"object has no attribute 'ctype'

[Jeffrey Kantor]

All …

I’m continuing to work with pyomo.kernel, and have come across the following error message. Do I need to somehow specify a data type for the summation expression?

Jeff

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
Input In [79], in <cell line: 26>()
     22         m.d[j] = m.z[j] == sum([m.a[i] for i in range(n)])
     24     return m
---> 26 svm_conic(X_train, y_train, 1)

Input In [79], in svm_conic(X, y, c)
     20 m.d = pmo.constraint_dict()
     21 for j in range(p + 1):
---> 22     m.d[j] = m.z[j] == sum([m.a[i] for i in range(n)])
     24 return m

File ~/opt/anaconda3/lib/python3.9/site-packages/pyomo/core/kernel/dict_container.py:78, in DictContainer.__setitem__(self, key, item)
     77 def __setitem__(self, key, item):
---> 78     if item.ctype is self.ctype:
     79         if item._parent is None:
     80             if key in self._data:

AttributeError: 'EqualityExpression' object has no attribute 'ctype'

import pyomo.kernel as pmo

def svm_conic(X, y, c):

    

    n, p = X.shape

    

    F = np.append(np.ones((n, 1)), X.to_numpy(), axis=1)

    m = pmo.block()

    

    m.r = pmo.variable()

    m.a = pmo.variable_dict()

    for i in range(n):

        m.a[i] = pmo.variable(domain_type=pmo.RealSet)

        

    m.z = pmo.variable_dict()

    for j in range(p + 1):

        m.z[j] = pmo.variable(domain_type=pmo.RealSet)

        

    m.d = pmo.constraint_dict()

    for j in range(p + 1):

        m.d[j] = m.z[j] == sum([m.a[i] for i in range(n)])

    return m

svm_conic(X_train, y_train, 1)

[Gabriel Hackebeil]

With kernel, assignments to the containers have to be done with explicit objects of the container type. In your case, you need to assign a ‘kernel.constraint(…)’ to m.d[j]. You just pass the equality expression into the constraint object as the sole argument, but there are more efficient ways to create it by using body and rhs keywords for constraint.

Gabe

On Apr 27, 2022, at 3:56 PM, Jeffrey Kantor <jeff....@gmail.com> wrote:

All …

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[Gabriel Hackebeil]

Adding to this… requiring explicit assignments makes it a bit easier to support other nice things. For instance you can assign `kernel.linear_constraint` objects, which make building the linear constraints of a model much faster. You can also assign a constraint container (e.g., to have a nested container structure) and other constraint subclasses.

Gabe

[Jeffrey Kantor]

Gabe,

Thanks, that’s perfectly clear and, in retrospect, makes sense.

Diving into this deeper, I’d like to use kernel.matrix_constraint to build a constraint  y = A x where x and y are both variable lists, and A is a matrix.  Perhaps I could do something like [A -I] * [x y] = 0, but I’m not understanding how to use the container objects to make this simple.  Any advice?

Jeff

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[Gabriel Hackebeil]

Here’s an example, Jeff. You can just treat it like any another kernel.constraint() and put it on a block or in a constraint container.

```

import pyomo.kernel as pmo
import numpy as np

m = pmo.block()
m.x = pmo.variable()
m.y = pmo.variable()

m.c = pmo.constraint_list()
for i in range(3):
    m.c.append(
        # this can be assigned to a block or put in a constraint container                                                      
        # it is a constraint_tuple                                                                                              
        pmo.matrix_constraint(
            np.random.rand(3, 2), # a sparse or dense matrix                                                                    
            x=[m.x, m.y], # also fine to use a variable_list()                                                                  
            lb=0, # a number or array                                                                                           
            ub=1, # a number or array                                                                                           
            #rhs=..., # same, but only use for equality constraint                                                              
            sparse=False, # set this if do not want to use scipy.sparse storage                                                 
        )
    )
```

Gabe

To view this discussion on the web visit https://groups.google.com/d/msgid/pyomo-forum/9A1FBCEA-1C51-4187-9D51-093305ED36FF%40gmail.com.

[Gabriel Hackebeil]

For your example, you will need to compose the final Ax = b matrix / rhs forms, but you can likely do that using scipy block-/sparse-matrix functionality without too much pain.

Gabe