python - Comparing numpy array of dtype object

Question

My question is "why?:"

aa[0]
array([[405, 162, 414, 0,
        array([list([1, 9, 2]), 18, (405, 18, 207), 64, 'Universal'],
      dtype=object),
        0, 0, 0]], dtype=object)

aaa
array([[405, 162, 414, 0,
        array([list([1, 9, 2]), 18, (405, 18, 207), 64, 'Universal'],
      dtype=object),
        0, 0, 0]], dtype=object)

np.array_equal(aaa,aa[0])
False

Those arrays are completly identical.

My minimal example doesn't reproduce this:

be=np.array([1],dtype=object)

be
array([1], dtype=object)

ce=np.array([1],dtype=object)

ce
array([1], dtype=object)

np.array_equal(be,ce)
True

Nor does this one:

ce=np.array([np.array([1]),'5'],dtype=object)

be=np.array([np.array([1]),'5'],dtype=object)

np.array_equal(be,ce)
True

However, to reproduce my problem try this:

be=np.array([[405, 162, 414, 0, np.array([list([1, 9, 2]), 18, (405, 18, 207), 64, 'Universal'],dtype=object),0, 0, 0]], dtype=object)

ce=np.array([[405, 162, 414, 0, np.array([list([1, 9, 2]), 18, (405, 18, 207), 64, 'Universal'],dtype=object),0, 0, 0]], dtype=object)

np.array_equal(be,ce)
False

np.array_equal(be[0],ce[0])
False

And I have no idea why those are not equal. And to add the bonus question, how do I compare them?

I need an efficient way to check if aaa is in the stack aa.

I'm not using aaa in aa because of DeprecationWarning: elementwise == comparison failed; this will raise an error in the future. and because it still returns False if anyone is wondering.

---

What else have I tried?:

np.equal(be,ce)
*** ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

np.all(be,ce)
*** TypeError: only integer scalar arrays can be converted to a scalar index

all(be,ce)
*** TypeError: all() takes exactly one argument (2 given)

all(be==ce)
*** TypeError: 'bool' object is not iterable

np.where(be==ce)
(array([], dtype=int64),)

And these, which I can't get to run in the console, all evaluate to False, some giving the deprecation warning:

import numpy as np

ce=np.array([[405, 162, 414, 0, np.array([list([1, 9, 2]), 18, (405, 18, 207), 64, 'Universal'],dtype=object),0, 0, 0]], dtype=object)

be=np.array([[405, 162, 414, 0, np.array([list([1, 9, 2]), 18, (405, 18, 207), 64, 'Universal'],dtype=object),0, 0, 0]], dtype=object)

print(np.any([bee in ce for bee in be]))

print(np.any([bee==cee for bee in be for cee in ce]))

print(np.all([bee in ce for bee in be]))

print(np.all([bee==cee for bee in be for cee in ce]))

And of course other questions telling me this should work...

Answer 1

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).