Is there a way to get an alias for a part of a list in python?
Specifically, I want the equivalent of this to happen:
>>> l=[1,2,3,4,5]
>>> a=l
>>> l[0]=10
>>> a
[10, 2, 3, 4, 5]
But what I get is this:
>>> l=[1,2,3,4,5]
>>> a=l[0:2]
>>> l[0]=10
>>> a
[1, 2]
If numpy is an option:
import numpy as np
l = np.array(l)
a = l[:2]
l[0] = 10
print(l)
print(a)
Output:
[10 2 3 4 5]
[10 2]
slicing with basic indexing returns a view object with numpy so any change are reflected in the view object
Or use a memoryview with an array.array:
from array import array
l = memoryview(array("l", [1, 2, 3, 4,5]))
a = l[:2]
l[0]= 10
print(l.tolist())
print(a.tolist())
Output:
[10, 2, 3, 4, 5]
[10, 2]
You could embed each element into its own mutable data structure (like a list).
>>> l=[1,2,3,4,5]
>>> l = [[item] for item in l]
>>> l
[[1], [2], [3], [4], [5]]
>>> a = l[:2]
>>> a
[[1], [2]]
>>> l[0][0] = 10
>>> l
[[10], [2], [3], [4], [5]]
>>> a
[[10], [2]]
However, I recommend trying to come up with a solution to your original issue (whatever that was) that doesn't create issues of its own.
What you're looking for is a view of the original list, so that any modifications are reflected in the original list. This is doable with the array in the numpy package:
>>> import numpy
>>> x = numpy.array([1, 2, 3])
>>> y = x[2:]
>>> y[0] = 999
>>> x
array([ 1, 2, 999])
list. If you had twolists of different sizes, how could they be the same list, i.e., have the same address?list, but that would be so hacky. You probably have an XY Problem.