Values of Basis Functions or their Derivatives
A set of basis functions are evaluated at a vector of argument values. If a linear differential object is provided, the values are the result of applying the the operator to each basis function.
eval.basis(evalarg, basisobj, Lfdobj=0, returnMatrix=FALSE)
## S3 method for class 'basisfd'
predict(object, newdata=NULL, Lfdobj=0,
returnMatrix=FALSE, ...)evalarg, newdata |
a vector of argument values at which the basis functiona is to be evaluated. |
basisobj |
a basis object defining basis functions whose values are to be computed. |
Lfdobj |
either a nonnegative integer or a linear differential. operator object. |
object |
an object of class |
... |
optional arguments for |
returnMatrix |
logical: If TRUE, a two-dimensional is returned using a special class from the Matrix package. |
If a linear differential operator object is supplied, the basis must be such that the highest order derivative can be computed. If a B-spline basis is used, for example, its order must be one larger than the highest order of derivative required.
a matrix of basis function values with rows corresponding to argument values and columns to basis functions.
predict.basisfd is a convenience wrapper for
eval.basis.
Ramsay, James O., and Silverman, Bernard W. (2006), Functional Data Analysis, 2nd ed., Springer, New York.
Ramsay, James O., and Silverman, Bernard W. (2002), Applied Functional Data Analysis, Springer, New York
##
## 1. B-splines
##
# The simplest basis currently available:
# a single step function
bspl1.1 <- create.bspline.basis(norder=1, breaks=0:1)
eval.bspl1.1 <- eval.basis(seq(0, 1, .2), bspl1.1)
# check
eval.bspl1.1. <- matrix(rep(1, 6), 6,
dimnames=list(NULL, 'bspl') )
all.equal(eval.bspl1.1, eval.bspl1.1.)
# The second simplest basis:
# 2 step functions, [0, .5], [.5, 1]
bspl1.2 <- create.bspline.basis(norder=1, breaks=c(0,.5, 1))
eval.bspl1.2 <- eval.basis(seq(0, 1, .2), bspl1.2)
# Second order B-splines (degree 1: linear splines)
bspl2.3 <- create.bspline.basis(norder=2, breaks=c(0,.5, 1))
eval.bspl2.3 <- eval.basis(seq(0, 1, .1), bspl2.3)
# 3 bases: order 2 = degree 1 = linear
# (1) line from (0,1) down to (0.5, 0), 0 after
# (2) line from (0,0) up to (0.5, 1), then down to (1,0)
# (3) 0 to (0.5, 0) then up to (1,1).
##
## 2. Fourier
##
# The false Fourier series with 1 basis function
falseFourierBasis <- create.fourier.basis(nbasis=1)
eval.fFB <- eval.basis(seq(0, 1, .2), falseFourierBasis)
# Simplest real Fourier basis with 3 basis functions
fourier3 <- create.fourier.basis()
eval.fourier3 <- eval.basis(seq(0, 1, .2), fourier3)
# 3 basis functions on [0, 365]
fourier3.365 <- create.fourier.basis(c(0, 365))
eval.F3.365 <- eval.basis(day.5, fourier3.365)
matplot(eval.F3.365, type="l")
# The next simplest Fourier basis (5 basis functions)
fourier5 <- create.fourier.basis(nbasis=5)
eval.F5 <- eval.basis(seq(0, 1, .1), fourier5)
matplot(eval.F5, type="l")
# A more complicated example
dayrng <- c(0, 365)
nbasis <- 51
norder <- 6
weatherBasis <- create.fourier.basis(dayrng, nbasis)
basisMat <- eval.basis(day.5, weatherBasis)
matplot(basisMat[, 1:5], type="l")
##
## 3. predict.basisfd
##
basisMat. <- predict(weatherBasis, day.5)
all.equal(basisMat, basisMat.)
##
## 4. 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 <- seq(July4.1776, Apr30.1789, length.out=14)
AmRevBases <- predict(BspRevolution, AmRevYears)
matplot(AmRevYears, AmRevBases, type='b')
# Image is correct, but
# matplot does not recogize the Date class of x
# POSIXct
AmRev.ct <- as.POSIXct1970(c('1776-07-04', '1789-04-30'))
BspRev.ct <- create.bspline.basis(AmRev.ct)
AmRevYrs.ct <- seq(AmRev.ct[1], AmRev.ct[2], length.out=14)
AmRevBas.ct <- predict(BspRev.ct, AmRevYrs.ct)
matplot(AmRevYrs.ct, AmRevBas.ct, type='b')
# Image is correct, but
# matplot does not recognize the POSIXct class of xPlease choose more modern alternatives, such as Google Chrome or Mozilla Firefox.