Comparing object ids of two numpy arrays

Question

I have been using numpy for quite a while but I stumbled upon one thing that I didn't understand fully:

a = np.ones(20)
b = np.zeros(10)

print(id(a)==id(b))          # prints False
print(id(a), id(b))          # prints (4591424976, 4590843504)

print(id(a[0])==id(b[0]))    # prints True 
print(id(a[0]), id(b[0]))    # prints (4588947064, 4588947064)

print(id(a[0]))              # 4588947184
print(id(b[0]))              # 4588947280

Can someone please explain the behavior observed in last four print statements? Also, I was aware of the fact that id gives you unique object id actually allocated in the memory but every time I run the last two print statements, I got different id values. Is this the expected behavior?

Answer 1

The short answer is that you should forget about relying on id to try and gain deep insight into the workings of python. Its output is affected by CPython implementation details, peephole optimizations and memory reuse. More often than not id is a red herring. This is especially true with numpy.

In your specific case only a and b exist as python objects. When you take an element, a[0], you instantiate a new python object, a scalar of type numpy.float64. These are new python objects and are thus given a new id, unless the interpreter realizes that you're trying to use this object twice (this is probably what's happening in your middle example, although I do find it surprising that two numpy.float64 objects with different values are given the same id. But the weird magic goes away if you assign a[0] and b[0] to proper names first, so this is probably due to some optimization). It could also happen that memory addresses get reused by the interpreter, giving you ids that have appeared before.

Just to see how pointless id is with numpy, even trivial views are new python objects with new ids, even though for all intents and purposes they are as good as the original:

>>> arr = np.arange(3)

>>> id(arr)
140649669302992

>>> id(arr[...])
140649669667056

And here's an example for id reuse in an interactive shell:

>>> id(np.arange(3))
139775926634896

>>> id(np.arange(3))
139775926672480

>>> id(np.arange(3))
139775926634896

Surely there's no such thing as int interning for numpy arrays, so the above is only due to the interpreter reusing ids. The fact that id returns a memory address is again just a CPython implementation detail. Forget about id.

The only thing you might want to use with numpy is numpy.may_share_memory and numpy.shares_memory.