Outer product in CVXPY












0















I have an objective function that depends on a sum of outer products. If I could use Numpy functions, I would write this as:



A = np.ones(m, n)

U = Variable(m, n)

objective = np.trace(sum([np.outer(A[i,:], U[i,:]) for i in range(m)]))


Of course np.outer doesn't work when U is a variable. Is there a cvxpy affine function that would implement this?






python numpy cvxpy







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    0















    I have an objective function that depends on a sum of outer products. If I could use Numpy functions, I would write this as:



    A = np.ones(m, n)
    
U = Variable(m, n)
    
objective = np.trace(sum([np.outer(A[i,:], U[i,:]) for i in range(m)]))


    Of course np.outer doesn't work when U is a variable. Is there a cvxpy affine function that would implement this?






    python numpy cvxpy







    share|improve this question

























      0












      0








      0








      I have an objective function that depends on a sum of outer products. If I could use Numpy functions, I would write this as:



      A = np.ones(m, n)
      
U = Variable(m, n)
      
objective = np.trace(sum([np.outer(A[i,:], U[i,:]) for i in range(m)]))


      Of course np.outer doesn't work when U is a variable. Is there a cvxpy affine function that would implement this?






      python numpy cvxpy







      share|improve this question














      I have an objective function that depends on a sum of outer products. If I could use Numpy functions, I would write this as:



      A = np.ones(m, n)
      
U = Variable(m, n)
      
objective = np.trace(sum([np.outer(A[i,:], U[i,:]) for i in range(m)]))


      Of course np.outer doesn't work when U is a variable. Is there a cvxpy affine function that would implement this?






      python numpy cvxpy




      python numpy cvxpy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 24 '18 at 7:37









      geoffn91geoffn91

      10413




      10413
























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          cvxpy has a kron function that works. Since it only works on 2d Variables and 2d-arrays, you have to reshape the vectors to (n, 1) vectors first:



          cvx.trace(sum([cvx.kron(A[i, :].reshape(n,1), cvx.reshape(U[i,:], (n,1))
          
               for i in range(m)]))





          share|improve this answer























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            1 Answer
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            active

            oldest

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            1














            cvxpy has a kron function that works. Since it only works on 2d Variables and 2d-arrays, you have to reshape the vectors to (n, 1) vectors first:



            cvx.trace(sum([cvx.kron(A[i, :].reshape(n,1), cvx.reshape(U[i,:], (n,1))
            
               for i in range(m)]))





            share|improve this answer




























              1














              cvxpy has a kron function that works. Since it only works on 2d Variables and 2d-arrays, you have to reshape the vectors to (n, 1) vectors first:



              cvx.trace(sum([cvx.kron(A[i, :].reshape(n,1), cvx.reshape(U[i,:], (n,1))
              
               for i in range(m)]))





              share|improve this answer


























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                1








                1







                cvxpy has a kron function that works. Since it only works on 2d Variables and 2d-arrays, you have to reshape the vectors to (n, 1) vectors first:



                cvx.trace(sum([cvx.kron(A[i, :].reshape(n,1), cvx.reshape(U[i,:], (n,1))
                
               for i in range(m)]))





                share|improve this answer













                cvxpy has a kron function that works. Since it only works on 2d Variables and 2d-arrays, you have to reshape the vectors to (n, 1) vectors first:



                cvx.trace(sum([cvx.kron(A[i, :].reshape(n,1), cvx.reshape(U[i,:], (n,1))
                
               for i in range(m)]))






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 24 '18 at 7:37









                geoffn91geoffn91

                10413




                10413






























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