For anyone stumbling upon this, the best way to apply an element-wise multiplication of n np.ndarray of shape (d, ) is to first np.vstack them and apply np.prod on the first axis:

>>> import numpy as np
>>>
>>> arrays = [
...   np.array([1, 2, 3]),
...   np.array([5, 8, 2]),
...   np.array([9, 2, 0]),
... ]
>>>
>>> print(np.prod(np.vstack(arrays), axis=0))
[45 32  0]

One way is to use the outer function of np.multiply (and transpose if you want the same order as in your question):

>>> np.multiply.outer(x, y).T
array([[3, 6],
       [4, 8]])

Most ufuncs in NumPy have this useful outer feature (add, subtract, divide, etc.). As @Akavall suggests, np.outer is equivalent for the multiplication case here.

Alternatively, np.einsum can perform the multiplication and transpose in one go:

>>> np.einsum('i,j->ji', x, y)
array([[3, 6],
       [4, 8]])

A third approach is to insert a new axis in one the arrays and then multiply, although this is a little more verbose:

>>> (x[:, np.newaxis] * y).T
array([[3, 6],
       [4, 8]])

For those interested in performance, here are the timings of the operations, from quickest to slowest, on two arrays of length 15:

In [70]: x = np.arange(15)
In [71]: y = np.arange(0, 30, 2)
In [72]: %timeit np.einsum('i,j->ji', x, y)
100000 loops, best of 3: 2.88 µs per loop
In [73]: %timeit np.multiply.outer(x, y).T
100000 loops, best of 3: 5.48 µs per loop
In [74]: %timeit (x[:, np.newaxis] * y).T
100000 loops, best of 3: 6.68 µs per loop
In [75]: %timeit np.outer(x, y).T
100000 loops, best of 3: 12.2 µs per loop