Lee*_*e B 6 python scheduling permutation timeslots timetable
希望你能帮我解决这个问题.这对工作没有帮助 - 这是一个非常努力工作的志愿者的慈善机构,他们真的可以使用一个比他们现有的更少混乱/烦人的时间表系统.
如果有人知道一个好的第三方应用程序(当然)自动化这个,那几乎同样好.只是...请不要建议随机的时间表,例如预订教室的东西,因为我认为他们不能这样做.
提前感谢阅读; 我知道这是一个很重要的帖子.我尽我所能尽力记录这一点,并表明我已经自己努力了.
我需要一个工人/时间段调度算法,它为工人生成轮班,符合以下标准:
输入数据
import datetime.datetime as dt
class DateRange:
def __init__(self, start, end):
self.start = start
self.end = end
class Shift:
def __init__(self, range, min, max):
self.range = range
self.min_workers = min
self.max_workers = max
tue_9th_10pm = dt(2009, 1, 9, 22, 0)
wed_10th_4am = dt(2009, 1, 10, 4, 0)
wed_10th_10am = dt(2009, 1, 10, 10, 0)
shift_1_times = Range(tue_9th_10pm, wed_10th_4am)
shift_2_times = Range(wed_10th_4am, wed_10th_10am)
shift_3_times = Range(wed_10th_10am, wed_10th_2pm)
shift_1 = Shift(shift_1_times, 2,3) # allows 3, requires 2, but only 2 available
shift_2 = Shift(shift_2_times, 2,2) # allows 2
shift_3 = Shift(shift_3_times, 2,3) # allows 3, requires 2, 3 available
shifts = ( shift_1, shift_2, shift_3 )
joe_avail = [ shift_1, shift_2 ]
bob_avail = [ shift_1, shift_3 ]
sam_avail = [ shift_2 ]
amy_avail = [ shift_2 ]
ned_avail = [ shift_2, shift_3 ]
max_avail = [ shift_3 ]
jim_avail = [ shift_3 ]
joe = Worker('joe', joe_avail)
bob = Worker('bob', bob_avail)
sam = Worker('sam', sam_avail)
ned = Worker('ned', ned_avail)
max = Worker('max', max_avail)
amy = Worker('amy', amy_avail)
jim = Worker('jim', jim_avail)
workers = ( joe, bob, sam, ned, max, amy, jim )
从上面来看,轮班和工人是要处理的两个主要输入变量
每个班次都需要最少和最大数量的工人.填补班次的最低要求对于成功至关重要,但如果所有其他方法都失败了,手动填充空白的旋转优于"错误":)主要的算法问题是,当有足够的空间时,应该没有不必要的空白工人可以.
理想情况下,班次的最大工人数量将被填补,但这是相对于其他约束的最低优先级,因此如果必须给出,则应该是这样.
灵活的约束
这些有点灵活,如果找不到"完美"的解决方案,它们的界限可以推动一点.这种灵活性应该是最后的手段,而不是随机利用.理想情况下,灵活性可配置为"fudge_factor"变量或类似变量.
很高兴,但没有必要
如果你能想出一个能够实现上述功能并包含其中任何/所有这些算法的算法,我会非常感激和感激.即使是单独执行这些位的附加脚本也会很棒.
重叠的转变.例如,能够指定两个同时发生的"前台"班次和"后台"班次是很好的.这可以通过用不同的移位数据单独调用程序来完成,除了在给定时间段内关于为多个班次安排人员的约束将被遗漏.
每个工人(而非全球)可指定的工人的最短重新安排时间.例如,如果Joe感到工作过度或处理个人问题,或者是初学者学习绳索,我们可能希望比其他工作人员更少地安排他.
当没有可用的工人适合时,选择员工以填补最小班次数字的一些自动/随机/公平方式.
处理突然取消的一些方法,只是填补空白而不重新安排其他班次.
可能该算法应该生成尽可能多的匹配解决方案,其中每个解决方案如下所示:
class Solution:
def __init__(self, shifts_workers):
"""shifts_workers -- a dictionary of shift objects as keys, and a
a lists of workers filling the shift as values."""
assert isinstance(dict, shifts_workers)
self.shifts_workers = shifts_workers
鉴于上述数据,这是针对单个解决方案的测试功能.我认为这是对的,但我也很感激对它的一些同行评审.
def check_solution(solution):
assert isinstance(Solution, solution)
def shift_check(shift, workers, workers_allowed):
assert isinstance(Shift, shift):
assert isinstance(list, workers):
assert isinstance(list, workers_allowed)
num_workers = len(workers)
assert num_workers >= shift.min_workers
assert num_workers <= shift.max_workers
for w in workers_allowed:
assert w in workers
shifts_workers = solution.shifts_workers
# all shifts should be covered
assert len(shifts_workers.keys()) == 3
assert shift1 in shifts_workers.keys()
assert shift2 in shifts_workers.keys()
assert shift3 in shifts_workers.keys()
# shift_1 should be covered by 2 people - joe, and bob
shift_check(shift_1, shifts_workers[shift_1], (joe, bob))
# shift_2 should be covered by 2 people - sam and amy
shift_check(shift_2, shifts_workers[shift_2], (sam, amy))
# shift_3 should be covered by 3 people - ned, max, and jim
shift_check(shift_3, shifts_workers[shift_3], (ned,max,jim))
我已经尝试用遗传算法实现这一点,但似乎无法将其调整得恰到好处,所以尽管基本原理似乎只适用于单班制,但它甚至无法解决几个班次和一些班次的简单案例工作人员.
我最近的尝试是生成每个可能的排列作为解决方案,然后减少不符合约束的排列.这似乎工作得更快,并且让我更进一步,但我使用python 2.6的itertools.product()来帮助生成排列,我无法完全正确.如果有很多错误,我不会感到惊讶,老实说,这个问题不适合我的脑袋:)
目前我的代码是两个文件:models.py和rota.py. models.py看起来像:
# -*- coding: utf-8 -*-
class Shift:
def __init__(self, start_datetime, end_datetime, min_coverage, max_coverage):
self.start = start_datetime
self.end = end_datetime
self.duration = self.end - self.start
self.min_coverage = min_coverage
self.max_coverage = max_coverage
def __repr__(self):
return "<Shift %s--%s (%r<x<%r)" % (self.start, self.end, self.min_coverage, self.max_coverage)
class Duty:
def __init__(self, worker, shift, slot):
self.worker = worker
self.shift = shift
self.slot = slot
def __repr__(self):
return "<Duty worker=%r shift=%r slot=%d>" % (self.worker, self.shift, self.slot)
def dump(self, indent=4, depth=1):
ind = " " * (indent * depth)
print ind + "<Duty shift=%s slot=%s" % (self.shift, self.slot)
self.worker.dump(indent=indent, depth=depth+1)
print ind + ">"
class Avail:
def __init__(self, start_time, end_time):
self.start = start_time
self.end = end_time
def __repr__(self):
return "<%s to %s>" % (self.start, self.end)
class Worker:
def __init__(self, name, availabilities):
self.name = name
self.availabilities = availabilities
def __repr__(self):
return "<Worker %s Avail=%r>" % (self.name, self.availabilities)
def dump(self, indent=4, depth=1):
ind = " " * (indent * depth)
print ind + "<Worker %s" % self.name
for avail in self.availabilities:
print ind + " " * indent + repr(avail)
print ind + ">"
def available_for_shift(self, shift):
for a in self.availabilities:
if shift.start >= a.start and shift.end <= a.end:
return True
print "Worker %s not available for %r (Availability: %r)" % (self.name, shift, self.availabilities)
return False
class Solution:
def __init__(self, shifts):
self._shifts = list(shifts)
def __repr__(self):
return "<Solution: shifts=%r>" % self._shifts
def duties(self):
d = []
for s in self._shifts:
for x in s:
yield x
def shifts(self):
return list(set([ d.shift for d in self.duties() ]))
def dump_shift(self, s, indent=4, depth=1):
ind = " " * (indent * depth)
print ind + "<ShiftList"
for duty in s:
duty.dump(indent=indent, depth=depth+1)
print ind + ">"
def dump(self, indent=4, depth=1):
ind = " " * (indent * depth)
print ind + "<Solution"
for s in self._shifts:
self.dump_shift(s, indent=indent, depth=depth+1)
print ind + ">"
class Env:
def __init__(self, shifts, workers):
self.shifts = shifts
self.workers = workers
self.fittest = None
self.generation = 0
class DisplayContext:
def __init__(self, env):
self.env = env
def status(self, msg, *args):
raise NotImplementedError()
def cleanup(self):
pass
def update(self):
pass
和rota.py看起来像:
#!/usr/bin/env python2.6
# -*- coding: utf-8 -*-
from datetime import datetime as dt
am2 = dt(2009, 10, 1, 2, 0)
am8 = dt(2009, 10, 1, 8, 0)
pm12 = dt(2009, 10, 1, 12, 0)
def duties_for_all_workers(shifts, workers):
from models import Duty
duties = []
# for all shifts
for shift in shifts:
# for all slots
for cov in range(shift.min_coverage, shift.max_coverage):
for slot in range(cov):
# for all workers
for worker in workers:
# generate a duty
duty = Duty(worker, shift, slot+1)
duties.append(duty)
return duties
def filter_duties_for_shift(duties, shift):
matching_duties = [ d for d in duties if d.shift == shift ]
for m in matching_duties:
yield m
def duty_permutations(shifts, duties):
from itertools import product
# build a list of shifts
shift_perms = []
for shift in shifts:
shift_duty_perms = []
for slot in range(shift.max_coverage):
slot_duties = [ d for d in duties if d.shift == shift and d.slot == (slot+1) ]
shift_duty_perms.append(slot_duties)
shift_perms.append(shift_duty_perms)
all_perms = ( shift_perms, shift_duty_perms )
# generate all possible duties for all shifts
perms = list(product(*shift_perms))
return perms
def solutions_for_duty_permutations(permutations):
from models import Solution
res = []
for duties in permutations:
sol = Solution(duties)
res.append(sol)
return res
def find_clashing_duties(duty, duties):
"""Find duties for the same worker that are too close together"""
from datetime import timedelta
one_day = timedelta(days=1)
one_day_before = duty.shift.start - one_day
one_day_after = duty.shift.end + one_day
for d in [ ds for ds in duties if ds.worker == duty.worker ]:
# skip the duty we're considering, as it can't clash with itself
if duty == d:
continue
clashes = False
# check if dates are too close to another shift
if d.shift.start >= one_day_before and d.shift.start <= one_day_after:
clashes = True
# check if slots collide with another shift
if d.slot == duty.slot:
clashes = True
if clashes:
yield d
def filter_unwanted_shifts(solutions):
from models import Solution
print "possibly unwanted:", solutions
new_solutions = []
new_duties = []
for sol in solutions:
for duty in sol.duties():
duty_ok = True
if not duty.worker.available_for_shift(duty.shift):
duty_ok = False
if duty_ok:
print "duty OK:"
duty.dump(depth=1)
new_duties.append(duty)
else:
print "duty **NOT** OK:"
duty.dump(depth=1)
shifts = set([ d.shift for d in new_duties ])
shift_lists = []
for s in shifts:
shift_duties = [ d for d in new_duties if d.shift == s ]
shift_lists.append(shift_duties)
new_solutions.append(Solution(shift_lists))
return new_solutions
def filter_clashing_duties(solutions):
new_solutions = []
for sol in solutions:
solution_ok = True
for duty in sol.duties():
num_clashing_duties = len(set(find_clashing_duties(duty, sol.duties())))
# check if many duties collide with this one (and thus we should delete this one
if num_clashing_duties > 0:
solution_ok = False
break
if solution_ok:
new_solutions.append(sol)
return new_solutions
def filter_incomplete_shifts(solutions):
new_solutions = []
shift_duty_count = {}
for sol in solutions:
solution_ok = True
for shift in set([ duty.shift for duty in sol.duties() ]):
shift_duties = [ d for d in sol.duties() if d.shift == shift ]
num_workers = len(set([ d.worker for d in shift_duties ]))
if num_workers < shift.min_coverage:
solution_ok = False
if solution_ok:
new_solutions.append(sol)
return new_solutions
def filter_solutions(solutions, workers):
# filter permutations ############################
# for each solution
solutions = filter_unwanted_shifts(solutions)
solutions = filter_clashing_duties(solutions)
solutions = filter_incomplete_shifts(solutions)
return solutions
def prioritise_solutions(solutions):
# TODO: not implemented!
return solutions
# prioritise solutions ############################
# for all solutions
# score according to number of staff on a duty
# score according to male/female staff
# score according to skill/background diversity
# score according to when staff last on shift
# sort all solutions by score
def solve_duties(shifts, duties, workers):
# ramify all possible duties #########################
perms = duty_permutations(shifts, duties)
solutions = solutions_for_duty_permutations(perms)
solutions = filter_solutions(solutions, workers)
solutions = prioritise_solutions(solutions)
return solutions
def load_shifts():
from models import Shift
shifts = [
Shift(am2, am8, 2, 3),
Shift(am8, pm12, 2, 3),
]
return shifts
def load_workers():
from models import Avail, Worker
joe_avail = ( Avail(am2, am8), )
sam_avail = ( Avail(am2, am8), )
ned_avail = ( Avail(am2, am8), )
bob_avail = ( Avail(am8, pm12), )
max_avail = ( Avail(am8, pm12), )
joe = Worker("joe", joe_avail)
sam = Worker("sam", sam_avail)
ned = Worker("ned", sam_avail)
bob = Worker("bob", bob_avail)
max = Worker("max", max_avail)
return (joe, sam, ned, bob, max)
def main():
import sys
shifts = load_shifts()
workers = load_workers()
duties = duties_for_all_workers(shifts, workers)
solutions = solve_duties(shifts, duties, workers)
if len(solutions) == 0:
print "Sorry, can't solve this. Perhaps you need more staff available, or"
print "simpler duty constraints?"
sys.exit(20)
else:
print "Solved. Solutions found:"
for sol in solutions:
sol.dump()
if __name__ == "__main__":
main()
在结果之前剪切调试输出,这当前给出:
Solved. Solutions found:
<Solution
<ShiftList
<Duty shift=<Shift 2009-10-01 02:00:00--2009-10-01 08:00:00 (2<x<3) slot=1
<Worker joe
<2009-10-01 02:00:00 to 2009-10-01 08:00:00>
>
>
<Duty shift=<Shift 2009-10-01 02:00:00--2009-10-01 08:00:00 (2<x<3) slot=1
<Worker sam
<2009-10-01 02:00:00 to 2009-10-01 08:00:00>
>
>
<Duty shift=<Shift 2009-10-01 02:00:00--2009-10-01 08:00:00 (2<x<3) slot=1
<Worker ned
<2009-10-01 02:00:00 to 2009-10-01 08:00:00>
>
>
>
<ShiftList
<Duty shift=<Shift 2009-10-01 08:00:00--2009-10-01 12:00:00 (2<x<3) slot=1
<Worker bob
<2009-10-01 08:00:00 to 2009-10-01 12:00:00>
>
>
<Duty shift=<Shift 2009-10-01 08:00:00--2009-10-01 12:00:00 (2<x<3) slot=1
<Worker max
<2009-10-01 08:00:00 to 2009-10-01 12:00:00>
>
>
>
>
好吧，我不知道特定的算法，但这是我会考虑的。
评估
无论采用哪种方法，您都需要一个函数来评估您的解决方案满足约束的程度。您可以采用“比较”方法（没有全局分数，而是比较两个解决方案的方法），但我建议进行评估。
真正好的是，如果您可以获得较短时间跨度的分数，例如每天，如果您可以从部分解决方案“预测”最终分数的范围（例如，仅前 3 个分数），这对算法确实很有帮助7 天）。这样，如果它已经太低而无法满足您的期望，您可以中断基于此部分解决方案的计算。
对称
很可能在这 200 个人中，您有相似的个人资料：即人们具有相同的特征（可用性、经验、意愿……）。如果你有两个具有相同个人资料的人，他们将可以互换：
从您的角度来看，实际上是相同的解决方案。
好处是，通常情况下，您的配置文件比人少，这对节省计算时间有很大帮助！
例如，假设您根据解决方案 1 生成了所有解决方案，那么无需根据解决方案 2 计算任何内容。
迭代
您可以考虑增量生成（例如一次 1 周），而不是立即生成整个计划。净收益是一周的复杂性降低了（可能性更少）。
然后，一旦你有了这周的时间，你就计算第二个，当然要小心地将第一个考虑到你的限制。
优点是您明确设计算法以考虑已使用的解决方案，这样对于下一个时间表生成，它将确保不会让一个人连续 24 小时工作！
序列化
您应该考虑解决方案对象的序列化（选择您的选择，pickle 对于 Python 来说非常好）。在生成新计划时，您将需要之前的计划，而且我敢打赌您不会为这 200 人手动输入它。
详尽无遗
现在，毕竟，我实际上赞成进行详尽的搜索，因为使用对称性和评估的可能性可能不会那么多（尽管问题仍然是 NP 完全的，但没有灵丹妙药）。
您可能愿意尝试回溯算法。
另外，您应该查看以下处理类似问题的链接：
两者都讨论了实施过程中遇到的问题，因此检查它们应该会对您有所帮助。