How do I generate PuLP variables and constrains without using exec?

Question

I have written the following Python Code using the PuLP Library for solving the Knapsack Problem using the Integer Programming formulation. I am using strings to generate the LpVariable commands and add the constraints and then executing them with eval. Is there a way to do this without using eval?

from pulp import *

#Knapsack problem

items = input ('Enter the number of items :')

items = int(items)

#print('Enter %d items one by one')

print ('Enter {0} items profit one by one'.format(items))

obj = []
weight = []
knapweight = 0


for i in range(0,items):
    print('Enter {0} item profit : '.format(i+1))
    obj.append(input())

for i in range(0, items):
    print('The profit at {0} is {1}'.format(i, obj[i]))

print ('\nEnter {0} items weights one by one'.format(items))


for i in range(0, items):
    print('Enter {0} item weight'.format(i+1))
    weight.append(input())

for i in range(0, items):
    print('The profit at {0} is {1}'.format(i, weight[i]))

print ('\nEnter the weight of the knapsack :')
knapweight = input()

print ('The weight of the knapsack is : {0}'.format(knapweight))


#generating variables
for i in range(0, items):
    str = 'x{0} = LpVariable("x{1}", cat=\'Binary\')'.format(i+1,i+1)
    print (str)
    exec(str)

prob = LpProblem('Knapsack', LpMaximize)

print ('\nDynamic Generaion\n')

#weight constraint generation
str = "prob += "

for i in range(0, items):
    if i == (items-1):
        str = str + weight[i] + '*x{0}'.format(i+1)
    else:
        str = str + weight[i] + '*x{0}'.format(i+1) + '+'

str = str + '<=' + knapweight
exec(str)
print(str)

#objective function generation
str = "prob += "

for i in range(0, items):
    if i == (items-1):
        str = str + obj[i] + '*x{0}'.format(i+1)
    else:
        str = str + obj[i] + '*x{0}'.format(i+1) + '+'

exec(str)
print(str)

status = prob.solve()
print(LpStatus[status])

print ('\nThe values of the variables : \n')

for i in range(0, items):
    print('x{0} = {1}'.format(i+1, value(eval('x{0}'.format(i+1)))))

Answer 1

The key is recognizing that it's okay to have an object -- say one in a list or a dictionary -- which you don't have explicitly bound to a name. You can make an object, append it to a list, and only ever refer to it as some_list[2]. Once you allow yourself that freedom, your code can become much simpler.

Here I've hardcoded the inputs, because that doesn't matter:

from pulp import *

objs = [2,3,2,5,3]
weights = [1,2,2,1,3]
knapweight = 5

prob = LpProblem('Knapsack', LpMaximize)
xs = [LpVariable("x{}".format(i+1), cat="Binary") for i in range(len(objs))]

# add objective
total_prof = sum(x * obj for x,obj in zip(xs, objs))
prob += total_prof

# add constraint
total_weight = sum(x * w for x,w in zip(xs, weights))
prob += total_weight <= knapweight

status = prob.solve()
print(LpStatus[status])
print("Objective value:", value(prob.objective))
print ('\nThe values of the variables : \n')
for v in prob.variables():
    print(v.name, "=", v.varValue)

which gives me

Optimal
Objective value: 10.0

The values of the variables : 

x1 = 1.0
x2 = 1.0
x3 = 0.0
x4 = 1.0
x5 = 0.0

Here I'm building the LpVariables in the list comprehension

xs = [LpVariable("x{}".format(i+1), cat="Binary") for i in range(len(objs))]

where the objects then just live in the list xs.

>>> xs
[x1, x2, x3, x4, x5]
>>> xs[3]
x4
>>> type(xs[3])
<class 'pulp.pulp.LpVariable'>