我希望创建一个分配来显示员工可上班的时间。与此图相似，在此链接staff distribution中可以找到该图。
为此，我创建了staff_availability_df，其中包含要选择的员工数量，该数量可在['Person']列中找到。他们可以工作的min - max小时，所获得的报酬是这样的。他们可以工作的可用时间分为小时数['Availability_Hr']，该小时数表示他们可以工作的时间，以小时表示。因此，第一个人是'8-18'，也就是8:00:00am - 18:00:00pm。 ['Availability_15min_Seg']基本上是相同的，但小时数分为4个部分。因此，第一个人是'1-41'，又是8:00:00am - 18:00:00pm。
注意：标准班次在8:00:00am - 3:30:00am之间运行，因此大约需要20个小时。staff_requirements_df在整个班次中显示Time以及我需要的必需People。
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.dates as dates
#This is the employee availability:
staff_availability = pd.DataFrame({
'Person' : ['C1','C2','C3','C4','C5','C6','C7','C8','C9','C10','C11'],
'MinHours' : [5,5,5,5,5,5,5,5,5,5,5],
'MaxHours' : [10,10,10,10,10,10,10,10,10,10,10],
'HourlyWage' : [26,26,26,26,26,26,26,26,26,26,26],
'Availability_Hr' : ['8-18','8-18','8-18','9-18','9-18','9-18','12-1','12-1','17-3','17-3','17-3'],
'Availability_15min_Seg' : ['1-41','1-41','1-41','5-41','5-41','5-41','17-69','17-69','37-79','37-79','37-79'],
})
#These are the staffing requirements:
staffing_requirements = pd.DataFrame({
'Time' : ['0/1/1900 8:00:00','0/1/1900 9:59:00','0/1/1900 10:00:00','0/1/1900 12:29:00','0/1/1900 12:30:00','0/1/1900 13:00:00','0/1/1900 13:02:00','0/1/1900 13:15:00','0/1/1900 13:20:00','0/1/1900 18:10:00','0/1/1900 18:15:00','0/1/1900 18:20:00','0/1/1900 18:25:00','0/1/1900 18:45:00','0/1/1900 18:50:00','0/1/1900 19:05:00','0/1/1900 19:07:00','0/1/1900 21:57:00','0/1/1900 22:00:00','0/1/1900 22:30:00','0/1/1900 22:35:00','1/1/1900 3:00:00','1/1/1900 3:05:00','1/1/1900 3:20:00','1/1/1900 3:25:00'],
'People' : [1,1,2,2,3,3,2,2,3,3,4,4,3,3,2,2,3,3,4,4,3,3,2,2,1],
})
我已使用以下功能将
8:00:00am - 3:30:00am之间的15分钟细分中的人员编制要求导出。每15分钟分配给string 'T'。所以T1 = 8:00:00am和T79 = 3:00:00amstaffing_requirements['Time'] = ['/'.join([str(int(x.split('/')[0])+1)] + x.split('/')[1:]) for x in staffing_requirements['Time']]
staffing_requirements['Time'] = pd.to_datetime(staffing_requirements['Time'], format='%d/%m/%Y %H:%M:%S')
formatter = dates.DateFormatter('%Y-%m-%d %H:%M:%S')
staffing_requirements = staffing_requirements.groupby(pd.Grouper(freq='15T',key='Time'))['People'].max().ffill()
staffing_requirements = staffing_requirements.reset_index(level=['Time'])
staffing_requirements.insert(2, 'T', range(1, 1 + len(staffing_requirements)))
staffing_requirements['T'] = 'T' + staffing_requirements['T'].astype(str)
st_req = staffing_requirements['People'].tolist()
[1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 2.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 4.0, 4.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 3.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 3.0, 2.0]
我希望使用这些功能来创建一个
linear programming matrix，它返回每个员工可工作时间的分布。但我希望使用15分钟的时间段和小时。例如注意：此导出将扩展到3:30am。因此它将包含79个细分。注意：要明确。我希望返回分发时间表，以便将来使用。不只是一个数字。
有一些使用
mixed-integer linear programming的人员可用example 1 example 2方法，但是他们使用的是封闭源代码软件。我希望将其翻译成Python。 最佳答案
对于整数编程而言，这确实是一项伟大的工作。您可以使用pulp，您首先需要通过命令行进行安装，例如pip install pulp
数据处理以确保成功
然后，首先确保您的DataFrames处于最佳形状，以便我们解决问题：
# Since timeslots for staffing start counting at 1, also make the
# DataFrame index start counting at 1
staffing_requirements.index = range(1, len(staffing_requirements) + 1)
print(staffing_requirements.tail())
staff_availability.set_index('Person')
staff_costs = staff_availability.set_index('Person')[['MinHours', 'MaxHours', 'HourlyWage']]
availability = staff_availability.set_index('Person')[['Availability_15min_Seg']]
availability[['first_15min', 'last_15min']] = availability['Availability_15min_Seg'].str.split('-', expand=True).astype(int)
availability_per_member = [pd.DataFrame(1, columns=[idx], index=range(row['first_15min'], row['last_15min']+1))
for idx, row in availability.iterrows()]
availability_per_member = pd.concat(availability_per_member, axis='columns').fillna(0).astype(int).stack()
availability_per_member.index.names = ['Timeslot', 'Person']
availability_per_member = (availability_per_member.to_frame()
.join(staff_costs[['HourlyWage']])
.rename(columns={0: 'Available'}))
现在，
availability_per_member是一个MultiIndex DataFrame，每个时隙每个人只有一行，表明他/她的可用性和工资：# Available HourlyWage
#Timeslot Person
#1 C1 1 26
# C2 1 26
# C3 1 26
# C4 0 26
# C5 0 26
另外，我们稍微修改了必要条件，以便实际上可以解决问题；为何需要这样做，请参阅附录
import numpy as np
np.random.seed(42)
staffing_requirements['People'] = np.random.randint(1, 4, size=len(staffing_requirements))
staff_costs['MinHours'] = 3
使用
pulp解决整数编程问题现在，我们可以开始工作了：设置问题以最小化成本为目标，并逐一添加您提到的约束，请参阅注释的代码。
staffed现在是一本纸浆字典，其中包含某人在特定时间段（0或1）的工作人员import pulp
prob = pulp.LpProblem('CreateStaffing', pulp.LpMinimize) # Minimize costs
timeslots = staffing_requirements.index
persons = availability_per_member.index.levels[1]
# A member is either staffed or is not at a certain timeslot
staffed = pulp.LpVariable.dicts("staffed",
((timeslot, staffmember) for timeslot, staffmember
in availability_per_member.index),
lowBound=0,
cat='Binary')
# Objective = cost (= sum of hourly wages)
prob += pulp.lpSum(
[staffed[timeslot, staffmember] * availability_per_member.loc[(timeslot, staffmember), 'HourlyWage']
for timeslot, staffmember in availability_per_member.index]
)
# Staff the right number of people
for timeslot in timeslots:
prob += (sum([staffed[(timeslot, person)] for person in persons])
== staffing_requirements.loc[timeslot, 'People'])
# Do not staff unavailable persons
for timeslot in timeslots:
for person in persons:
if availability_per_member.loc[(timeslot, person), 'Available'] == 0:
prob += staffed[timeslot, person] == 0
# Do not underemploy people
for person in persons:
prob += (sum([staffed[(timeslot, person)] for timeslot in timeslots])
>= staff_costs.loc[person, 'MinHours']*4) # timeslot is 15 minutes => 4 timeslots = hour
# Do not overemploy people
for person in persons:
prob += (sum([staffed[(timeslot, person)] for timeslot in timeslots])
<= staff_costs.loc[person, 'MaxHours']*4) # timeslot is 15 minutes => 4 timeslots = hour
然后，让
pulp解决问题就可以了：prob.solve()
print(pulp.LpStatus[prob.status])
output = []
for timeslot, staffmember in staffed:
var_output = {
'Timeslot': timeslot,
'Staffmember': staffmember,
'Staffed': staffed[(timeslot, staffmember)].varValue,
}
output.append(var_output)
output_df = pd.DataFrame.from_records(output)#.sort_values(['timeslot', 'staffmember'])
output_df.set_index(['Timeslot', 'Staffmember'], inplace=True)
if pulp.LpStatus[prob.status] == 'Optimal':
print(output_df)
现在，这将返回一个DataFrame
output_df，每个时隙和每个人都包含他们是否配备了人员：# Staffed
#Timeslot Staffmember
#1 C1 1.0
# C2 1.0
# C3 1.0
# C4 0.0
# C5 0.0
# C6 0.0
# C7 0.0
# C8 0.0
# C9 0.0
# C10 0.0
# C11 0.0
#2 C1 1.0
# C2 1.0
我已经修改了http://benalexkeen.com/linear-programming-with-python-and-pulp-part-5/的代码，这对制浆和线性编程是一个很好的教程，因此请务必将其检出。
附录：您的要求不可行。
根据您的条件，这实际上将返回
'Infeasible'。很容易看出为什么是这样的：您会看到所需的工作人员比最近几个时隙中可用的工作人员多。该图是由以下人员创建的：
fig, ax = plt.subplots()
staffing_requirements.plot(y='People', ax=ax, label='Required', drawstyle='steps-mid')
availability_per_member.groupby(level='Timeslot')['Available'].sum().plot(ax=ax,
label='Available', drawstyle='steps-mid')
plt.legend()
关于python - 创建可用值的分布-Python，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55330016/