英文:
Trying to Optimize Process Using Linear Programming. Getting error about: IndexError: index 1 is out of bounds for axis 0 with size 1
问题
在这行代码中：
prob += lpSum([w[j] * y[j] for j in range(n)])
你遇到了这个错误：
IndexError: index 1 is out of bounds for axis 0 with size 1
这个错误是因为在 w 中的数据是一个一维数组，但你尝试访问索引1，而实际上只有一个元素，索引应为0。你可以尝试将 w 定义为一个标量值而不是一个数组，或者根据你的需求调整代码来解决这个错误。
英文:
I am trying to optimize worker's schedules, based on the following dataframe.
Time Windows Shift 1 Shift 2 Shift 3 Shift 4 Workers Required
0 6:00 - 9:00 1 0 0 1 55.0
1 9:00 - 12:00 1 0 0 0 46.0
2 12:00 - 15:00 1 1 0 0 59.0
3 15:00 - 18:00 0 1 0 0 23.0
4 18:00 - 21:00 0 1 1 0 60.0
5 21:00 - 24:00 0 0 1 0 38.0
6 24:00 - 3:00 0 0 1 1 20.0
7 3:00 - 6:00 0 0 0 1 30.0
8 Wage_Rate 135 140 190 188 0.0
First (create dataframe):
import pandas as pd
df = pd.read_clipboard(sep='\\s+')
df = pd.DataFrame(df)
Here is the code that I am testing.
import pandas as pd
import pulp
from pulp import LpMaximize, LpMinimize, LpProblem, LpStatus, lpSum, LpVariable
import numpy as np
df = df.fillna(0).applymap(lambda x: 1 if x == "X" else x)
df.set_index('Time Windows')
a = df.drop(columns=["Workers Required"]).values
a
df.drop(df.tail(1).index,inplace=True)
print(df.shape)
df = df.fillna(0).applymap(lambda x: 1 if x == "X" else x)
print(df.shape)
a = df.to_numpy()
a
# number of shifts
n = a.shape[0]
# number of time windows
T = a.shape[0]
# number of workers required per time window
d = df["Workers Required"].values
# wage rate per shift
#Get last row of dataframe
last_row = df.iloc[-1:,1:]
#Get last row of dataframe as numpy array
w = last_row.to_numpy()
w
# Decision variables
y = LpVariable.dicts("num_workers", list(range(n)), lowBound=0, cat="Integer")
y
# Create problem
prob = LpProblem("scheduling_workers", LpMinimize)
prob += lpSum([w[j] * y[j] for j in range(n)])
for t in range(T):
prob += lpSum([a[t, j] * y[j] for j in range(n)]) >= d[t]
prob.solve()
print("Status:", LpStatus[prob.status])
for shift in range(n):
print(f"The number of workers needed for shift {shift} is {int(y[shift].value())} workers")
When I get to this line:
prob += lpSum([w[j] * y[j] for j in range(n)])
I get this error.
Traceback (most recent call last):
Cell In[197], line 1
prob += lpSum([w[j] * y[j] for j in range(n)])
Cell In[197], line 1 in <listcomp>
prob += lpSum([w[j] * y[j] for j in range(n)])
IndexError: index 1 is out of bounds for axis 0 with size 1
The example I am trying to follow is from the link below.
https://towardsdatascience.com/how-to-solve-a-staff-scheduling-problem-with-python-63ae50435ba4
答案1
得分: 1
你的问题主要是因为对 Pandas 的误用。正确使用切片，它将正常工作：
from io import StringIO
import pandas as pd
import pulp
with StringIO(
'''Time Windows,Shift 1,Shift 2,Shift 3,Shift 4,Workers Required
6:00 - 9:00, 1, 0, 0, 1, 55.0
9:00 - 12:00, 1, 0, 0, 0, 46.0
12:00 - 15:00, 1, 1, 0, 0, 59.0
15:00 - 18:00, 0, 1, 0, 0, 23.0
18:00 - 21:00, 0, 1, 1, 0, 60.0
21:00 - 24:00, 0, 0, 1, 0, 38.0
24:00 - 3:00, 0, 0, 1, 1, 20.0
3:00 - 6:00, 0, 0, 0, 1, 30.0
Wage_Rate, 135, 140, 190, 188, 0.0''') as f:
df = pd.read_csv(f, skipinitialspace=True, index_col='Time Windows')
is_shift = df.columns.str.startswith('Shift')
is_wage = df.index == 'Wage_Rate'
shifts = df.loc[~is_wage, is_shift]
wage_rate = df.loc[is_wage, is_shift].squeeze()
workers_req = df.loc[~is_wage, 'Workers Required']
workers_per_shift = pulp.LpVariable.dicts(name='workers_per', indices=shifts.columns, lowBound=0, cat=pulp.LpInteger)
prob = pulp.LpProblem(name='scheduling_workers', sense=pulp.LpMinimize)
prob.objective = pulp.lpDot(wage_rate, workers_per_shift.values())
for (time, shift), worker_req in zip(shifts.iterrows(), workers_req):
prob.addConstraint(name=f'workers_min_{time}', constraint=pulp.lpDot(shift, workers_per_shift.values()) >= worker_req)
print(prob)
prob.solve()
for k, v in workers_per_shift.items():
print(f'{k} has {v.value():.0f} workers')
scheduling_workers:
MINIMIZE
135*workers_per_Shift_1 + 140*workers_per_Shift_2 + 190*workers_per_Shift_3 + 188*workers_per_Shift_4 + 0
SUBJECT TO
workers_min_6:00___9:00: workers_per_Shift_1 + workers_per_Shift_4 >= 55
workers_min_9:00___12:00: workers_per_Shift_1 >= 46
workers_min_12:00___15:00: workers_per_Shift_1 + workers_per_Shift_2 >= 59
workers_min_15:00___18:00: workers_per_Shift_2 >= 23
workers_min_18:00___21:00: workers_per_Shift_2 + workers_per_Shift_3 >= 60
workers_min_21:00___24:00: workers_per_Shift_3 >= 38
workers_min_24:00___3:00: workers_per_Shift_3 + workers_per_Shift_4 >= 20
workers_min_3:00___6:00: workers_per_Shift_4 >= 30
VARIABLES
0 <= workers_per_Shift_1 Integer
0 <= workers_per_Shift_2 Integer
0 <= workers_per_Shift_3 Integer
0 <= workers_per_Shift_4 Integer
欢迎使用 CBC MILP 求解器
版本：2.10.3
构建日期：2019年12月15日
命令行 - C:\Users\gtoom\src\stackexchange\.venv\lib\site-packages\pulp\solverdir\cbc\win\cbc.exe C:\Users\gtoom\AppData\Local\Temp759563a6c3439b8221b857b9f617af-pulp.mps timeMode elapsed branch printingOptions all solution C:\Users\gtoom\AppData\Local\Temp759563a6c3439b8221b857b9f617af-pulp.sol (默认策略 1)
第2行 NAME MODEL
第3行 ROWS
第13行 COLUMNS
第38行 RHS
第47行 BOUNDS
第52行 ENDATA
问题 MODEL 共有 8 行，4 列，12 个元素
Coin0008I 以 0 个错误读取模型
将 timeMode 选项从 CPU 更改为经过的时间
连续目标值为 22290 - 0.00 秒
Cgl0004I 处理后的模型有 0 行，0 列（0 个整数（其中 0 个是二进制））和 0 个元素
Cbc3007W 没有整数变量 - 无事可做
根节点处的切割将目标从 22290 更改为 -1.79769e+308
探测尝试了 0 次，并在添加切割回合后创建了 0 个切割，其中 0 个活跃（0.000 秒）
Gomory 尝试了 0 次，并在添加切割回合后创建了 0 个切割，其中 0 个活跃（0.000 秒）
尝试了 0 次的背包问题，并创建了 0 个切割，其中 0 个在添加切割回合后是活跃的（0.000 秒）
尝试了 0 次的团问题，并创建了 0 个切割，其中 0 个在添加切割回合后是活跃的（0.000 秒）
尝试了 0 次的混合整数舍入问题，并创建了 0 个切割，其中 0 个在添加切割回合后是活跃的（0.000 秒）
FlowCover 尝试了 0 次，并创建了 0
<details>
<summary>英文:</summary>
Your problems mostly come from misuse of Pandas. Use sane slicing, and it works fine:
```python
from io import StringIO
import pandas as pd
import pulp
with StringIO(
'''Time Windows,Shift 1,Shift 2,Shift 3,Shift 4,Workers Required
6:00 - 9:00, 1, 0, 0, 1, 55.0
9:00 - 12:00, 1, 0, 0, 0, 46.0
12:00 - 15:00, 1, 1, 0, 0, 59.0
15:00 - 18:00, 0, 1, 0, 0, 23.0
18:00 - 21:00, 0, 1, 1, 0, 60.0
21:00 - 24:00, 0, 0, 1, 0, 38.0
24:00 - 3:00, 0, 0, 1, 1, 20.0
3:00 - 6:00, 0, 0, 0, 1, 30.0
Wage_Rate, 135, 140, 190, 188, 0.0''') as f:
df = pd.read_csv(f, skipinitialspace=True, index_col='Time Windows')
is_shift = df.columns.str.startswith('Shift')
is_wage = df.index == 'Wage_Rate'
shifts = df.loc[~is_wage, is_shift]
wage_rate = df.loc[is_wage, is_shift].squeeze()
workers_req = df.loc[~is_wage, 'Workers Required']
workers_per_shift = pulp.LpVariable.dicts(name='workers_per', indices=shifts.columns, lowBound=0, cat=pulp.LpInteger)
prob = pulp.LpProblem(name='scheduling_workers', sense=pulp.LpMinimize)
prob.objective = pulp.lpDot(wage_rate, workers_per_shift.values())
for (time, shift), worker_req in zip(shifts.iterrows(), workers_req):
prob.addConstraint(name=f'workers_min_{time}', constraint=pulp.lpDot(shift, workers_per_shift.values()) >= worker_req)
print(prob)
prob.solve()
for k, v in workers_per_shift.items():
print(f'{k} has {v.value():.0f} workers')
scheduling_workers:
MINIMIZE
135*workers_per_Shift_1 + 140*workers_per_Shift_2 + 190*workers_per_Shift_3 + 188*workers_per_Shift_4 + 0
SUBJECT TO
workers_min_6:00___9:00: workers_per_Shift_1 + workers_per_Shift_4 >= 55
workers_min_9:00___12:00: workers_per_Shift_1 >= 46
workers_min_12:00___15:00: workers_per_Shift_1 + workers_per_Shift_2 >= 59
workers_min_15:00___18:00: workers_per_Shift_2 >= 23
workers_min_18:00___21:00: workers_per_Shift_2 + workers_per_Shift_3 >= 60
workers_min_21:00___24:00: workers_per_Shift_3 >= 38
workers_min_24:00___3:00: workers_per_Shift_3 + workers_per_Shift_4 >= 20
workers_min_3:00___6:00: workers_per_Shift_4 >= 30
VARIABLES
0 <= workers_per_Shift_1 Integer
0 <= workers_per_Shift_2 Integer
0 <= workers_per_Shift_3 Integer
0 <= workers_per_Shift_4 Integer
Welcome to the CBC MILP Solver
Version: 2.10.3
Build Date: Dec 15 2019
command line - C:\Users\gtoom\src\stackexchange\.venv\lib\site-packages\pulp\solverdir\cbc\win\cbc.exe C:\Users\gtoom\AppData\Local\Temp759563a6c3439b8221b857b9f617af-pulp.mps timeMode elapsed branch printingOptions all solution C:\Users\gtoom\AppData\Local\Temp759563a6c3439b8221b857b9f617af-pulp.sol (default strategy 1)
At line 2 NAME MODEL
At line 3 ROWS
At line 13 COLUMNS
At line 38 RHS
At line 47 BOUNDS
At line 52 ENDATA
Problem MODEL has 8 rows, 4 columns and 12 elements
Coin0008I MODEL read with 0 errors
Option for timeMode changed from cpu to elapsed
Continuous objective value is 22290 - 0.00 seconds
Cgl0004I processed model has 0 rows, 0 columns (0 integer (0 of which binary)) and 0 elements
Cbc3007W No integer variables - nothing to do
Cuts at root node changed objective from 22290 to -1.79769e+308
Probing was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Gomory was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Knapsack was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Clique was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
MixedIntegerRounding2 was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
FlowCover was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
TwoMirCuts was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
ZeroHalf was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Result - Optimal solution found
Objective value: 22290.00000000
Enumerated nodes: 0
Total iterations: 0
Time (CPU seconds): 0.00
Time (Wallclock seconds): 0.00
Option for printingOptions changed from normal to all
Total time (CPU seconds): 0.01 (Wallclock seconds): 0.01
Shift 1 has 46 workers
Shift 2 has 23 workers
Shift 3 has 38 workers
Shift 4 has 30 workers
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论