几天来,我一直在努力简化约束。 还在学习Python。 请帮助并提前感谢您。
Employees=['Paul', 'Ben', 'Nasim', 'Ceci', 'Victoria',]
Days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday',
'Saturday', 'Sunday']
avail = pulp.LpVariable.dicts("on_off", ((employee, day) for
employee in Employees for day in Days), cat="Binary")
requests={"Paul": {"Monday":1, "Tuesday":1, "Wednesday":1,
"Thursday":1, "Friday":1, "Saturday":1, "Sunday":1},
"Ben": {"Monday":1, "Tuesday":1, "Wednesday":1, "Thursday":1,
"Friday":1, "Saturday":1, "Sunday":0},
"Nasim": {"Monday":1, "Tuesday":1, "Wednesday":1, "Thursday":1,
"Friday":1, "Saturday":1, "Sunday":1},
"Ceci": {"Monday":1, "Tuesday":1, "Wednesday":1, "Thursday":1,
"Friday":1,"Saturday":1, "Sunday":0},
"Victoria": {"Monday":1, "Tuesday":1, "Wednesday":0, "Thursday":0,
"Friday":0, "Saturday":0, "Sunday":0}}
for employee, day in avail:
prob += avail[employee, day] == [requests[i][j] for i in
requests for j in requests[i]]
例如,前几个约束:
"_C13: on_off_('Paul',_'Monday') = 28
_C14: on_off_('Paul',_'Tuesday') = 28
_C15: on_off_('Paul',_'Wednesday') = 28
_C16: on_off_('Paul',_'Thursday') = 28
_C17: on_off_('Paul',_'Friday') = 28
_C18: on_off_('Paul',_'Saturday') = 28"
("28"恰好是我嵌套词典中所有 1 和 0 的总和。
相反,我喜欢让每个变量与我的嵌套字典中的 1 和 0 匹配。
"_C13: on_off_('Paul',_'Monday') = 1
_C14: on_off_('Paul',_'Tuesday') = 1
_C15: on_off_('Paul',_'Wednesday') = 1
_C16: on_off_('Paul',_'Thursday') = 1
_C17: on_off_('Paul',_'Friday') = 1
_C18: on_off_('Paul',_'Saturday') = 1"
在 == 符号之后,您希望字典中的值为 1 或 0,因此您应该使用:
for employee, day in avail:
prob += avail[employee, day] == requests[employee][day]
对于请求[员工],您可以为该员工选择字典
requests['Paul'] = {"Monday":1, "Tuesday":1, "Wednesday":1,
"Thursday":1, "Friday":1, "Saturday":1, "Sunday":1}
然后通过添加 [day],您可以从该词典中选择日期:
requests['Paul']['Monday'] = 1