This function performs kfold crossvalidation to choose the value of
the regularization parameter lambda for a trend filtering problem,
given the computed solution path. This function only applies to trend
filtering objects with identity predictor matrix (no X
passed).
1 2 3 
object 
the solution path object, of class "trendfilter", as returned by the

k 
an integer indicating the number of folds to split the data into. Must be between 2 and n2 (n being the number of observations), default is 5. It is generally not a good idea to pass a value of k much larger than 10 (say, on the scale of n); see "Details" below. 
mode 
a character string, either "lambda" or "df". Specifying "lambda" means that the crossvalidation error will be computed and reported at each value of lambda that appears as a knot in the solution path. Specifying "df" means that the crossvalidation error will be computed and reported for every of degrees of freedom value (actually, estimate) incurred along the solution path. In the case that the same degrees of freedom value is visited multiple times, the model with the most regularization (smallest value of lambda) is considered. Default is "lambda". 
approx 
a logical variable indicating if the approximate solution path
should be used (with no dual coordinates leaving the boundary).
Default is 
rtol 
a numeric variable giving the relative tolerance used in the calculation of the hitting and leaving times. A larger value is more conservative, and may cause the algorithm to miss some hitting or leaving events (do not change unless you know what you're getting into!). Defaultis 1e7. 
btol 
similar to 
verbose 
a logical variable indicating if progress should be reported after each knot in the path. 
For trend filtering (with an identity predictor matrix), the folds
for kfold crossvalidation are chosen by placing every kth point into
the same fold. (Here the points are implicitly ordered according to their
underlying positionsâ€”either assumed to be evenly spaced, or explicitly
passed through the pos
argument.)
The first and last points are not included in any fold and are always
included in building the predictive model. As an example,
with n=15 data points and k=4 folds, the points are assigned to folds
in the following way:
x 1 2 3 4 1 2 3 4 1 2 3 4 1 x
where x indicates no assignment. Therefore, the folds are not
random and running cv.trendfilter
twice will give the same
result. In the calculation of the crossvalidated error, the
predicted value at a point is given by the average of the fits at this
point's two neighbors (guaranteed to be in a different fold).
Running crossvalidation in modes "lambda" and "df" often yields very similar results. The mode "df" simply gives an alternative parametrization for the sequence of crossvalidated models and can be more convenient for some applications; if you are confused about its function, simply leave the mode equal to "lambda".
Returns and object of class "cv.trendfilter", a list with the following components:
err 
a numeric vector of crossvalidated errors. 
se 
a numeric vector of standard errors (standard deviations of the crossvalidation error estimates). 
mode 
a character string indicating the mode, either "lambda" or "df". 
lambda 
if 
lambda.min 
if 
lambda.1se 
if 
df 
if 
df.min 
if 
df.1se 
if 
i.min 
the index of the model minimizing the crossvalidation error. 
i.1se 
the index of the model chosen by the one standard error rule. 
call 
the matched call. 
trendfilter
, plot.cv.trendfilter
,
plot.trendfilter
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  # Constant trend filtering (the 1d fused lasso)
set.seed(0)
n = 50
beta0 = rep(sample(1:10,5),each=n/5)
y = beta0 + rnorm(n,sd=0.8)
a = fusedlasso1d(y)
plot(a)
# Choose lambda by 5fold crossvalidation
cv = cv.trendfilter(a)
plot(cv)
plot(a,lambda=cv$lambda.min,main="Minimal CV error")
plot(a,lambda=cv$lambda.1se,main="One standard error rule")
## Not run:
# Cubic trend filtering
set.seed(0)
n = 100
beta0 = numeric(100)
beta0[1:40] = (1:4020)^3
beta0[40:50] = 60*(40:5050)^2 + 60*100+20^3
beta0[50:70] = 20*(50:7050)^2 + 60*100+20^3
beta0[70:100] = 1/6*(70:100110)^3 + 1/6*40^3 + 6000
beta0 = beta0
beta0 = (beta0min(beta0))*10/diff(range(beta0))
y = beta0 + rnorm(n)
a = trendfilter(y,ord=3,maxsteps=150)
plot(a,nlam=5)
# Choose lambda by 5fold crossvalidation
cv = cv.trendfilter(a)
plot(cv)
plot(a,lambda=cv$lambda.min,main="Minimal CV error")
plot(a,lambda=cv$lambda.1se,main="One standard error rule")
## End(Not run)

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