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oldpar <- par(no.readonly=TRUE)
###
###
###   vector response with functional explanatory variable  
###
###

#  data are in Canadian Weather object
#  print the names of the data
print(names(CanadianWeather))
#  set up log10 of annual precipitation for 35 weather stations
annualprec <- 
  log10(apply(CanadianWeather$dailyAv[,,"Precipitation.mm"], 2,sum))
# The simplest 'fRegress' call is singular with more bases
# than observations, so we use only 25 basis functions, for this example
smallbasis  <- create.fourier.basis(c(0, 365), 25)
# The covariate is the temperature curve for each station.
tempfd <- 
  smooth.basis(day.5, CanadianWeather$dailyAv[,,"Temperature.C"], smallbasis)$fd
##
## formula interface:  specify the model by a formula, the method
## fRegress.formula automatically sets up the regression coefficient functions,
## a constant function for the intercept, 
## and a higher dimensional function
## for the inner product with temperature
##

precip.Temp1 <- fRegress(annualprec ~ tempfd, method="fRegress")

#  the output is a list with class name fRegress, display names
names(precip.Temp1)
#[c1] "yvec"           "xfdlist"        "betalist"       "betaestlist"    "yhatfdobj"     
# [6] "Cmat"           "Dmat"           "Cmatinv"        "wt"             "df"            
#[11] "GCV"            "OCV"            "y2cMap"         "SigmaE"         "betastderrlist"
#[16] "bvar"           "c2bMap"       

#  the vector of fits to the data is object  precip.Temp1$yfdPar,
#  but since the dependent variable is a vector, so is the fit
annualprec.fit1 <- precip.Temp1$yhatfdobj
#  plot the data and the fit
plot(annualprec.fit1, annualprec, type="p", pch="o")
lines(annualprec.fit1, annualprec.fit1, lty=2)
#  print root mean squared error
RMSE <- round(sqrt(mean((annualprec-annualprec.fit1)^2)),3)
print(paste("RMSE =",RMSE))
#  plot the estimated regression function
plot(precip.Temp1$betaestlist[[2]])
#  This isn't helpful either, the coefficient function is too
#  complicated to interpret.
#  display the number of basis functions used:
print(precip.Temp1$betaestlist[[2]]$fd$basis$nbasis)
#  25 basis functions to fit 35 values, no wonder we over-fit the data

##
## Get the default setup and modify it
## the "model" value of the method argument causes the analysis
## to produce a list vector of arguments for calling the
## fRegress function
##

precip.Temp.mdl1 <- fRegress(annualprec ~ tempfd, method="model")
# First confirm we get the same answer as above by calling
# function fRegress() with these arguments:
precip.Temp.m <- do.call('fRegress', precip.Temp.mdl1)

all.equal(precip.Temp.m, precip.Temp1)


#  set up a smaller basis for beta2 than for temperature so that we
#  get a more parsimonious fit to the data

nbetabasis2 <- 21  #  not much less, but we add some roughness penalization
betabasis2  <- create.fourier.basis(c(0, 365), nbetabasis2)
betafd2     <- fd(rep(0, nbetabasis2), betabasis2)
# add smoothing
betafdPar2  <- fdPar(betafd2, lambda=10)

# replace the regress coefficient function with this fdPar object

precip.Temp.mdl2 <- precip.Temp.mdl1
precip.Temp.mdl2[['betalist']][['tempfd']] <- betafdPar2

# Now do re-fit the data

precip.Temp2 <- do.call('fRegress', precip.Temp.mdl2)

# Compare the two fits:
#  degrees of freedom
precip.Temp1[['df']] # 26
precip.Temp2[['df']] # 22
#  root-mean-squared errors:
RMSE1 <- round(sqrt(mean(with(precip.Temp1, (yhatfdobj-yvec)^2))),3)
RMSE2 <- round(sqrt(mean(with(precip.Temp2, (yhatfdobj-yvec)^2))),3)
print(c(RMSE1, RMSE2))
#  display further results for the more parsimonious model
annualprec.fit2 <- precip.Temp2$yhatfdobj
plot(annualprec.fit2, annualprec, type="p", pch="o")
lines(annualprec.fit2, annualprec.fit2, lty=2)
#  plot the estimated regression function
plot(precip.Temp2$betaestlist[[2]])
#  now we see that it is primarily the temperatures in the
#  early winter that provide the fit to log precipitation by temperature

##
## Manual construction of xfdlist and betalist
##

xfdlist <- list(const=rep(1, 35), tempfd=tempfd)

# The intercept must be constant for a scalar response
betabasis1 <- create.constant.basis(c(0, 365))
betafd1    <- fd(0, betabasis1)
betafdPar1 <- fdPar(betafd1)

betafd2     <- fd(matrix(0,7,1), create.bspline.basis(c(0, 365),7))
# convert to an fdPar object
betafdPar2  <- fdPar(betafd2)

betalist <- list(const=betafdPar1, tempfd=betafdPar2)

precip.Temp3   <- fRegress(annualprec, xfdlist, betalist)
annualprec.fit3 <- precip.Temp3$yhatfdobj
#  plot the data and the fit
plot(annualprec.fit3, annualprec, type="p", pch="o")
lines(annualprec.fit3, annualprec.fit3)
plot(precip.Temp3$betaestlist[[2]])

###
###
###  functional response with vector explanatory variables  
###
###

##
## simplest:  formula interface
##

daybasis65 <- create.fourier.basis(rangeval=c(0, 365), nbasis=65,
                                   axes=list('axesIntervals'))
Temp.fd <- with(CanadianWeather, smooth.basisPar(day.5,
                                                 dailyAv[,,'Temperature.C'], daybasis65)$fd)
TempRgn.f <- fRegress(Temp.fd ~ region, CanadianWeather)

##
## Get the default setup and possibly modify it
##

TempRgn.mdl <- fRegress(Temp.fd ~ region, CanadianWeather, method='model')

# make desired modifications here
# then run

TempRgn.m <- do.call('fRegress', TempRgn.mdl)

# no change, so match the first run

all.equal(TempRgn.m, TempRgn.f)


##
## More detailed set up
##

region.contrasts <- model.matrix(~factor(CanadianWeather$region))
rgnContr3 <- region.contrasts
dim(rgnContr3) <- c(1, 35, 4)
dimnames(rgnContr3) <- list('', CanadianWeather$place, c('const',
                                                         paste('region', c('Atlantic', 'Continental', 'Pacific'), sep='.')) )

const365 <- create.constant.basis(c(0, 365))
region.fd.Atlantic <- fd(matrix(rgnContr3[,,2], 1), const365)
# str(region.fd.Atlantic)
region.fd.Continental <- fd(matrix(rgnContr3[,,3], 1), const365)
region.fd.Pacific <- fd(matrix(rgnContr3[,,4], 1), const365)
region.fdlist <- list(const=rep(1, 35),
                      region.Atlantic=region.fd.Atlantic,
                      region.Continental=region.fd.Continental,
                      region.Pacific=region.fd.Pacific)
# str(TempRgn.mdl$betalist)

###
###
###  functional response with functional explanatory variable  
###
###

##
##  predict knee angle from hip angle;  
##     from demo('gait', package='fda')

##
## formula interface
##
gaittime   <- as.matrix((1:20)/21)
gaitrange  <- c(0,20)
gaitbasis  <- create.fourier.basis(gaitrange, nbasis=21)
gaitnbasis <- gaitbasis$nbasis
gaitcoef   <- matrix(0,gaitnbasis,dim(gait)[2])
harmaccelLfd <- vec2Lfd(c(0, (2*pi/20)^2, 0), rangeval=gaitrange)
gaitfd     <- smooth.basisPar(gaittime, gait, gaitbasis, 
                              Lfdobj=harmaccelLfd, lambda=1e-2)$fd
hipfd  <- gaitfd[,1]
kneefd <- gaitfd[,2]

knee.hip.f <- fRegress(kneefd ~ hipfd)

##
## manual set-up
##

#  set up the list of covariate objects
const  <- rep(1, dim(kneefd$coef)[2])
xfdlist  <- list(const=const, hipfd=hipfd)

beta0 <- with(kneefd, fd(gaitcoef, gaitbasis, fdnames))
beta1 <- with(hipfd,  fd(gaitcoef, gaitbasis, fdnames))

betalist  <- list(const=fdPar(beta0), hipfd=fdPar(beta1))

fRegressout <- fRegress(kneefd, xfdlist, betalist)
par(oldpar)