使用列表在 python pulp 中加载数据
Load data in python pulp using lists
我有以下python纸浆代码：
import pulp
# Create a LP Minimization problem
Lp_prob = pulp.LpProblem('Problem', pulp.LpMaximize)
# Create problem Variables
a01 = pulp.LpVariable("a01", 0, 1, cat="Integer") # Create a variable x >= 0
x31 = pulp.LpVariable("x31", 0, 1, cat="Integer") # Create a variable x >= 0
c97 = pulp.LpVariable("c97", 0, 1, cat="Integer") # Create a variable x >= 0
ab12 = pulp.LpVariable("ab12", 0, 1, cat="Integer") # Create a variable x >= 0
eaz = pulp.LpVariable("eaz", 0, 1, cat="Integer") # Create a variable x >= 0
f12 = pulp.LpVariable("f12", 0, 1, cat="Integer") # Create a variable x >= 0
# Objective Function
Lp_prob += 100*a01 + 110*x31 + 150*c97 + 90*ab12 + 50*eaz + 60*f12
# Constraints:
Lp_prob += 3*a01 + 10*x31 + 10*c97 + 2*ab12 + 4*eaz + 6*f12 <= 25
Lp_prob += 4*a01 + 9*x31 + 3*c97 + 9*ab12 + 1*eaz + 4*f12 <= 21
Lp_prob += 6*a01 + 2*x31 + 9*c97 + 4*ab12 + 5*eaz + 5*f12 <= 22
Lp_prob += 10*a01 + 4*x31 + 10*c97 + 8*ab12 + 8*eaz + 9*f12 <= 34
# Display the problem
print(Lp_prob)
status = Lp_prob.solve() # Solver
print(pulp.LpStatus[status]) # The solution status
# Printing the final solution
print(pulp.value(a01), pulp.value(x31), pulp.value(c97), pulp.value(ab12),
pulp.value(eaz), pulp.value(f12), pulp.value(Lp_prob.objective))
我想做的是：
1-) 在列表中加载 a01、x31、c97、ab12、eaz、f12，然后将它们加载到 pulp 中。原因是此列表将包含大约 20,000 个项目。它会将 #Create 问题变量减少到一两行代码。
2-) 加载列表中的所有约束编号，并将它们加载到 pulp 中。原因是此列表将包含大约 1,000 个项目。它会将#Constraint 减少到一两行代码。
3-) 也将纸浆溶液列入列表。
原矩阵为：
p q r s Revenue
a 3 4 6 10 100
b 10 9 2 4 110
c 10 3 9 10 150
d 2 9 4 8 90
e 4 1 5 8 50
f 6 4 5 9 60
Quantity available 25 21 22 34
您可以使用 for 循环和 pulp.lpSum 方法的组合来完成您想要的：
import pulp
import pandas as pd
# List of tuples
names = ['a01', 'x31', 'c97', 'ab12', 'eaz', 'f12']
data = [(3, 4, 6, 10, 100),
(10, 9, 2, 4, 110),
(10, 3, 9, 10, 150),
(2, 9, 4, 8, 90),
(4, 1, 5, 8, 50),
(6, 4, 5, 9, 60)]
qty = [25, 21, 22, 34]
# Create DataFrame object from list of tuples
data_frame = pd.DataFrame(data,
columns=['p' , 'q', 'r' , 's', 'rev'],
index=names)
# Create a LP Minimization problem
Lp_prob = pulp.LpProblem('Problem', pulp.LpMaximize)
# Create problem Variables
list_of_vars = [pulp.LpVariable(name, 0, 1, cat="Integer") for name in names]
# Objective Function
Lp_prob += pulp.lpSum([data_frame.rev[i]*list_of_vars[i] for i in range(len(data_frame))])
# Constraints:
for i in range(len(qty)):
Lp_prob += pulp.lpSum([data_frame.iloc[j,i] * list_of_vars[j]
for j in range(len(list_of_vars))]) <= qty[i]
# Display the problem
print(Lp_prob)
status = Lp_prob.solve() # Solver
print(pulp.LpStatus[status]) # The solution status
# Printing all the vars:
for v in Lp_prob.variables ():
print (v.name, "=", v.varValue)
如果您的数据发生变化并允许您添加新的物料清单、项目、比例等，此方法将起作用。您只需将您的 csv 映射到 boms 列表：
编辑：添加了 csv 加载，我的 csv 看起来像这样，但您可以更改列 headers 或交换您的实际数据，它们只需要匹配 available_parts 键。
csv 格式：
import csv
import pulp
def open_csv(path):
'''return a list of dictionaries
'''
with open(path, 'r') as file:
reader = csv.DictReader(file)
return [dict(row) for row in reader]
# load the bom item data
boms = open_csv('./boms.csv')
# define the available parts, these will be matched with columns from the csv so they must match
available_parts = {'p': 25, 'q': 21, 'r': 22, 's': 34}
# {LpVariable: item} dictionary for easy referencing later
variables = {pulp.LpVariable(i['name'], 0, 1, cat='Integer'): i
for i in boms}
prob = pulp.LpProblem('Problem', pulp.LpMaximize)
# add variable*revenue for each variable (dict we creted earlier makes this easier)
prob += pulp.lpSum([variable * float(item['revenue'])
for variable, item in variables.items()])
for part, amount in available_parts.items():
# lp_sum of the boms each part in the available parts list
maximum = pulp.lpSum([float(bom[part]) * prob.variablesDict()[bom['name']]
for bom in boms])
# add the part
prob += maximum <= amount, f'max_{part}'
# Display the problem
print(prob)
status = prob.solve() # Solver
print(pulp.LpStatus[status]) # The solution status
# Printing the final solution
output = {v.name: v.varValue for v in prob.variables()}
print(output)
输出：
Problem:
MAXIMIZE
100*a01 + 90*ab12 + 150*c97 + 50*eaz + 60*f12 + 110*x31 + 0
SUBJECT TO
max_p: 3 a01 + 2 ab12 + 10 c97 + 4 eaz + 6 f12 + 10 x31 <= 25
max_q: 4 a01 + 9 ab12 + 3 c97 + eaz + 4 f12 + 9 x31 <= 21
max_r: 6 a01 + 4 ab12 + 9 c97 + 5 eaz + 5 f12 + 2 x31 <= 22
max_s: 10 a01 + 8 ab12 + 10 c97 + 8 eaz + 9 f12 + 4 x31 <= 34
VARIABLES
0 <= a01 <= 1 Integer
0 <= ab12 <= 1 Integer
0 <= c97 <= 1 Integer
0 <= eaz <= 1 Integer
0 <= f12 <= 1 Integer
0 <= x31 <= 1 Integer
Optimal
{'a01': 1.0, 'ab12': 0.0, 'c97': 1.0, 'eaz': 0.0, 'f12': 0.0, 'x31': 1.0}
360.0
我有以下python纸浆代码：
import pulp
# Create a LP Minimization problem
Lp_prob = pulp.LpProblem('Problem', pulp.LpMaximize)
# Create problem Variables
a01 = pulp.LpVariable("a01", 0, 1, cat="Integer") # Create a variable x >= 0
x31 = pulp.LpVariable("x31", 0, 1, cat="Integer") # Create a variable x >= 0
c97 = pulp.LpVariable("c97", 0, 1, cat="Integer") # Create a variable x >= 0
ab12 = pulp.LpVariable("ab12", 0, 1, cat="Integer") # Create a variable x >= 0
eaz = pulp.LpVariable("eaz", 0, 1, cat="Integer") # Create a variable x >= 0
f12 = pulp.LpVariable("f12", 0, 1, cat="Integer") # Create a variable x >= 0
# Objective Function
Lp_prob += 100*a01 + 110*x31 + 150*c97 + 90*ab12 + 50*eaz + 60*f12
# Constraints:
Lp_prob += 3*a01 + 10*x31 + 10*c97 + 2*ab12 + 4*eaz + 6*f12 <= 25
Lp_prob += 4*a01 + 9*x31 + 3*c97 + 9*ab12 + 1*eaz + 4*f12 <= 21
Lp_prob += 6*a01 + 2*x31 + 9*c97 + 4*ab12 + 5*eaz + 5*f12 <= 22
Lp_prob += 10*a01 + 4*x31 + 10*c97 + 8*ab12 + 8*eaz + 9*f12 <= 34
# Display the problem
print(Lp_prob)
status = Lp_prob.solve() # Solver
print(pulp.LpStatus[status]) # The solution status
# Printing the final solution
print(pulp.value(a01), pulp.value(x31), pulp.value(c97), pulp.value(ab12),
pulp.value(eaz), pulp.value(f12), pulp.value(Lp_prob.objective))
我想做的是：
1-) 在列表中加载 a01、x31、c97、ab12、eaz、f12，然后将它们加载到 pulp 中。原因是此列表将包含大约 20,000 个项目。它会将 #Create 问题变量减少到一两行代码。
2-) 加载列表中的所有约束编号，并将它们加载到 pulp 中。原因是此列表将包含大约 1,000 个项目。它会将#Constraint 减少到一两行代码。
3-) 也将纸浆溶液列入列表。
原矩阵为：
p q r s Revenue
a 3 4 6 10 100
b 10 9 2 4 110
c 10 3 9 10 150
d 2 9 4 8 90
e 4 1 5 8 50
f 6 4 5 9 60
Quantity available 25 21 22 34
您可以使用 for 循环和 pulp.lpSum 方法的组合来完成您想要的：
import pulp
import pandas as pd
# List of tuples
names = ['a01', 'x31', 'c97', 'ab12', 'eaz', 'f12']
data = [(3, 4, 6, 10, 100),
(10, 9, 2, 4, 110),
(10, 3, 9, 10, 150),
(2, 9, 4, 8, 90),
(4, 1, 5, 8, 50),
(6, 4, 5, 9, 60)]
qty = [25, 21, 22, 34]
# Create DataFrame object from list of tuples
data_frame = pd.DataFrame(data,
columns=['p' , 'q', 'r' , 's', 'rev'],
index=names)
# Create a LP Minimization problem
Lp_prob = pulp.LpProblem('Problem', pulp.LpMaximize)
# Create problem Variables
list_of_vars = [pulp.LpVariable(name, 0, 1, cat="Integer") for name in names]
# Objective Function
Lp_prob += pulp.lpSum([data_frame.rev[i]*list_of_vars[i] for i in range(len(data_frame))])
# Constraints:
for i in range(len(qty)):
Lp_prob += pulp.lpSum([data_frame.iloc[j,i] * list_of_vars[j]
for j in range(len(list_of_vars))]) <= qty[i]
# Display the problem
print(Lp_prob)
status = Lp_prob.solve() # Solver
print(pulp.LpStatus[status]) # The solution status
# Printing all the vars:
for v in Lp_prob.variables ():
print (v.name, "=", v.varValue)
如果您的数据发生变化并允许您添加新的物料清单、项目、比例等，此方法将起作用。您只需将您的 csv 映射到 boms 列表：
编辑：添加了 csv 加载，我的 csv 看起来像这样，但您可以更改列 headers 或交换您的实际数据，它们只需要匹配 available_parts 键。
csv 格式：
import csv
import pulp
def open_csv(path):
'''return a list of dictionaries
'''
with open(path, 'r') as file:
reader = csv.DictReader(file)
return [dict(row) for row in reader]
# load the bom item data
boms = open_csv('./boms.csv')
# define the available parts, these will be matched with columns from the csv so they must match
available_parts = {'p': 25, 'q': 21, 'r': 22, 's': 34}
# {LpVariable: item} dictionary for easy referencing later
variables = {pulp.LpVariable(i['name'], 0, 1, cat='Integer'): i
for i in boms}
prob = pulp.LpProblem('Problem', pulp.LpMaximize)
# add variable*revenue for each variable (dict we creted earlier makes this easier)
prob += pulp.lpSum([variable * float(item['revenue'])
for variable, item in variables.items()])
for part, amount in available_parts.items():
# lp_sum of the boms each part in the available parts list
maximum = pulp.lpSum([float(bom[part]) * prob.variablesDict()[bom['name']]
for bom in boms])
# add the part
prob += maximum <= amount, f'max_{part}'
# Display the problem
print(prob)
status = prob.solve() # Solver
print(pulp.LpStatus[status]) # The solution status
# Printing the final solution
output = {v.name: v.varValue for v in prob.variables()}
print(output)
输出：
Problem:
MAXIMIZE
100*a01 + 90*ab12 + 150*c97 + 50*eaz + 60*f12 + 110*x31 + 0
SUBJECT TO
max_p: 3 a01 + 2 ab12 + 10 c97 + 4 eaz + 6 f12 + 10 x31 <= 25
max_q: 4 a01 + 9 ab12 + 3 c97 + eaz + 4 f12 + 9 x31 <= 21
max_r: 6 a01 + 4 ab12 + 9 c97 + 5 eaz + 5 f12 + 2 x31 <= 22
max_s: 10 a01 + 8 ab12 + 10 c97 + 8 eaz + 9 f12 + 4 x31 <= 34
VARIABLES
0 <= a01 <= 1 Integer
0 <= ab12 <= 1 Integer
0 <= c97 <= 1 Integer
0 <= eaz <= 1 Integer
0 <= f12 <= 1 Integer
0 <= x31 <= 1 Integer
Optimal
{'a01': 1.0, 'ab12': 0.0, 'c97': 1.0, 'eaz': 0.0, 'f12': 0.0, 'x31': 1.0}
360.0