "x", ylab = "y", zlab = "z")
     contour(ans)
   }