KeyError: 0e

Answered

Maide Ceren Kaymaz

April 29, 2023 17:24 Edited

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Hello, I define the binary variables x and y as follows:

y = g.addVars(10, 5, GRB.BINARY, name='y_tl')
x = g.addVars(10, GRB.BINARY, name='x_l')

Here is the constraint I'd like to define:

and the formulation I did in Gurobi is:
task_assignment=g.addConstrs((y.sum('*',l)<= x[l]*1000 for l in locations), name="task_assignment")

but it's giving error:

KeyError: 0

What could be the problem here?

Thanks in advance!

 

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• Python
• Error

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• Official comment

  

  Simranjit Kaur

  • Gurobi Staff

  November 13, 2026 01:44

  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.
• 

  Riley Clement

  • Gurobi Staff

  April 30, 2023 11:14

  Hi Maide,

  If you were to print the keys of your y tupledict you would see the following

  >> print(y.keys())
  <gurobi.tuplelist (50 tuples, 3 values each):
   ( 0 , 0 , B )
   ( 0 , 1 , B )
   ( 0 , 2 , B )
   ( 0 , 3 , B )
  ...

  There is a third index with values "B" that you did not intend and this will give you a clue as to what is happening.

  Let's review the method signature for addVars from the reference manual:

  addVars ( *indices, lb=0.0, ub=float('inf'), obj=0.0, vtype=GRB.CONTINUOUS, name="" ) 

  The * has special meaning here and allows an arbitrary number of arguments, to index the variables, to be passed to the function.  However this means that every parameter that comes after it must have arguments passed via keyword.  This is why GRB.BINARY ( = "B") is being interpreted as another indice.

  So to fix this pass the vtype via keyword, i.e. vtype=GRB.BINARY (or vtype="B").

  - Riley

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