On 2023-01-10 20:41, Jen Kris via Python-list wrote:
Thanks for your comments. I'd like to make one small point. You say:
"Assignment in Python is a matter of object references. It's not
"conform them as long as they remain equal". You'll have to think in
terms of object references the entire way."
But where they have been set to the same object, an operation on one will
affect the other as long as they are equal (in Python). So I will have to
conform them in those cases because Python will reflect any math operation in
both the array and the matrix.
It's not a 2D matrix, it's a 1D list containing references to 1D lists,
each of which contains references to Python ints.
In CPython, references happen to be pointers, but that's just an
implementation detail.
Jan 10, 2023, 12:28 by ros...@gmail.com:
On Wed, 11 Jan 2023 at 07:14, Jen Kris via Python-list
<python-list@python.org> wrote:
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:
That's not an optimization; what you've done is set arr1 to be a
reference to that object.
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.
Yep, you have just set arr1 to be a completely different object.
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?
Assignment in Python is a matter of object references. It's not
"conform them as long as they remain equal". You'll have to think in
terms of object references the entire way.
ChrisA
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