Is there a way to check if NumPy arrays share the same data?

My impression is that in NumPy, two arrays can share the same memory. Take the following example:

import numpy as np
a=np.arange(27)
b=a.reshape((3,3,3))
a[0]=5000
print (b[0,0,0]) #5000

#Some tests:
a.data is b.data #False
a.data == b.data #True

c=np.arange(27)
c[0]=5000
a.data == c.data #True ( Same data, not same memory storage ), False positive

So clearly b didn't make a copy of a; it just created some new meta-data and attached it to the same memory buffer that a is using. Is there a way to check if two arrays reference the same memory buffer?

My first impression was to use a.data is b.data, but that returns false. I can do a.data == b.data which returns True, but I don't think that checks to make sure a and b share the same memory buffer, only that the block of memory referenced by a and the one referenced by b have the same bytes.

Solution 1:

You can use the base attribute to check if an array shares the memory with another array:

>>> import numpy as np
>>> a = np.arange(27)
>>> b = a.reshape((3,3,3))
>>> b.base is a
True
>>> a.base is b
False

Not sure if that solves your problem. The base attribute will be None if the array owns its own memory. Note that an array's base will be another array, even if it is a subset:

>>> c = a[2:]
>>> c.base is a
True

Solution 2:

I think jterrace's answer is probably the best way to go, but here is another possibility.

def byte_offset(a):
    """Returns a 1-d array of the byte offset of every element in `a`.
    Note that these will not in general be in order."""
    stride_offset = np.ix_(*map(range,a.shape))
    element_offset = sum(i*s for i, s in zip(stride_offset,a.strides))
    element_offset = np.asarray(element_offset).ravel()
    return np.concatenate([element_offset + x for x in range(a.itemsize)])

def share_memory(a, b):
    """Returns the number of shared bytes between arrays `a` and `b`."""
    a_low, a_high = np.byte_bounds(a)
    b_low, b_high = np.byte_bounds(b)

    beg, end = max(a_low,b_low), min(a_high,b_high)

    if end - beg > 0:
        # memory overlaps
        amem = a_low + byte_offset(a)
        bmem = b_low + byte_offset(b)

        return np.intersect1d(amem,bmem).size
    else:
        return 0

Example:

>>> a = np.arange(10)
>>> b = a.reshape((5,2))
>>> c = a[::2]
>>> d = a[1::2]
>>> e = a[0:1]
>>> f = a[0:1]
>>> f = f.reshape(())
>>> share_memory(a,b)
80
>>> share_memory(a,c)
40
>>> share_memory(a,d)
40
>>> share_memory(c,d)
0
>>> share_memory(a,e)
8
>>> share_memory(a,f)
8

Here is a plot showing the time for each share_memory(a,a[::2]) call as a function of the number of elements in a on my computer.

Solution 3:

To solve the problem exactly, you can use

import numpy as np

a=np.arange(27)
b=a.reshape((3,3,3))

# Checks exactly by default
np.shares_memory(a, b)

# Checks bounds only
np.may_share_memory(a, b)

Both np.may_share_memory and np.shares_memory take an optional max_work argument that lets you decide how much effort to put in to ensure no false positives. This problem is NP-complete, so always finding the correct answer can be quite computationally expensive.