Comparing numpy array of dtype object

To make an element-wise comparison between the arrays, you can use numpy.equal() with the keyword argument dtype=numpy.object as in :

In [60]: np.equal(be, ce, dtype=np.object)
Out[60]: 
array([[True, True, True, True,
        array([ True,  True,  True,  True,  True]), True, True, True]],
      dtype=object)

P.S. checked using NumPy version 1.15.2 and Python 3.6.6

edit

From the release notes for 1.15,

https://docs.scipy.org/doc/numpy-1.15.1/release.html#comparison-ufuncs-accept-dtype-object-overriding-the-default-bool

Comparison ufuncs accept dtype=object, overriding the default bool

This allows object arrays of symbolic types, which override == and 
other operators to return expressions, to be compared elementwise with 
np.equal(a, b, dtype=object).

To complement @kmario23's answer, what about doing

def wrpr(bools):
    try:
      # ints  = bools.flatten().prod()
        fltn_bools = np.hstack(bools)
    except: # should not pass silently.
        fltn_bools = np.array(wrpr(a) for a in bools)        
    ints = fltn_bools.prod()
    if isinstance(ints, np.ndarray):
        return wrpr(ints)
    return bool(ints)

And finally,

>>> wrpr(np.equal(ce, be, dtype=np.object))
True

Checked using (numpy1.15.1 & Python 3.6.5) & (numpy1.15.1 & Python 2.7.13).

---

But still, as commented here

NumPy is designed for rigid multidimensional grids of numbers. Trying to get anything but a rigid multidimensional grid is going to be painful. (@user2357112, Jul 31 '17 at 23:10)

and/or

Moral of the story: Don't use dtype=object arrays. They are stunted Python lists, with worse performance characteristics, and numpy is not designed to handle the case of sequence-like containers within these object arrays. (@juanpa.arrivillaga, Jul 31 '17 at 23:38)