使用log cdf在cvxopt(python)中定义一个函数

Question

我想将目标函数定义为：-sum(log(normcdf(x)))，其中normcdf对x的每个组件进行操作。 看起来cvxpy已经实现了它，但我想坚持使用 Python cvxopt。 有什么建议吗？

***** Example python code to make this question clearer:
from cvxopt import spmatrix, log
from cvxopt.modeling import variable, op, sum
# A is m x n matrix of type 'cvxopt.base.spmatrix' (not included here to save space)
# a_hat is n x 1 vector of type 'cvxopt.modeling.variable
a_hat = variable(n)
# constraints
c1 = (a_hat >= 0)
c2 = (a_hat <= 0)
#valid objective and optimization problem
f = -sum(A*a_hat)
op(f, [c1, c2]).solve()
# desired objective
# f = -sum(log( "cdf of each element of (A*a_hat)" ))
# this doesn't work either (because log 'argument must be a number of dense matrix')
# f = -sum(log(A*a_hat))

Answer 1

我找到了这样做的方法：需要计算自己的梯度和黑森并使用cvxopt.cp（下面，G和h是约束，为了清楚起见省略）。

def myFunc(x=None, z=None):
  if x is None: return 0, matrix(0.2, (n,1))
  y = (1/sigma)*A*x
  f = -sum(log(matrix(stats.norm.cdf(y))))
  r = matrix(stats.norm.pdf(y)/stats.norm.cdf(y))
  gradf = -A.T * r
  if z is None: return f, gradf.T
  H = A.T * spdiag( (1.0/sigma) * z[0] * r**2 + mul(r,y)) * A
  return f, gradf.T, H
xlb = solvers.cp(myFunc, G = G, h = h)