tupledict.prod() with partial dictionary key

Answered

Panggah Prabawa

• March 25, 2024 09:13
• Edited

Hi,

Similar to that in the documentation, suppose that I have 3-D variable x:

# My trial:
x = m.addVars([(1,2,0), (1,3,0), (2,3,0)]) # The indices are generated from tuplelist, in which the last index can span to T.
coeff = dict([((1,2), 2.0), ((1,3), 2.1), ((2,3), 3.3)])
expr = x.prod(coeff,1,'*','*')

# The result:
<gurobi.LinExpr: 0.0>

# Desired result:
2.0 x[1,2,0] + 2.1 x[1,3,0] + 3.3 x[2,3,0]

Suppose that the variables \( x \) vary with t and the \( coeff \) does not, how do I get a workaround?

Thank you!

Related to

• Model Building

1

• 

  Jaromił Najman

  Gurobi Staff

  • March 19, 2024 11:09

  You are getting a \(0\) result, because the tuple key for your variables are of 3 while the ones for coeff are of length 2.

  You can do

  # My trial:
  x = m.addVars([(1,2,0), (1,3,0), (2,3,0)]) # The indices are generated from tuplelist, in which the last index can span to T.
  m.update()
  coeff = dict([((1,2,0), 2.0), ((1,3,0), 2.1), ((2,3,0), 3.3)])
  expr = x.prod(coeff,'*',3,'*')
  
  print(expr)
  
  > 2.1 C1 + 3.3 C2

  The pattern states to use only the entries with \(3\) in the second place.

  0
• 

  Panggah Prabawa

  • March 20, 2024 03:20

  

  

  

  Thank you for your response.

  I know that is a way to do that.
  Let's say that the index is a tuple (i,j,t), where t < T.
  I maintain parameter coeff to have 2 key lengths to avoid iterating coeff over T.
  This is a simple reproducible code, where the actual simulation has more than 100 (i,j) pairs and T > 100.

  So, is it still impossible to use variable x and coeff with different key lengths?
  Is there any other workaround? Or is it okay to just copy the coeff T times?

  0
• 

  Jaromił Najman

  Gurobi Staff

  • March 20, 2024 11:53
  • Edited

  So, is it still impossible to use variable x and coeff with different key lengths?

  No, this is not possible.

  Is there any other workaround? Or is it okay to just copy the coeff T times?

  There may be a workaround but I am not aware of one. Copying coeff T times should work well enough.

  1
• 

  Riley Clement

  Gurobi Staff

  • March 20, 2024 13:09
  • Edited

  Hi Panggah,

  Jarek has given me the green light to post some creative solutions if you want to keep your data in its current format, which does have it's advantages - not sure if speed is one of them though.

  # import gurobipy as gp
  
  gp.quicksum(coeff[k1,k2]*v for (k1,k2,_),v in x.items())
  
  gp.quicksum(coeff*x.sum(*k,"*") for k,coeff in coeff.items())
  
  gp.quicksum(coeff*x.sum(k1,k2,"*") for (k1,k2),coeff in coeff.items())
  
  gp.tupledict({k:x.sum(*k,"*") for k in coeff.keys()}).prod(coeff)

  - Riley

  1
• 

  Panggah Prabawa

  • March 21, 2024 07:03

  

  

  

  Thank you for your answers, Jaromil and Riley.
  I will test the creation speeds of those options. Very interesting.

   

  0
• 

  Riley Clement

  Gurobi Staff

  • March 21, 2024 07:08

  No problem, my advice for comparison is to use the following approach as an example:

  import numpy as np
  import timeit
  
  
  setup = """
  import gurobipy as gp
  m = gp.Model()
  x = m.addVar()
  y = m.addVar()
  """
  
  
  statement1= """
  m.addConstr(x <= 2)
  """
  
  statement2= """
  m.addLConstr(x,"<=",2)
  """
  
  factor = np.divide(
      np.min(timeit.repeat(stmt=statement1, setup=setup, number=1, repeat=1000)),
      np.min(timeit.repeat(stmt=statement2, setup=setup, number=1, repeat=1000)),
  )

  It runs the code many times, then takes the minimum for comparison, to reduce the effect of noise on your machine.

  - Riley

  1

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