Last chance! 50% off unlimited learning
Sale ends in
Evaluate a monotone functional data object at specified argument values, or evaluate a derivative of the functional object.
eval.monfd(evalarg, Wfdobj, Lfdobj=int2Lfd(0), returnMatrix=FALSE)
# S3 method for monfd
predict(object, newdata=NULL, Lfdobj=0, returnMatrix=FALSE, ...)
# S3 method for monfd
fitted(object, ...)
# S3 method for monfd
residuals(object, ...)a vector of argument values at which the functional data object is to be evaluated.
an object of class fd that defines the monotone function
to be evaluated. Only univariate functions are permitted.
a nonnegative integer specifying a derivative to be evaluated. At this time of writing, permissible derivative values are 0, 1, 2, or 3. A linear differential operator is not allowed.
an object of class monfd that defines the monotone function
to be evaluated. Only univariate functions are permitted.
logical: If TRUE, a two-dimensional is returned using a special class from the Matrix package.
optional arguments required by predict; not currently used.
a matrix containing the monotone function values. The first dimension
corresponds to the argument values in evalarg and the second to
replications.
A monotone function data object $h(t)$ is defined by $h(t) = [D^{-1}
exp Wfdobj](t)$. In this equation, the operator $D^{-1}$ means
taking the indefinite integral of the function to which it applies.
Note that this equation implies that the monotone function has a value
of zero at the lower limit of the arguments. To actually fit monotone
data, it will usually be necessary to estimate an intercept and a
regression coefficient to be applied to $h(t)$, usually with the least
squares regression function lsfit. The function Wfdobj
that defines the monotone function is usually estimated by monotone
smoothing function smooth.monotone.
eval.monfd only computes the standardized monotone form.
predict.monfd computes the scaled version using
with(object, beta[1] + beta[2]*eval.monfd(...)) if Lfdobj = 0
or beta[2]*eval.monfd(...) if Lfdobj > 0.
# NOT RUN {
# Estimate the acceleration functions for growth curves
# See the analyses of the growth data.
# Set up the ages of height measurements for Berkeley data
age <- c( seq(1, 2, 0.25), seq(3, 8, 1), seq(8.5, 18, 0.5))
# Range of observations
rng <- c(1,18)
# First set up a basis for monotone smooth
# We use b-spline basis functions of order 6
# Knots are positioned at the ages of observation.
norder <- 6
nage <- length(age)
nbasis <- nage + norder - 2
wbasis <- create.bspline.basis(rng, nbasis, norder, age)
# starting values for coefficient
cvec0 <- matrix(0,nbasis,1)
Wfd0 <- fd(cvec0, wbasis)
# set up functional parameter object
Lfdobj <- 3 # penalize curvature of acceleration
lambda <- 10^(-0.5) # smoothing parameter
growfdPar <- fdPar(Wfd0, Lfdobj, lambda)
# Smooth the data for the first girl
hgt1 <- growth$hgtf[,1]
# set conv = 0.1 and iterlim=1 to reduce the compute time
# required for this test on CRAN;
# We would not do this normally.
result <- smooth.monotone(age, hgt1, growfdPar, conv=0.1,
iterlim=1)
# Extract the functional data object and regression
# coefficients
Wfd <- result$Wfdobj
beta <- result$beta
# Evaluate the fitted height curve over a fine mesh
agefine <- seq(1,18,len=60)
hgtfine <- beta[1] + beta[2]*eval.monfd(agefine, Wfd)
# Plot the data and the curve
plot(age, hgt1, type="p")
lines(agefine, hgtfine)
# Evaluate the acceleration curve
accfine <- beta[2]*eval.monfd(agefine, Wfd, 2)
# Plot the acceleration curve
plot(agefine, accfine, type="l")
lines(c(1,18),c(0,0),lty=4)
##
## using predict.monfd
##
hgtfit <- with(result, beta[1]+beta[2]*eval.monfd(argvals, Wfdobj))
hgtfit. <- fitted(result)
# }
# NOT RUN {
all.equal(hgtfit, hgtfit.)
# }
# NOT RUN {
accfine. <- predict(result, agefine, Lfdobj=2)
# }
# NOT RUN {
all.equal(accfine, accfine.)
# }
# NOT RUN {
growthResid <- resid(result)
# }
# NOT RUN {
all.equal(growthResid, with(result, y-hgtfit.))
# }
# NOT RUN {
# }
Run the code above in your browser using DataLab