What are the rules for comparing numpy arrays using ==?

NumPy tries to broadcast the two arrays to compatible shapes before comparison. If the broadcasting fails, False is currently returned. In the future,

The equality operator == will in the future raise errors like np.equal if broadcasting or element comparisons, etc. fails.

Otherwise, a boolean array resulting from the element-by-element comparison is returned. For example, since x and np.array([1]) are broadcastable, an array of shape (10,) is returned:

In [49]: np.broadcast(x, np.array([1])).shape
Out[49]: (10,)

Since x and np.array([[1,3],[2]]) are not broadcastable, False is returned by x == np.array([[1,3],[2]]).

In [50]: np.broadcast(x, np.array([[1,3],[2]])).shape
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-50-56e4868cd7f7> in <module>()
----> 1 np.broadcast(x, np.array([[1,3],[2]])).shape

ValueError: shape mismatch: objects cannot be broadcast to a single shape

It's possible that what's confusing you is that:

1. some broadcasting is going on.

2. you appear to have an older version of numpy.

---

x == np.array([[1],[2]])

is broadcasting. It compares x to each of the first and second arrays; as they are scalars, broadcasting implies that it compares each element of x to each of the scalars.

---

However, each of

x == np.array([1,2])

and

x == np.array([[1,3],[2]])

can't be broadcast. By me, with numpy 1.10.4, this gives

/usr/local/bin/ipython:1: DeprecationWarning: elementwise == comparison failed; this will raise an error in the future.
#!/usr/bin/python
False