Can I produce the result of np.outer using np.dot?

Question

I am trying to improve my understanding of numpy functions. I understand the behaviour of numpy.dot. I'd like to understand the behaviour of numpy.outer in terms of numpy.dot.

Based on this Wikipedia article https://en.wikipedia.org/wiki/Outer_product I'd expect for array_equal to return True in the following code. However it does not.

X = np.matrix([
    [1,5],
    [5,9],
    [4,1]
])

r1 = np.outer(X,X)
r2 = np.dot(X, X.T)
np.array_equal(r1, r2)

How can I assign r2 so that np.array_equal returns True? Also, why does numpy's implementation of np.outer not match the definition of outer multiplication on Wikipedia?

Using numpy 1.9.2

Answer 1

In [303]: X=np.array([[1,5],[5,9],[4,1]])
In [304]: X
Out[304]: 
array([[1, 5],
       [5, 9],
       [4, 1]])
In [305]: np.inner(X,X)
Out[305]: 
array([[ 26,  50,   9],
       [ 50, 106,  29],
       [  9,  29,  17]])
In [306]: np.dot(X,X.T)
Out[306]: 
array([[ 26,  50,   9],
       [ 50, 106,  29],
       [  9,  29,  17]])

The Wiki outer link mostly talks about vectors, 1d arrays. Your X is 2d.

In [310]: x=np.arange(3)
In [311]: np.outer(x,x)
Out[311]: 
array([[0, 0, 0],
       [0, 1, 2],
       [0, 2, 4]])
In [312]: np.inner(x,x)
Out[312]: 5
In [313]: np.dot(x,x)   # same as inner
Out[313]: 5
In [314]: x[:,None]*x[None,:]   # same as outer
Out[314]: 
array([[0, 0, 0],
       [0, 1, 2],
       [0, 2, 4]])

Notice that the Wiki outer does not involve summation. Inner does, in this example 5 is the sum of the 3 diagonal values of the outer.

dot also involves summation - all the products followed summation along a specific axis.

Some of the wiki outer equations use explicit indices. The einsum function can implement these calculations.

In [325]: np.einsum('ij,kj->ik',X,X)
Out[325]: 
array([[ 26,  50,   9],
       [ 50, 106,  29],
       [  9,  29,  17]])
In [326]: np.einsum('ij,jk->ik',X,X.T)
Out[326]: 
array([[ 26,  50,   9],
       [ 50, 106,  29],
       [  9,  29,  17]])
In [327]: np.einsum('i,j->ij',x,x)
Out[327]: 
array([[0, 0, 0],
       [0, 1, 2],
       [0, 2, 4]])
In [328]: np.einsum('i,i->',x,x)
Out[328]: 5

As mentioned in the comment, np.outer uses ravel, e.g.

return a.ravel()[:, newaxis]*b.ravel()[newaxis,:]

This the same broadcasted multiplication that I demonstrated earlier for x.

Answer 2

numpy.outer only works for 1-d vectors, not matrices. But for the case of 1-d vectors, there is a relation.

If

import numpy as np
A = np.array([1.0,2.0,3.0])

then this

np.matrix(A).T.dot(np.matrix(A))

should be the same as this

np.outer(A,A)