如何修复 Python（或工具）中的“警告：

Question

我正在尝试使用 python 3.7.1、spider 和 or-tools 解决优化问题。目前，我想使用约束将对象分为 3 个不同的类。首先，我尝试使用以下方法解决它：    #solver = pywraplp.Solver('LinearExample',    #                           pywraplp.Solver.GLOP_LINEAR_PROGRAMMING)我得到了一些结果但不是预期的结果，因为 xA xB xC 应该是 3 个二进制向量。那又怎样，我将这两行替换为整数问题来解决问题，这在我看来更合乎逻辑，通过：    solver = pywraplp.Solver('SolveIntegerProblem',                          pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)当我运行代码时，会打开一个带有消息的窗口：它已停止工作，然后我收到以下警告：    "An error ocurred while starting the kernel"    WARNING: Logging before InitGoogleLogging() is written to STDERR    F0327 09:54:41.733001 3784 map_util.h:126] Check failed:   collection‑>insert(value_type(key, data)).second duplicate key: xA    *** Check failure stack trace: ***然后我必须关闭控制台我不明白为什么问题似乎是 x... 而不是“LinearExample”

Answer 1

根据@CodyGray 的要求，以下是工作代码。它定义了 14 * 3 = 42 个变量。OP 的代码在 for 循环中仅定义了一个 ( xA) 或三个变量 ( xA, xB, xC)，这可能导致错误：duplicate key: xA.

from __future__ import print_function

import pandas as pd

from ortools.linear_solver import pywraplp

def main():

  solver = pywraplp.Solver('SolveIntegerProblem',

                          pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)

  d = {

        'A': [19286.0, 23786.0, 9822.0, 5054.0, 97466.0, 728998.0, 275708.0, 

              4576.0, 67284.0, 385582.0, 13450.0, 43271.0, 44601.0, 88372.0],

        'B': [12073.0, 21563.0, 13077.0, 6407.0, 91850.0, 557996.0, 206372.0,

              2812.0, 52362.0, 244102.0, 11225.0, 50612.0, 49299.0, 76099.0],

        'C': [12048.0, 42648.0, 35491.0, 19800.0, 117602.0, 643498.0, 232377.0,

              5217.0, 79200.0, 234259.0, 19296.0, 114048.0, 100725.0, 130911.0]

      }

  coeff = pd.DataFrame(data=d)

  c = {

        'A': [11503, 10638, 1984, 364, 15022, 40343, 41478,

              238, 3528, 51649, 5759, 5305, 7883, 301],

        'B': [1783, 2047, 425, 88, 2306, 6261, 6423,

              51, 610, 7976, 1034, 1021, 1443, 537],

        'C': [128, 250, 61, 15, 161, 453, 461,

              8, 60, 566, 111, 125, 161, 57]

      }

  weight = pd.DataFrame(data=c)

  nb_obj=len(coeff['A'])

  xA = [solver.IntVar(0.0, 1.0, 'xA{:02d}'.format(i)) for i in range(nb_obj)]

  xB = [solver.IntVar(0.0, 1.0, 'xB{:02d}'.format(i)) for i in range(nb_obj)]

  xC = [solver.IntVar(0.0, 1.0, 'xC{:02d}'.format(i)) for i in range(nb_obj)]

  # total weight per class is limited 

  solver.Add(sum(xA * weight.A) <= 80000)

  solver.Add(sum(xB * weight.B) <= 15000)

  solver.Add(sum(xC * weight.C) <= 1500)

  # number of object in each class is limited

  solver.Add(sum(xA) <= 3)

  solver.Add(sum(xB) <= 6) 

  solver.Add(sum(xC) <= 5)

  # 1 object can only belong to a single class

  for i in range (nb_obj):

    solver.Add(xA[i] + xB[i] + xC[i] == 1)

  objective = solver.Objective()

  for i in range(nb_obj):

    objective.SetCoefficient(xA[i], coeff.A[i])

    objective.SetCoefficient(xB[i], coeff.B[i])

    objective.SetCoefficient(xC[i], coeff.C[i])

  objective.SetMaximization()

  print('Number of variables =', solver.NumVariables())

  print('Number of constraints =', solver.NumConstraints())

  # Solve the problem and print the solution.

  result_status = solver.Solve()

  # The problem has an optimal solution.

  assert result_status == pywraplp.Solver.OPTIMAL

  # The objective value of the solution.

  print('Optimal objective value = %d' % solver.Objective().Value())

  print()

  # The value of each variable in the solution.

  for i in range(nb_obj):

    print("Obj {:02d}:".format(i), xA[i].solution_value(), xB[i].solution_value(), xC[i].solution_value())

if __name__ == '__main__':

  main()

结果是：

Number of variables = 42

Number of constraints = 20

Optimal objective value = 1840645

Obj 00: 1.0 0.0 0.0

Obj 01: 0.0 1.0 0.0

Obj 02: 0.0 1.0 0.0

Obj 03: 0.0 1.0 0.0

Obj 04: 0.0 1.0 0.0

Obj 05: 0.0 0.0 1.0

Obj 06: 0.0 0.0 1.0

Obj 07: 0.0 1.0 0.0

Obj 08: 1.0 0.0 0.0

Obj 09: 1.0 0.0 0.0

Obj 10: 0.0 1.0 0.0

Obj 11: 0.0 0.0 1.0

Obj 12: 0.0 0.0 1.0

Obj 13: 0.0 0.0 1.0