基本应用函数和基本函数CVXR

Question

有没有办法使用base::apply和CVXR elementwise函数将elementwise函数应用于矩阵的每一列？假设向量的长度与矩阵的行数相匹配，我希望我的目标函数能够如下所示：

new_matrix <- apply(constant_matrix, 2, function(x) x * variable_vector)
objective <- sum_entries( max_entries( as.matrix(new_matrix), axis = 2 ) )

我在解决这个问题时遇到了以下错误:min(常量)错误:参数的'type‘(列表)无效

(如果我的代码不清楚，我的目标是将一个变量向量乘以一个常量矩阵的每一列，然后得到每个缩放列的最大值，然后对所有这些最大值求和。)

谢谢!

Answer 1

下面我给出了一个使用base::apply和CVXR的逐个元素乘法的例子以及输出。有关更复杂的示例，请参阅Pliable Lasso。

library(CVXR)
library(magrittr)

constant_matrix <- base::matrix(1:20, nrow = 4)
variable_vector <- CVXR::Variable(4, pos = TRUE)

constant_matrix %>%
    base::apply(MARGIN = 2, FUN = function(x) CVXR::multiply(lh_exp = x, rh_exp = variable_vector)) %>%
    base::lapply(max) ->
    column_max_list

objective <- base::Reduce(f = '+', x = column_max_list)
prob <- CVXR::Problem(CVXR::Minimize(objective), constraints = list(variable_vector >= 2))

result  <- CVXR::solve(prob, verbose = TRUE)
result$value
result$getValue(variable_vector)

下面是输出：

-----------------------------------------------------------------
           OSQP v0.6.0  -  Operator Splitting QP Solver
              (c) Bartolomeo Stellato,  Goran Banjac
        University of Oxford  -  Stanford University 2019
-----------------------------------------------------------------
problem:  variables n = 9, constraints m = 28
          nnz(P) + nnz(A) = 48
settings: linear system solver = qdldl,
          eps_abs = 1.0e-05, eps_rel = 1.0e-05,
          eps_prim_inf = 1.0e-04, eps_dual_inf = 1.0e-04,
          rho = 1.00e-01 (adaptive),
          sigma = 1.00e-06, alpha = 1.60, max_iter = 10000
          check_termination: on (interval 25),
          scaling: on, scaled_termination: off
          warm start: on, polish: on, time_limit: off

iter   objective    pri res    dua res    rho        time
   1  -1.0811e+02   1.94e+01   1.56e+01   1.00e-01   6.07e-05s
 200   1.1999e+02   3.92e-04   4.69e-04   1.00e-01   2.31e-04s
 300   1.2000e+02   1.46e-05   2.92e-06   1.00e-01   2.86e-04s

status:               solved
solution polish:      unsuccessful
number of iterations: 300
optimal objective:    120.0005
run time:             3.11e-04s
optimal rho estimate: 3.06e-01

> result$value
[1] 120.0005
> result$getValue(variable_vector)
         [,1]
[1,] 2.248055
[2,] 2.149815
[3,] 2.057720
[4,] 2.000007
`