Description Usage Arguments Details Value Author(s) See Also Examples
Evaluate a functional data object at specified argument values, or evaluate a derivative or the result of applying a linear differential operator to the functional object.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 | eval.fd(evalarg, fdobj, Lfdobj=0, returnMatrix=FALSE)
## S3 method for class 'fd'
predict(object, newdata=NULL, Lfdobj=0, returnMatrix=FALSE,
...)
## S3 method for class 'fdPar'
predict(object, newdata=NULL, Lfdobj=0,
returnMatrix=FALSE, ...)
## S3 method for class 'fdSmooth'
predict(object, newdata=NULL, Lfdobj=0,
returnMatrix=FALSE, ...)
## S3 method for class 'fdSmooth'
fitted(object, returnMatrix=FALSE, ...)
## S3 method for class 'fdSmooth'
residuals(object, returnMatrix=FALSE, ...)
|
evalarg, newdata |
a vector or matrix of argument values at which the functional data object is to be evaluated. If a matrix with more than one column, the number of columns must match ncol(dfobj[['coefs']]). |
fdobj |
a functional data object to be evaluated. |
Lfdobj |
either a nonnegative integer or a linear differential operator object. If present, the derivative or the value of applying the operator is evaluated rather than the functions themselves. |
object |
an object of class |
returnMatrix |
logical: Should a 2-dimensional array to be returned using a special class from the Matrix package if appropriate? |
... |
optional arguments for |
eval.fd
evaluates Lfdobj
of fdobj
at
evalarg
.
predict.fd
is a convenience wrapper for
eval.fd
. If newdata
is NULL and
fdobj[['basis']][['type']]
is bspline
, newdata
=
unique(knots(fdojb,interior=FALSE))
; otherwise, newdata
= fdobj[['basis']][['rangeval']]
.
predict.fdSmooth
, fitted.fdSmooth
and
residuals.fdSmooth
are other wrappers for eval.fd
.
an array of 2 or 3 dimensions containing the function
values. The first dimension corresponds to the argument values in
evalarg
,
the second to replications, and the third if present to functions.
Soren Hosgaard wrote an initial version of predict.fdSmooth
,
fitted.fdSmooth
, and residuals.fdSmooth
.
getbasismatrix
,
eval.bifd
,
eval.penalty
,
eval.monfd
,
eval.posfd
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 | ##
## eval.fd
##
# set up the fourier basis
daybasis <- create.fourier.basis(c(0, 365), nbasis=65)
# Make temperature fd object
# Temperature data are in 12 by 365 matrix tempav
# See analyses of weather data.
# Set up sampling points at mid days
# Convert the data to a functional data object
tempfd <- smooth.basis(day.5, CanadianWeather$dailyAv[,,"Temperature.C"],
daybasis)$fd
# set up the harmonic acceleration operator
Lbasis <- create.constant.basis(c(0, 365))
Lcoef <- matrix(c(0,(2*pi/365)^2,0),1,3)
bfdobj <- fd(Lcoef,Lbasis)
bwtlist <- fd2list(bfdobj)
harmaccelLfd <- Lfd(3, bwtlist)
# evaluate the value of the harmonic acceleration
# operator at the sampling points
Ltempmat <- eval.fd(day.5, tempfd, harmaccelLfd)
# Confirm that it still works with
# evalarg = a matrix with only one column
# when fdobj[['coefs']] is a matrix with multiple columns
Ltempmat. <- eval.fd(matrix(day.5, ncol=1), tempfd, harmaccelLfd)
# confirm that the two answers are the same
all.equal(Ltempmat, Ltempmat.)
# Plot the values of this operator
matplot(day.5, Ltempmat, type="l")
##
## predict.fd
##
predict(tempfd) # end points only at 35 locations
str(predict(tempfd, day.5)) # 365 x 35 matrix
str(predict(tempfd, day.5, harmaccelLfd))
# cublic splie with knots at 0, .5, 1
bspl3 <- create.bspline.basis(c(0, .5, 1))
plot(bspl3) # 5 bases
fd.bspl3 <- fd(c(0, 0, 1, 0, 0), bspl3)
pred3 <- predict(fd.bspl3)
pred3. <- matrix(c(0, .5, 0), 3)
dimnames(pred3.) <- list(NULL, 'reps 1')
all.equal(pred3, pred3.)
pred.2 <- predict(fd.bspl3, c(.2, .8))
pred.2. <- matrix(.176, 2, 1)
dimnames(pred.2.) <- list(NULL, 'reps 1')
all.equal(pred.2, pred.2.)
##
## predict.fdSmooth
##
lipSm9 <- smooth.basisPar(liptime, lip, lambda=1e-9)$fd
plot(lipSm9)
##
## with evalarg of class 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)
(AmRevLinear <- as.numeric(AmRevYears-July4.1776))
fitLin <- smooth.basis(AmRevYears, AmRevLinear, BspRevolution)
AmPred <- predict(fitLin, AmRevYears)
# 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)
(AmRevLin.ct <- as.numeric(AmRevYrs.ct-AmRev.ct[2]))
fitLin.ct <- smooth.basis(AmRevYrs.ct, AmRevLin.ct, BspRev.ct)
AmPred.ct <- predict(fitLin.ct, AmRevYrs.ct)
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