pyomo +网状错误6句柄无效

Question

我正在尝试进行pyomo优化，但收到错误消息[Error 6] The handle is invalid。不确定如何解释它，环顾四周似乎与特权有关，但我不太了解。

在下面找到完整的错误跟踪，以及一个再现它的玩具示例。

完整的错误跟踪：

py_run_file_impl（文件，本地，转换）中的错误：ApplicationError：无法执行命令：'C：\ Users \ xxx \ AppData \ Local \ Continuum \ anaconda3 \ envs \ lucy \ Library \ bin \ ipopt.exe c：\ users \ xxx \ appdata \ local \ temp \ tmpp2hmid.pyomo.nl -AMPL'错误消息：[错误6]句柄无效

详细的追溯：文件“ C：\ Users \ xxx \ AppData \ Local \ CONTIN〜1 \ ANACON〜1 \ envs \ lucy \ lib \ site-packages \ pyomo \ opt \ base \ solvers中的文件“”，第46行。 py”，在第578行中，解决_status = self._apply_solver（）文件“ C：\ Users \ xxx \ AppData \ Local \ CONTIN〜1 \ ANACON〜1 \ envs \ lucy \ lib \ site-packages \ pyomo \ opt \ “ solver \ shellcmd.py”，行246，在_apply_solver self._rc中，self._log = self._execute_command（self._command）文件“ C：\ Users \ xxx \ AppData \ Local \ CONTIN〜1 \ ANACON〜1 \ envs \ lucy \ lib \ site-packages \ pyomo \ opt \ solver \ shellcmd.py”，第309行，位于_execute_command tee tee = self._tee文件“ C：\ Users \ xxx \ AppData \ Local \ CONTIN〜1 \ ANACON〜1 \ envs \ lucy \ lib \ site-packages \ pyutilib \ subprocess \ processmngr.py“，行660，在run_command中

基于此的可复制示例。

纯python代码（当我在conda称为“ lucy” 的环境中以python运行它时，它可以工作）：

from pyomo.environ import *
infinity = float('inf')

model = AbstractModel()

# Foods
model.F = Set()
# Nutrients
model.N = Set()

# Cost of each food
model.c    = Param(model.F, within=PositiveReals)
# Amount of nutrient in each food
model.a    = Param(model.F, model.N, within=NonNegativeReals)
# Lower and upper bound on each nutrient
model.Nmin = Param(model.N, within=NonNegativeReals, default=0.0)
model.Nmax = Param(model.N, within=NonNegativeReals, default=infinity)
# Volume per serving of food
model.V    = Param(model.F, within=PositiveReals)
# Maximum volume of food consumed
model.Vmax = Param(within=PositiveReals)

# Number of servings consumed of each food
model.x = Var(model.F, within=NonNegativeIntegers)

# Minimize the cost of food that is consumed
def cost_rule(model):
    return sum(model.c[i]*model.x[i] for i in model.F)
model.cost = Objective(rule=cost_rule)

# Limit nutrient consumption for each nutrient
def nutrient_rule(model, j):
    value = sum(model.a[i,j]*model.x[i] for i in model.F)
    return model.Nmin[j] <= value <= model.Nmax[j]
model.nutrient_limit = Constraint(model.N, rule=nutrient_rule)

# Limit the volume of food consumed
def volume_rule(model):
    return sum(model.V[i]*model.x[i] for i in model.F) <= model.Vmax
model.volume = Constraint(rule=volume_rule)

opt = SolverFactory('ipopt')
instance = model.create_instance('diet.dat')
results = opt.solve(instance, tee=False)
results

在R中运行它的代码reticulate非常简单：

library(reticulate)
use_condaenv(condaenv = "lucy")
py_run_file("../pyomo_scripts/test.py")

最后为了完整起见，这是diet.dat文件（必须与python / R文件位于同一路径）：

param:  F:                          c     V  :=
  "Cheeseburger"                 1.84   4.0  
  "Ham Sandwich"                 2.19   7.5  
  "Hamburger"                    1.84   3.5  
  "Fish Sandwich"                1.44   5.0  
  "Chicken Sandwich"             2.29   7.3  
  "Fries"                         .77   2.6  
  "Sausage Biscuit"              1.29   4.1  
  "Lowfat Milk"                   .60   8.0 
  "Orange Juice"                  .72  12.0 ;

param Vmax := 75.0;

param:  N:       Nmin   Nmax :=
        Cal      2000      .
        Carbo     350    375
        Protein    55      .
        VitA      100      .
        VitC      100      .
        Calc      100      .
        Iron      100      . ;

param a:
                               Cal  Carbo Protein   VitA   VitC  Calc  Iron :=
  "Cheeseburger"               510     34     28     15      6    30    20
  "Ham Sandwich"               370     35     24     15     10    20    20
  "Hamburger"                  500     42     25      6      2    25    20
  "Fish Sandwich"              370     38     14      2      0    15    10
  "Chicken Sandwich"           400     42     31      8     15    15     8
  "Fries"                      220     26      3      0     15     0     2
  "Sausage Biscuit"            345     27     15      4      0    20    15
  "Lowfat Milk"                110     12      9     10      4    30     0
  "Orange Juice"                80     20      1      2    120     2     2 ;

---

评论后编辑：

这些都是版本pyomo和ipopt

pyomo                     5.6.4                    py36_0    conda-forge
pyomo.extras              3.3                 py36_182212    conda-forge
ipopt                     3.11.1                        2    conda-forge

我已经继承了R中的大量代码，并pyomo通过系统调用完成了优化。我正在尝试通过使用来改进它，reticulate从而避免写入和读取文件，并且拥有更多的控制权...如果仍然可以在python中进行系统调用，则使用不会带来太多收益reticulate。

谢谢。

Answer 1

如果可以执行python版本，请尝试使用以下代码以管理权限进行r会话

library("reticulate")


##-- your directory containing 'diet.py' and 'diet.dat'
setwd("D:/project/Dropbox/lectures/2104xxx scg_opt/src/02"")



##-- execute code
a <- py_run_file("diet.py",local=T)
a$results