How to store a solution from gurobi optimizer in a list in python ?
AnsweredI have an assignment problem where I used gurobi to optimise my model. I have results in this format when I use the optimize() function :
assign[287,128]:1.0 assign[292,129]:1.0 assign[271,130]:1.0 assign[272,131]:1.0 assign[270,132]:1.0
How do I store them in list ?
new_list = [("287, "128") , ("292", "129")..........]
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Official comment
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Do you want the list to contain the indices of the \( \texttt{assign} \) variables for which the corresponding \( \texttt{assign} \) variable takes a value of 1 at the optimal solution? Assuming you defined your \( \texttt{assign} \) variables like
assign = m.addVars(...)
you can do this with a list comprehension:
new_list = [k for k, v in assign.items() if v.X > 0.5]
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Hi Eli,
I used the list comprehension as you said and my combinations have been stored. Thank you for you help.
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Hi Eli,
I have another question please. How I store the costs related to each variable.
By using multidict(), I am attributing costs to each variable as:
combinations, ms = multidict({k: v for k, v in zip(list3, Costs)})
x = m.addVars(combinations, name='assign')
when I print ms I have this:
('287','128'): 2 ('292','129'): 3 ('271','130'): 1 ('272','131']: 0
etc....
After running the model I want to store the delays assigned. How is it possible ?0 -
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Could you clarify what the delays are in your code, and what format you're trying to store them in?
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I have two lists:
Costs = [2, 3, 1, 0...]
list3 = [('287','128'), ('292','129'), ('271','130'), ('272','131')...]I am using this code below to assign the cost to each combination. (I have more than thousands combination and associated cost, I just wrote few of them in this example)
combinations, ms = multidict({k: v for k, v in zip(list3, Costs)})If I print ms I will have:
{('287','128'): 2, ('292','129'): 3, ('271','130'): 1, ('272','131'): 0...}When I run my model I have this result :
assign[287,128]:1.0 assign[292,129]:1.0 assign[271,130]:1.0 assign[272,131]:1.0
But it does not show the costs associated with every variable assigned after the optimisation, and I would like to store the cost in a new list as :
new_cost = [2, 3, 1, 0]
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The construction is basically the same, with the extra step of mapping the keys to the appropriate costs using the \( \texttt{ms} \) dictionary:
cost_list = [ms[k] for k, v in x.items() if v.X > 0.5]
Also, I don't see a reason to use multidict here, since you are just constructing a single dictionary. You could instead write
ms = {k: v for k, v in zip(list3, Costs)}to obtain the same \( \texttt{ms} \) dictionary. Then, you can use the \( \texttt{list3} \) list instead of the \( \texttt{combinations} \) list, since these should be the same.
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