To clarify how Python handles two equal objects

[Jen Kris]

I am writing a spot speedup in assembly language for a short but computation-intensive Python loop, and I discovered something about Python array handling that I would like to clarify.  

For a simplified example, I created a matrix mx1 and assigned the array arr1 to the third row of the matrix:

mx1 = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
arr1 = mx1[2]

The pointers to these are now the same:

ida = id(mx1[2]) - 140260325306880
idb = id(arr1) - 140260325306880

That’s great because when I encounter this in assembly or C, I can just borrow the pointer to row 3 for the array arr1, on the assumption that they will continue to point to the same object.  Then when I do any math operations in arr1 it will be reflected in both arrays because they are now pointing to the same array:

arr1[0] += 2
print(mx1[2]) - [9, 8, 9]
print(arr1) - [9, 8, 9]

Now mx1 looks like this:

[ 1, 2, 3 ]
[ 4, 5, 6 ]
[ 9, 8, 9 ]

and it stays that way for remaining iterations.  

But on the next iteration we assign arr1 to something else:

arr1 = [ 10, 11, 12 ]
idc = id(arr1) – 140260325308160
idd = id(mx1[2]) – 140260325306880

Now arr1 is no longer equal to mx1[2], and any subsequent operations in arr1 will not affect mx1.  So where I’m rewriting some Python code in a low level language, I can’t assume that the two objects are equal because that equality will not remain if either is reassigned.  So if I do some operation on one array I have to conform the two arrays for as long as they remain equal, I can’t just do it in one operation because I can’t rely on the objects remaining equal. 

Is my understanding of this correct?  Is there anything I’m missing? 

Thanks very much. 

Jen