# NOT RUN {
# Simulate 200 curves with pre-curve sample sizes ranging from 1 to 10
# Make curves with odd-numbered IDs have an x-distribution that is random
# uniform [0,1] and those with even-numbered IDs have an x-dist. that is
# half as wide but still centered at 0.5. Shift y values higher with
# increasing IDs
set.seed(1)
N <- 200
nc <- sample(1:10, N, TRUE)
id <- rep(1:N, nc)
x <- y <- id
for(i in 1:N) {
x[id==i] <- if(i %% 2) runif(nc[i]) else runif(nc[i], c(.25, .75))
y[id==i] <- i + 10*(x[id==i] - .5) + runif(nc[i], -10, 10)
}
w <- curveRep(x, y, id, kxdist=2, p=10)
w
par(ask=TRUE, mfrow=c(4,5))
plot(w) # show everything, profiles going across
par(mfrow=c(2,5))
plot(w,1) # show n=1 results
# Use a color assignment table, assigning low curves to green and
# high to red. Unique curve (subject) IDs are the names of the vector.
cols <- c(rep('green', N/2), rep('red', N/2))
names(cols) <- as.character(1:N)
plot(w, 3, idcol=cols)
par(ask=FALSE, mfrow=c(1,1))
plot(w, 1, 'lattice') # show n=1 results
plot(w, 3, 'lattice') # show n=4-5 results
plot(w, 3, 'lattice', idcol=cols) # same but different color mapping
plot(w, 3, 'lattice', m=1) # show a single "representative" curve
# Show median, 10th, and 90th percentiles of supposedly representative curves
plot(w, 3, 'lattice', m='quantiles', probs=c(.5,.1,.9))
# Same plot but with much less grouping of x variable
plot(w, 3, 'lattice', m='quantiles', probs=c(.5,.1,.9), nx=2)
# Smooth data before profiling. This allows later plotting to plot
# smoothed representative curves rather than raw curves (which
# specifying smooth=TRUE to curveRep would do, if curveSmooth was not used)
d <- curveSmooth(x, y, id)
w <- with(d, curveRep(x, y, id))
# Example to show that curveRep can cluster profiles correctly when
# there is no noise. In the data there are four profiles - flat, flat
# at a higher mean y, linearly increasing then flat, and flat at the
# first height except for a sharp triangular peak
set.seed(1)
x <- 0:100
m <- length(x)
profile <- matrix(NA, nrow=m, ncol=4)
profile[,1] <- rep(0, m)
profile[,2] <- rep(3, m)
profile[,3] <- c(0:3, rep(3, m-4))
profile[,4] <- c(0,1,3,1,rep(0,m-4))
col <- c('black','blue','green','red')
matplot(x, profile, type='l', col=col)
xeval <- seq(0, 100, length.out=5)
s <- x <!-- %in% xeval -->
matplot(x[s], profile[s,], type='l', col=col)
id <- rep(1:100, each=m)
X <- Y <- id
cols <- character(100)
names(cols) <- as.character(1:100)
for(i in 1:100) {
s <- id==i
X[s] <- x
j <- sample(1:4,1)
Y[s] <- profile[,j]
cols[i] <- col[j]
}
table(cols)
yl <- c(-1,4)
w <- curveRep(X, Y, id, kn=1, kxdist=1, k=4)
plot(w, 1, 'lattice', idcol=cols, ylim=yl)
# Found 4 clusters but two have same profile
w <- curveRep(X, Y, id, kn=1, kxdist=1, k=3)
plot(w, 1, 'lattice', idcol=cols, freq=cols, plotfreq=TRUE, ylim=yl)
# Incorrectly combined black and red because default value p=5 did
# not result in different profiles at x=xeval
w <- curveRep(X, Y, id, kn=1, kxdist=1, k=4, p=40)
plot(w, 1, 'lattice', idcol=cols, ylim=yl)
# Found correct clusters because evaluated curves at 40 equally
# spaced points and could find the sharp triangular peak in profile 4
# }
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