Since slice does shallow copies why can it be used update the original mutable collection?

Asked Apr 26 '20 at 17:43

Active Apr 26 '20 at 17:43

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I know that slice operator does a shallow copy.

But when I used the slice operator on a list of integers and mutate the slice then the mutation was reflected in the original list. Since integers are immutable, the result confused me.

x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print("Before slicing...")
print(x)
print("Slicing...")
x[1::2]=[0]*len(x[1::2])
print("After slicing...")
print(x)
x[2:2]=[1]
print(x)

The output was

Before slicing...
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Slicing...
After slicing...
[1, 0, 3, 0, 5, 0, 7, 0, 9, 0]
[1, 0, 1, 3, 0, 5, 0, 7, 0, 9, 0]

Since x[1::2] and x[2:2] does a shallow copy why does a change to it reflected in the original list?

asked Apr 26 '20 at 17:43

Neminda Prabhashwara

Because you explicitly put that slice back into the original list. – jonrsharpe Apr 26 '20 at 17:47
@jonrsharpe What does that mean I didn't assign it to original list? – Neminda Prabhashwara Apr 26 '20 at 17:50
2

Yes you did, that's what the slice notation on the left hand side of the = means. See https://stackoverflow.com/questions/10623302/how-assignment-works-with-python-list-slice – jonrsharpe Apr 26 '20 at 17:51
@jonrsharpe So when used on the left hand side of the `=` operator, does slice operator behaves as an indexing for the original list? – Neminda Prabhashwara Apr 26 '20 at 17:54
@jonrsharpe Thank you very much that post solved it. – Neminda Prabhashwara Apr 26 '20 at 17:58

Since slice does shallow copies why can it be used update the original mutable collection?

0 Answers0