优化算法、约束条件和Optapy评分计算配置
优化算法、约束条件和Optapy评分计算配置
提问于 2022-02-23 13:52:18
我在python中使用Optapy库，并使用GitHub上的学校时间表实例作为基础。关于库的配置，我有几个问题：
如何选择优化算法(例如，禁忌搜索或模拟annealing)?
- How，做选项法，计算解决方案的分数)？我是否可以选择更改python中的分数计算类型？
- 如何确定除硬约束或软约束外的每个约束的权重？
我看了OptaPlanner用户指南，但是我不知道如何在python上实现它。
感谢你的指导。
回答 1
Stack Overflow用户
回答已采纳
发布于 2022-02-23 14:29:50
OptaPy可以使用编程API进行配置。可以在optapy.config包中找到配置类。特别是，通过withPhases选择优化算法。
import optapy.config
solver_config = optapy.config.solver.SolverConfig().withEntityClasses(get_class(Lesson)) \
.withSolutionClass(get_class(TimeTable)) \
.withConstraintProviderClass(get_class(define_constraints)) \
.withTerminationSpentLimit(Duration.ofSeconds(30)) \
.withPhases([
optapy.config.constructionheuristic.ConstructionHeuristicPhaseConfig(),
optapy.config.localsearch.LocalSearchPhaseConfig()
.withAcceptorConfig(optapy.config.localsearch.decider.acceptor.LocalSearchAcceptorConfig()
.withSimulatedAnnealingStartingTemperature("0hard/0soft"))
])(上述配置模拟退火)。
最近添加了@easy_score_calculator和@incremental_score_calculator装饰器，它们允许您分别定义EasyScoreCalculator或IncrementalScoreCalculator。例如，(EasyScoreCalculator，最大化价值)：
@optapy.easy_score_calculator
def my_score_calculator(solution: Solution):
total_score = 0
for entity in solution.entity_list:
total_score += 0 if entity.value is None else entity.value
return optapy.score.SimpleScore.of(total_score)
solver_config = optapy.config.solver.SolverConfig()
termination_config = optapy.config.solver.termination.TerminationConfig()
termination_config.setBestScoreLimit('9')
solver_config.withSolutionClass(optapy.get_class(Solution)) \
.withEntityClasses(optapy.get_class(Entity)) \
.withEasyScoreCalculatorClass(optapy.get_class(my_score_calculator)) \
.withTerminationConfig(termination_config)或使用IncrementalScoreCalculator (NQueens)：
@optapy.incremental_score_calculator
class IncrementalScoreCalculator:
score: int
row_index_map: dict
ascending_diagonal_index_map: dict
descending_diagonal_index_map: dict
def resetWorkingSolution(self, working_solution: Solution):
n = working_solution.n
self.row_index_map = dict()
self.ascending_diagonal_index_map = dict()
self.descending_diagonal_index_map = dict()
for i in range(n):
self.row_index_map[i] = list()
self.ascending_diagonal_index_map[i] = list()
self.descending_diagonal_index_map[i] = list()
if i != 0:
self.ascending_diagonal_index_map[n - 1 + i] = list()
self.descending_diagonal_index_map[-i] = list()
self.score = 0
for queen in working_solution.queen_list:
self.insert(queen)
def beforeEntityAdded(self, entity: any):
pass
def afterEntityAdded(self, entity: any):
self.insert(entity)
def beforeVariableChanged(self, entity: any, variableName: str):
self.retract(entity)
def afterVariableChanged(self, entity: any, variableName: str):
self.insert(entity)
def beforeEntityRemoved(self, entity: any):
self.retract(entity)
def afterEntityRemoved(self, entity: any):
pass
def insert(self, queen: Queen):
row = queen.row
if row is not None:
row_index = queen.row
row_index_list = self.row_index_map[row_index]
self.score -= len(row_index_list)
row_index_list.append(queen)
ascending_diagonal_index_list = self.ascending_diagonal_index_map[queen.getAscendingDiagonalIndex()]
self.score -= len(ascending_diagonal_index_list)
ascending_diagonal_index_list.append(queen)
descending_diagonal_index_list = self.descending_diagonal_index_map[queen.getDescendingDiagonalIndex()]
self.score -= len(descending_diagonal_index_list)
descending_diagonal_index_list.append(queen)
def retract(self, queen: Queen):
row = queen.row
if row is not None:
row_index = queen.row
row_index_list = self.row_index_map[row_index]
row_index_list.remove(queen)
self.score += len(row_index_list)
ascending_diagonal_index_list = self.ascending_diagonal_index_map[queen.getAscendingDiagonalIndex()]
ascending_diagonal_index_list.remove(queen)
self.score += len(ascending_diagonal_index_list)
descending_diagonal_index_list = self.descending_diagonal_index_map[queen.getDescendingDiagonalIndex()]
descending_diagonal_index_list.remove(queen)
self.score += len(descending_diagonal_index_list)
def calculateScore(self) -> optapy.score.SimpleScore:
return optapy.score.SimpleScore.of(self.score)
solver_config = optapy.config.solver.SolverConfig()
termination_config = optapy.config.solver.termination.TerminationConfig()
termination_config.setBestScoreLimit('0')
solver_config.withSolutionClass(optapy.get_class(Solution)) \
.withEntityClasses(optapy.get_class(Queen)) \
.withScoreDirectorFactory(optapy.config.score.director.ScoreDirectorFactoryConfig() \
.withIncrementalScoreCalculatorClass(optapy.get_class(IncrementalScoreCalculator))) \
.withTerminationConfig(termination_config)如果您所说的权重是指ConstraintConfiguration (它允许您在每个问题上定义自定义约束权重)，那么它还没有通过OptaPy公开。如果您的意思是如何使约束权重更多/更少，可以将第二个参数更改为penalize/reward (如果常量)，或者添加第三个参数来计算约束乘数(第二个参数将被乘以)，如下所示：
def undesired_day_for_employee(constraint_factory: ConstraintFactory):
return constraint_factory.forEach(shift_class) \
.join(availability_class, [Joiners.equal(lambda shift: shift.employee,
lambda availability: availability.employee),
Joiners.equal(lambda shift: shift.start.date(),
lambda availability: availability.date)
]) \
.filter(lambda shift, availability: availability.availability_type == AvailabilityType.UNDESIRED) \
.penalize('Undesired day for employee', HardSoftScore.ofSoft(2),
lambda shift, availability: get_shift_duration_in_minutes(shift))(此约束对员工在不想要的一天工作的每分钟进行2次软处罚)
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/71238147
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