# We compute a solution path of the generalized lasso dual problem:
#
# \hat{u}(\lambda) =
# \argmin_u \|y - D^T u\|_2^2 \rm{s.t.} \|\u\|_\infty \leq \lambda
#
# where D is m x n. Here there is no assumption on D, and we use a
# fresh SVD at each iteration (computationally naive but stable).
#
# Note: the df estimates at each lambda_k can be thought of as the df
# for all solutions corresponding to lambda in (lambda_k,lambda_{k-1}),
# the open interval to the *right* of the current lambda_k.
dualpathSvd <- function(y, D, approx=FALSE, maxsteps=2000, minlam=0,
rtol=1e-7, btol=1e-7, verbose=FALSE, object=NULL) {
# If we are starting a new path
if (is.null(object)) {
m = nrow(D)
n = ncol(D)
# Compute the dual solution at infinity, and
# find the first critical point
sv = svdsolve(t(D),y,rtol) # SVD solver
uhat = sv$x # Dual solution
q = sv$q # Rank of D
ihit = which.max(abs(uhat)) # Hitting coordinate
hit = abs(uhat[ihit]) # Critical lambda
s = Sign(uhat[ihit]) # Sign
if (verbose) {
cat(sprintf("1. lambda=%.3f, adding coordinate %i, |B|=%i...",
hit,ihit,1))
}
# Now iteratively find the new dual solution, and
# the next critical point
# Things to keep track of, and return at the end
buf = min(maxsteps,1000)
u = matrix(0,m,buf) # Dual solutions
lams = numeric(buf) # Critical lambdas
h = logical(buf) # Hit or leave?
df = numeric(buf) # Degrees of freedom
lams[1] = hit
h[1] = TRUE
df[1] = n-q
u[,1] = uhat
# Other things to keep track of, but not return
r = 1 # Size of boundary set
B = ihit # Boundary set
I = Seq(1,m)[-ihit] # Interior set
Ds = D[ihit,]*s # Vector t(D[B,])%*%s
D1 = D[-ihit,,drop=FALSE] # Matrix D[I,]
D2 = D[ihit,,drop=FALSE] # Matrix D[B,]
k = 2 # What step are we at?
}
# If iterating an already started path
else {
# Grab variables needed to construct the path
lambda = NULL
for (j in 1:length(object)) {
if (names(object)[j] != "pathobjs") {
assign(names(object)[j], object[[j]])
}
}
for (j in 1:length(object$pathobjs)) {
assign(names(object$pathobjs)[j], object$pathobjs[[j]])
}
lams = lambda
}
tryCatch({
while (k<=maxsteps && lams[k-1]>=minlam) {
##########
# Check if we've reached the end of the buffer
if (k > length(lams)) {
buf = length(lams)
lams = c(lams,numeric(buf))
h = c(h,logical(buf))
df = c(df,numeric(buf))
u = cbind(u,matrix(0,m,buf))
}
##########
# If the interior is empty, then nothing will hit
if (r==m) {
a = b = numeric(0)
hit = 0
q = 0
}
# Otherwise, find the next hitting time
else {
sv = svdsolve(t(D1),cbind(y,Ds),rtol)
a = as.numeric(sv$x[,1])
b = as.numeric(sv$x[,2])
q = sv$q
shits = Sign(a)
hits = a/(b+shits)
# Make sure none of the hitting times are larger
# than the current lambda (precision issue)
hits[hits>lams[k-1]+btol] = 0
hits[hits>lams[k-1]] = lams[k-1]
ihit = which.max(hits)
hit = hits[ihit]
shit = shits[ihit]
}
##########
# If nothing is on the boundary, then nothing will leave
# Also, skip this if we are in "approx" mode
if (r==0 || approx) {
leave = 0
}
# Otherwise, find the next leaving time
else {
c = s*(D2%*%(y-t(D1)%*%a))
d = s*(D2%*%(Ds-t(D1)%*%b))
leaves = c/d
# c must be negative
leaves[c>=0] = 0
# Make sure none of the leaving times are larger
# than the current lambda (precision issue)
leaves[leaves>lams[k-1]+btol] = 0
leaves[leaves>lams[k-1]] = lams[k-1]
ileave = which.max(leaves)
leave = leaves[ileave]
}
##########
# Stop if the next critical point is negative
if (hit<=0 && leave<=0) break
# If a hitting time comes next
if (hit > leave) {
# Record the critical lambda and solution
lams[k] = hit
h[k] = TRUE
df[k] = n-q
uhat = numeric(m)
uhat[B] = hit*s
uhat[I] = a-hit*b
u[,k] = uhat
# Update all of the variables
r = r+1
B = c(B,I[ihit])
I = I[-ihit]
Ds = Ds + D1[ihit,]*shit
s = c(s,shit)
D2 = rbind(D2,D1[ihit,])
D1 = D1[-ihit,,drop=FALSE]
if (verbose) {
cat(sprintf("\n%i. lambda=%.3f, adding coordinate %i, |B|=%i...",
k,hit,B[r],r))
}
}
# Otherwise a leaving time comes next
else {
# Record the critical lambda and solution
lams[k] = leave
h[k] = FALSE
df[k] = n-q
uhat = numeric(m)
uhat[B] = leave*s
uhat[I] = a-leave*b
u[,k] = uhat
# Update all of the variables
r = r-1
I = c(I,B[ileave])
B = B[-ileave]
Ds = Ds - D2[ileave,]*s[ileave]
s = s[-ileave]
D1 = rbind(D1,D2[ileave,])
D2 = D2[-ileave,,drop=FALSE]
if (verbose) {
cat(sprintf("\n%i. lambda=%.3f, deleting coordinate %i, |B|=%i...",
k,leave,I[m-r],r))
}
}
# Step counter
k = k+1
}
}, error = function(err) {
err$message = paste(err$message,"\n(Path computation has been terminated;",
" partial path is being returned.)",sep="")
warning(err)})
# Trim
lams = lams[Seq(1,k-1)]
h = h[Seq(1,k-1)]
df = df[Seq(1,k-1),drop=FALSE]
u = u[,Seq(1,k-1),drop=FALSE]
# Save needed elements for continuing the path
pathobjs = list(type="svd", r=r, B=B, I=I, approx=approx,
k=k, df=df, D1=D1, D2=D2, Ds=Ds, ihit=ihit, m=m, n=n, h=h,
rtol=rtol, btol=btol, s=s, y=y)
# If we reached the maximum number of steps
if (k>maxsteps) {
if (verbose) {
cat(sprintf("\nReached the maximum number of steps (%i),",maxsteps))
cat(" skipping the rest of the path.")
}
completepath = FALSE
}
# If we reached the minimum lambda
else if (lams[k-1]<minlam) {
if (verbose) {
cat(sprintf("\nReached the minimum lambda (%.3f),",minlam))
cat(" skipping the rest of the path.")
}
completepath = FALSE
}
# Otherwise, note that we completed the path
else completepath = TRUE
if (verbose) cat("\n")
colnames(u) = as.character(round(lams,3))
beta = y-t(D)%*%u
return(list(lambda=lams,beta=beta,fit=beta,u=u,hit=h,df=df,y=y,
completepath=completepath,bls=y,pathobjs=pathobjs))
}
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