L1-范数正则化问题定义如下：
minimize || A*x - b ||_2^2 + || x ||_1
但在我的情况下，我想解决这种形式的问题，而不是通常的 L1 范数正则化最小二乘问题：
minimize || A*x - b ||_2^2 + || W*G*x ||_1
由于我的表 W 和 G 没有像 AI 一样的尺寸，因此无法更改我的变量并解决这样的问题
minimize || A*x/(W*B) - b ||_2^2 + || x ||_1
这样我就可以使用互联网上可用的求解器之一。所以我找到了一个解决上述L1平方问题的方程，如：
minimize || A*x - b ||_2^2 + || x ||_1
minimize || A*x - b ||_2^2 + e'*u
subject to -u <= x <= u
据我所知，如果我使用 -u <= W G x <= ui 可以解决问题。但我无法得到我应该完全适应代码的内容。任何人都可以帮忙吗？代码如下（取自CVXOPT）
from cvxopt import matrix, spdiag, mul, div, sqrt, normal, setseed
from cvxopt import blas, lapack, solvers, sparse, spmatrix
import math
def l1regls(A, b):
"""
Returns the solution of l1-norm regularized least-squares problem
minimize || A*x - b ||_2^2 + || x ||_1.
"""
m, n = A.size
q = matrix(1.0, (2*n,1))
q[:n] = -2.0 * A.T * b
def P(u, v, alpha = 1.0, beta = 0.0 ):
"""
v := alpha * 2.0 * [ A'*A, 0; 0, 0 ] * u + beta * v
"""
v *= beta
v[:n] += alpha * 2.0 * A.T * (A * u[:n])
def G(u, v, alpha=1.0, beta=0.0, trans='N'):
"""
v := alpha*[I, -I; -I, -I] * u + beta * v (trans = 'N' or 'T')
"""
v *= beta
v[:n] += alpha*(u[:n] - u[n:])
v[n:] += alpha*(-u[:n] - u[n:])
h = matrix(0.0, (2*n,1))
# Customized solver for the KKT system
#
# [ 2.0*A'*A 0 I -I ] [x[:n] ] [bx[:n] ]
# [ 0 0 -I -I ] [x[n:] ] = [bx[n:] ].
# [ I -I -D1^-1 0 ] [zl[:n]] [bzl[:n]]
# [ -I -I 0 -D2^-1 ] [zl[n:]] [bzl[n:]]
#
# where D1 = W['di'][:n]**2, D2 = W['di'][:n]**2.
#
# We first eliminate zl and x[n:]:
#
# ( 2*A'*A + 4*D1*D2*(D1+D2)^-1 ) * x[:n] =
# bx[:n] - (D2-D1)*(D1+D2)^-1 * bx[n:] +
# D1 * ( I + (D2-D1)*(D1+D2)^-1 ) * bzl[:n] -
# D2 * ( I - (D2-D1)*(D1+D2)^-1 ) * bzl[n:]
#
# x[n:] = (D1+D2)^-1 * ( bx[n:] - D1*bzl[:n] - D2*bzl[n:] )
# - (D2-D1)*(D1+D2)^-1 * x[:n]
#
# zl[:n] = D1 * ( x[:n] - x[n:] - bzl[:n] )
# zl[n:] = D2 * (-x[:n] - x[n:] - bzl[n:] ).
#
# The first equation has the form
#
# (A'*A + D)*x[:n] = rhs
#
# and is equivalent to
#
# [ D A' ] [ x:n] ] = [ rhs ]
# [ A -I ] [ v ] [ 0 ].
#
# It can be solved as
#
# ( A*D^-1*A' + I ) * v = A * D^-1 * rhs
# x[:n] = D^-1 * ( rhs - A'*v ).
S = matrix(0.0, (m,m))
Asc = matrix(0.0, (m,n))
v = matrix(0.0, (m,1))
def Fkkt(W):
# Factor
#
# S = A*D^-1*A' + I
#
# where D = 2*D1*D2*(D1+D2)^-1, D1 = d[:n]**-2, D2 = d[n:]**-2.
d1, d2 = W['di'][:n]**2, W['di'][n:]**2
# ds is square root of diagonal of D
ds = math.sqrt(2.0) * div( mul( W['di'][:n], W['di'][n:]),
sqrt(d1+d2) )
d3 = div(d2 - d1, d1 + d2)
# Asc = A*diag(d)^-1/2
Asc = A * spdiag(ds**-1)
# S = I + A * D^-1 * A'
blas.syrk(Asc, S)
S[::m+1] += 1.0
lapack.potrf(S)
def g(x, y, z):
x[:n] = 0.5 * ( x[:n] - mul(d3, x[n:]) +
mul(d1, z[:n] + mul(d3, z[:n])) - mul(d2, z[n:] -
mul(d3, z[n:])) )
x[:n] = div( x[:n], ds)
# Solve
#
# S * v = 0.5 * A * D^-1 * ( bx[:n] -
# (D2-D1)*(D1+D2)^-1 * bx[n:] +
# D1 * ( I + (D2-D1)*(D1+D2)^-1 ) * bzl[:n] -
# D2 * ( I - (D2-D1)*(D1+D2)^-1 ) * bzl[n:] )
blas.gemv(Asc, x, v)
lapack.potrs(S, v)
# x[:n] = D^-1 * ( rhs - A'*v ).
blas.gemv(Asc, v, x, alpha=-1.0, beta=1.0, trans='T')
x[:n] = div(x[:n], ds)
# x[n:] = (D1+D2)^-1 * ( bx[n:] - D1*bzl[:n] - D2*bzl[n:] )
# - (D2-D1)*(D1+D2)^-1 * x[:n]
x[n:] = div( x[n:] - mul(d1, z[:n]) - mul(d2, z[n:]), d1+d2 )\
- mul( d3, x[:n] )
# zl[:n] = D1^1/2 * ( x[:n] - x[n:] - bzl[:n] )
# zl[n:] = D2^1/2 * ( -x[:n] - x[n:] - bzl[n:] ).
z[:n] = mul( W['di'][:n], x[:n] - x[n:] - z[:n] )
z[n:] = mul( W['di'][n:], -x[:n] - x[n:] - z[n:] )
return g
return solvers.coneqp(P, q, G, h, kktsolver = Fkkt)['x'][:n]
任何人都可以帮忙吗？先感谢您