CODEWITHSUNDEEP
pythoncplexdocplex
zhongyao zhang
zhongyao zhang

Reputation: 21

how to set a four dimension variable in docplex with python?

I want to set a binary four-dimensions variable, like X[a][b][c][d], bur docplex only have function binary_var_cube to set a three-dimension. How can I to create a four-dimension? I have found that someone use this to create a three-dimension and said it could extend to more dimensions.But it isn't useful.

binary_var_dict((a,b,c) for a in ... for b in ... for c in ...)

Upvotes: 0

Views: 1465

Answers (2)

Alex Fleischer
Alex Fleischer

Reputation: 10062

Let me share a tiny example:

from docplex.mp.model import Model

# Data

r=range(1,3)

i=[(a,b,c,d) for a in r for b in r for c in r for d in r]

print(i)

mdl = Model(name='model')

#decision variables
mdl.x=mdl.integer_var_dict(i,name="x")

# Constraint
for it in i:
    mdl.add_constraint(mdl.x[it] == it[0]+it[1]+it[2]+it[3], 'ct')

mdl.solve()

# Dislay solution
for it in i:
    print(" x ",it," --> ",mdl.x[it].solution_value);

which gives

[(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 2, 1), (1, 1, 2, 2), (1, 2, 1, 1), (1, 2, 1, 2), (1, 2, 2, 1), (1, 2, 2, 2), (2, 1, 1, 1), (2, 1, 1, 2), (2, 1, 2, 1), (2, 1, 2, 2), (2, 2, 1, 1), (2, 2, 1, 2), (2, 2, 2, 1), (2, 2, 2, 2)]
 x  (1, 1, 1, 1)  -->  4.0
 x  (1, 1, 1, 2)  -->  5.0
 x  (1, 1, 2, 1)  -->  5.0
 x  (1, 1, 2, 2)  -->  6.0
 x  (1, 2, 1, 1)  -->  5.0
 x  (1, 2, 1, 2)  -->  6.0
 x  (1, 2, 2, 1)  -->  6.0
 x  (1, 2, 2, 2)  -->  7.0
 x  (2, 1, 1, 1)  -->  5.0
 x  (2, 1, 1, 2)  -->  6.0
 x  (2, 1, 2, 1)  -->  6.0
 x  (2, 1, 2, 2)  -->  7.0
 x  (2, 2, 1, 1)  -->  6.0
 x  (2, 2, 1, 2)  -->  7.0
 x  (2, 2, 2, 1)  -->  7.0
 x  (2, 2, 2, 2)  -->  8.0

Upvotes: 2

Xavier Nodet
Xavier Nodet

Reputation: 5105

Here's an answer from Daniel Junglas, copied almost verbatim from https://developer.ibm.com/answers/questions/385771/decision-matrices-with-more-than-3-variables-for-d/

You can use tuples of any arity as keys to access the variables from a dictionary:

 x = m.binary_var_dict((i, l, t, v, r)
                       for i in types
                       for l in locations
                       for t in times
                       for v in vehicles
                       for r in routes)

You can then access the variables using:

for i in types:
    for l in locations:
       for t in times:
          for v in vehicles:
             for r in routes:
                print x[i, l, t, v, r]

You could as well use:

 x = [[[[[m.binary_var('%d_%d_%d_%d_%d' % (i, l, t, v, r))
          for r in routes]
         for v in vehicles]
        for t in times]
       for l in locations]
      for i in types]

 for i in types:
    for l in locations:
       for t in times:
          for v in vehicles:
             for r in routes:
                print x[i][l][t][v][r]

but this method doesn't support sparse dimensions, and requires many more brackets to express.

Upvotes: 2

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