Does Slicing a List in Python Create Copies of the Underlying Objects?

Slicing a List Without Generating a Copy in Python

The Problem:
Given a list of integers, generate all sublists of the form L[k:] for k in [0, len(L)-1] efficiently without creating copies.

The Answer:
Slicing a list in Python does not generate copies of the underlying objects, only references to them.

Understanding Reference Copying

To demonstrate this, consider three integer objects with the same value:

a = [1000 + 1, 1000 + 1, 1000 + 1]

They have distinct object IDs, indicating that they are separate objects:

map(id, a)  # returns [140502922988976, 140502922988952, 140502922988928]

Slicing these objects results in new list objects with the same references, showing that no new objects have been created:

b = a[1:3]
map(id, b)  # returns [140502922988952, 140502922988928]

This behavior applies to both immutable (e.g., integers) and mutable (e.g., lists) values.

Overhead Considerations

While slicing does not generate new objects, it does incur some memory overhead due to copying references and maintaining list metadata. For example, a list with N elements has a memory overhead of 72 bytes.

Views and numpy Arrays

If memory optimization is crucial, consider using numpy arrays instead of lists, as slicing numpy arrays shares memory between the slice and the original array. However, this approach requires careful handling to avoid inadvertent modifications.

Conclusion

Slicing lists in Python is an efficient operation that does not generate copies of the underlying objects. This behavior should be taken into account when considering performance optimizations.

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