我想要多个列表指向 python 中的相同 memory

Question

I know this doesn't work, but it illustrates my goal. If I have a list C, and I set A=C, then they point to the same object, which is what I want. But if I set A to a subset of C, then it creates a new list. I don't want that. I want it to point to the same data. Eg:

C = [1,2,3,4,5,6]
A = C[3:6]
C[4] = 9
print(C)
print(A)

I would like to see [4,9,6] for A

Answer 1

Python lists don't do that. It's something numpy arrays do:

In [123]: import numpy as np

In [124]: c = np.array([1,2,3,4,5,6])

In [125]: c
Out[125]: array([1, 2, 3, 4, 5, 6])

In [126]: a = c[3:6]

In [127]: a
Out[127]: array([4, 5, 6])

In [128]: c[4] = 9

In [129]: c
Out[129]: array([1, 2, 3, 4, 9, 6])

In [130]: a
Out[130]: array([4, 9, 6])

You could do that with Python lists if each element was also a list or any mutable object:

>>> c = [[1], [2], [3], [4], [5], [6]]  # list of lists
>>> a = c[3:6]
>>> a
[[4], [5], [6]]
>>> c[4].append(9)
>>> c
[[1], [2], [3], [4], [5, 9], [6]]
>>> a  # has also changed
[[4], [5, 9], [6]]
>>> c[4].pop(0)  # remove the first element of c[4]
5
>>> a
[[4], [9], [6]]
>>>

But you can't "overwrite" the value of c[4], only change it. Hence, works only with mutable objects that are themselves being changed.

>>> c[4] = 0
>>> c
[[1], [2], [3], [4], 0, [6]]
>>> a  # does not have the 0, retains previous value
[[4], [9], [6]]
>>>