Python: create sublist without copying

Question

I have a question about how to create a sublist (I hope this is the right term to use) from a given list without copying.

It seems that slicing can create sublists, but

Answer 1

numpy's array objects support this notion of creating interdependent sub-lists, by having slicing return views rather than copies of the data.

Altering the original numpy array will alter the views created from the array, and changes to any of the views will also be reflected in the original array. Especially for large data sets, views are a great way of cutting data in different ways, while saving on memory.

>>> import numpy as np
>>> array1 = np.array([1, 2, 3, 4])
>>> view1 = array1[1:]
>>> view1
array([2, 3, 4])
>>> view1[1] = 5
>>> view1
array([2, 5, 4])
>>> array1
array([1, 2, 5, 4]) # Notice that the change to view1 has been reflected in array1

For further reference, see the numpy documentation on views as well as this SO post.

Answer 2

There is no way to do this with built in Python data structures. However, I created a class that does what you need. I don't guarantee it to be bug-free, but it should get you started.

from itertools import islice

class SubLister(object):
    def __init__(self, base=[], start=0, end=None):
        self._base = base
        self._start = start
        self._end = end

    def __len__(self):
        if self._end is None:
            return len(self._base) - self._start
        return self._end - self._start

    def __getitem__(self, index):
        self._check_end_range(index)
        return self._base[index + self._start]

    def __setitem__(self, index, value):
        self._check_end_range(index, "list assignment index out of range")
        self._base[index + self._start] = value

    def __delitem__(self, index):
        self._check_end_range(index, "list assignment index out of range")
        del self._base[index + self._start]

    def __iter__(self):
        return islice(self._base, self._start, self._end)

    def __str__(self):
        return str(self._base[self._start:self._end])

    def __repr__(self):
        return repr(self._base[self._start:self._end])

    # ...etc...

    def get_sublist(self, start=0, end=None):
        return SubLister(base=self._base, start=start, end=end)

    def _check_end_range(self, index, msg="list index out of range"):
        if self._end is not None and index >= self._end - self._start:
            raise IndexError(msg)

Example:

>>> from sublister import SubLister
>>> base = SubLister([1, 2, 3, 4, 5])
>>> a = base.get_sublist(0, 2)
>>> b = base.get_sublist(1)

>>> base
[1, 2, 3, 4, 5]
>>> a
[1, 2]
>>> b
[2, 3, 4, 5]
>>> len(base)
5
>>> len(a)
2
>>> len(b)
4

>>> base[1] = 'ref'
>>> base
[1, 'ref', 3, 4, 5]
>>> a
[1, 'ref']
>>> b
['ref', 3, 4, 5]

Answer 3

There is no built-in way to do this. You could create your own list-like class that takes a reference to a list and reimplements all of the list accessor methods to operate on it.

Answer 4

you can't if you slice a to get b.

All slice operations return a new list containing the requested elements. This means that the following slice returns a new (shallow) copy of the list [1]

[1] https://docs.python.org/2/tutorial/introduction.html