Fast outer tensor product in numpy












0















I have two numpy arrays: 



x of shape ((d1,...,d_m)) 

y of shape ((e_1,...e_n))


I would like to form the outer tensor product, that is the numpy array 



z of shape ((d1,...,d_m,e_1,...,e_n))


such that 



z[i_1,...,i_n,i_{n+1}...,i_{m+n}] == x[i_1,...i_m]*y[i_{m+1},...,i_{m+n}]


I have to perform the above outer multiplication several times so I would like to speed this up as much as possible.






python arrays numpy linear-algebra







share|improve this question























  • insert dummy data with your expected result to understand better what do you mean

    – Francesco Nazzaro
    May 26 '16 at 15:19
















0















I have two numpy arrays: 



x of shape ((d1,...,d_m)) 

y of shape ((e_1,...e_n))


I would like to form the outer tensor product, that is the numpy array 



z of shape ((d1,...,d_m,e_1,...,e_n))


such that 



z[i_1,...,i_n,i_{n+1}...,i_{m+n}] == x[i_1,...i_m]*y[i_{m+1},...,i_{m+n}]


I have to perform the above outer multiplication several times so I would like to speed this up as much as possible.






python arrays numpy linear-algebra







share|improve this question























  • insert dummy data with your expected result to understand better what do you mean

    – Francesco Nazzaro
    May 26 '16 at 15:19














0












0








0


1






I have two numpy arrays: 



x of shape ((d1,...,d_m)) 

y of shape ((e_1,...e_n))


I would like to form the outer tensor product, that is the numpy array 



z of shape ((d1,...,d_m,e_1,...,e_n))


such that 



z[i_1,...,i_n,i_{n+1}...,i_{m+n}] == x[i_1,...i_m]*y[i_{m+1},...,i_{m+n}]


I have to perform the above outer multiplication several times so I would like to speed this up as much as possible.






python arrays numpy linear-algebra







share|improve this question














I have two numpy arrays: 



x of shape ((d1,...,d_m)) 

y of shape ((e_1,...e_n))


I would like to form the outer tensor product, that is the numpy array 



z of shape ((d1,...,d_m,e_1,...,e_n))


such that 



z[i_1,...,i_n,i_{n+1}...,i_{m+n}] == x[i_1,...i_m]*y[i_{m+1},...,i_{m+n}]


I have to perform the above outer multiplication several times so I would like to speed this up as much as possible.






python arrays numpy linear-algebra




python arrays numpy linear-algebra






share|improve this question













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asked May 26 '16 at 15:10









factfact

13129




13129













  • insert dummy data with your expected result to understand better what do you mean

    – Francesco Nazzaro
    May 26 '16 at 15:19



















  • insert dummy data with your expected result to understand better what do you mean

    – Francesco Nazzaro
    May 26 '16 at 15:19

















insert dummy data with your expected result to understand better what do you mean

– Francesco Nazzaro
May 26 '16 at 15:19





insert dummy data with your expected result to understand better what do you mean

– Francesco Nazzaro
May 26 '16 at 15:19












2 Answers
2






active

oldest

votes


















2














An alternative to outer is to explicitly expand the dimensions. For 1d arrays this would be



x[:,None]*y   # y[None,:] is automatic.


For 10x10 arrays, and generalizing the dimension expansion, I get the same times



In [74]: timeit x[[slice(None)]*x.ndim + [None]*y.ndim] * y

10000 loops, best of 3: 53.6 µs per loop



In [75]: timeit np.multiply.outer(x,y)

10000 loops, best of 3: 52.6 µs per loop


So outer does save some coding, but the basic broadcasted multiplication is the same.






share|improve this answer































    5














    You want np.multiply.outer:



    z = np.multiply.outer(x, y)





    share|improve this answer
























    • That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

      – fact
      May 26 '16 at 15:50











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    An alternative to outer is to explicitly expand the dimensions. For 1d arrays this would be



    x[:,None]*y   # y[None,:] is automatic.


    For 10x10 arrays, and generalizing the dimension expansion, I get the same times



    In [74]: timeit x[[slice(None)]*x.ndim + [None]*y.ndim] * y
    
10000 loops, best of 3: 53.6 µs per loop
    

    
In [75]: timeit np.multiply.outer(x,y)
    
10000 loops, best of 3: 52.6 µs per loop


    So outer does save some coding, but the basic broadcasted multiplication is the same.






    share|improve this answer




























      2














      An alternative to outer is to explicitly expand the dimensions. For 1d arrays this would be



      x[:,None]*y   # y[None,:] is automatic.


      For 10x10 arrays, and generalizing the dimension expansion, I get the same times



      In [74]: timeit x[[slice(None)]*x.ndim + [None]*y.ndim] * y
      
10000 loops, best of 3: 53.6 µs per loop
      

      
In [75]: timeit np.multiply.outer(x,y)
      
10000 loops, best of 3: 52.6 µs per loop


      So outer does save some coding, but the basic broadcasted multiplication is the same.






      share|improve this answer


























        2












        2








        2







        An alternative to outer is to explicitly expand the dimensions. For 1d arrays this would be



        x[:,None]*y   # y[None,:] is automatic.


        For 10x10 arrays, and generalizing the dimension expansion, I get the same times



        In [74]: timeit x[[slice(None)]*x.ndim + [None]*y.ndim] * y
        
10000 loops, best of 3: 53.6 µs per loop
        

        
In [75]: timeit np.multiply.outer(x,y)
        
10000 loops, best of 3: 52.6 µs per loop


        So outer does save some coding, but the basic broadcasted multiplication is the same.






        share|improve this answer













        An alternative to outer is to explicitly expand the dimensions. For 1d arrays this would be



        x[:,None]*y   # y[None,:] is automatic.


        For 10x10 arrays, and generalizing the dimension expansion, I get the same times



        In [74]: timeit x[[slice(None)]*x.ndim + [None]*y.ndim] * y
        
10000 loops, best of 3: 53.6 µs per loop
        

        
In [75]: timeit np.multiply.outer(x,y)
        
10000 loops, best of 3: 52.6 µs per loop


        So outer does save some coding, but the basic broadcasted multiplication is the same.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered May 26 '16 at 16:33









        hpauljhpaulj

        112k780148




        112k780148

























            5














            You want np.multiply.outer:



            z = np.multiply.outer(x, y)





            share|improve this answer
























            • That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

              – fact
              May 26 '16 at 15:50
















            5














            You want np.multiply.outer:



            z = np.multiply.outer(x, y)





            share|improve this answer
























            • That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

              – fact
              May 26 '16 at 15:50














            5












            5








            5







            You want np.multiply.outer:



            z = np.multiply.outer(x, y)





            share|improve this answer













            You want np.multiply.outer:



            z = np.multiply.outer(x, y)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered May 26 '16 at 15:43









            EricEric

            66.6k31169278




            66.6k31169278













            • That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

              – fact
              May 26 '16 at 15:50



















            • That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

              – fact
              May 26 '16 at 15:50

















            That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

            – fact
            May 26 '16 at 15:50





            That's it! Am I right to assume that since ufunc is part of numpy, it calls C code and there won't be a significantly faster way to do this?

            – fact
            May 26 '16 at 15:50


















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