model.addVars

Answered

Emma Innocente

March 12, 2024 09:07

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Hello, 

I have two lists of tuples, as follows : 

L = [('BE212', 'BE352', 0, 0), ('BE212', 'BE211', 0, 0), ('BE242', 'BE100', 0, 0), ('BE236', 'BE232', 0, 0), ('BE212', 'BE343', 0, 0)]
R = [('BE310', 'BE242', 0), ('BE334', 'BE332', 0), ('BE257', 'BE254', 0), ('BE341', 'BE241', 0), ('BE223', 'BE332', 0), ('BE258', 'BE256', 0), ('BE310', 'BE352', 0), ('BE242', 'BE100', 0), ('BE212', 'BE236', 0), ('BE235', 'BE234', 0)]

Using Gurobipy, I would like to create a binary variable x_rl with the index r being the tuples in R and the index l being the tuples in L. I want to be able to reference my variables with the tuples directly, and not through for instance x_13 with r=1 standing for the first element in R and l=3 standing for the thirdh elemnt in L. I want to be able to reference my viriable as x_('BE310', 'BE242', 0)('BE242', 'BE100', 0, 0). 

The following doesn't work :

x = model.addVars(R, L, vtype=gp.GRB.BINARY)

Thank you for your help ! 

0

• Official comment

  

  Simranjit Kaur

  • Gurobi Staff

  November 13, 2026 01:28

  This post is more than two years old. Some information may not be up to date. For current information, please check the Gurobi Documentation or Knowledge Base. If you need more help, please create a new post in the community forum, or try Gurobot, our chatbot interface offering instant, expert-level support.
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  Jonasz Staszek

  • Community Moderator

  March 12, 2024 12:01

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  You can do that by looping over both your lists:

  x = gp.tupledict()
  for l in L:
      for r in R:
          x[l,r] = m.addVar(vtype="B", name=f"x_{l}{r}")

  The reasons why your code doesn't work can be found here.

  Hope this helps.

  Best regards
  Jonasz

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• 

  Riley Clement

  • Gurobi Staff

  March 12, 2024 12:09 Edited

  You can also appeal to itertools.product to create a single index set, but this will run a bit slower than the solution from Jonasz

  import itertools
  x = m.addVars(itertools.product(R,L), vtype="B", name="x_")

   - Riley

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