Values of Basis Functions or their Derivatives
Evaluate a set of basis functions or their derivatives at a set of argument values.
getbasismatrix(evalarg, basisobj, nderiv=0, returnMatrix=FALSE)
evalarg |
a vector of arguments values. |
basisobj |
a basis object. |
nderiv |
a nonnegative integer specifying the derivative to be evaluated. |
returnMatrix |
logical: If TRUE, a two-dimensional is returned using a special class from the Matrix package. |
a matrix of basis function or derivative values. Rows correspond to argument values and columns to basis functions.
##
## Minimal example: a B-spline of order 1, i.e., a step function
## with 0 interior knots:
##
bspl1.1 <- create.bspline.basis(norder=1, breaks=0:1)
m <- getbasismatrix(seq(0, 1, .2), bspl1.1)
# check
m. <- matrix(rep(1, 6), 6,
dimnames=list(NULL, 'bspl') )
all.equal(m, m.)
##
## Date and POSIXct
##
# Date
July4.1776 <- as.Date('1776-07-04')
Apr30.1789 <- as.Date('1789-04-30')
AmRev <- c(July4.1776, Apr30.1789)
BspRevolution <- create.bspline.basis(AmRev)
AmRevYears <- as.numeric(seq(July4.1776, Apr30.1789, length.out=14))
AmRevMatrix <- getbasismatrix(AmRevYears, BspRevolution)
matplot(AmRevYears, AmRevMatrix, type='b')
# POSIXct
AmRev.ct <- as.POSIXct1970(c('1776-07-04', '1789-04-30'))
BspRev.ct <- create.bspline.basis(AmRev.ct)
AmRevYrs.ct <- as.numeric(seq(AmRev.ct[1], AmRev.ct[2], length.out=14))
AmRevMat.ct <- getbasismatrix(AmRevYrs.ct, BspRev.ct)
matplot(AmRevYrs.ct, AmRevMat.ct, type='b')Please choose more modern alternatives, such as Google Chrome or Mozilla Firefox.