Section 9.5 List slices

The slice operator also works on lists:

If you omit the first index, the slice starts at the beginning. If you omit the second, the slice goes to the end. So if you omit both, the slice is a copy of the whole list.

>>> t[:]
['a', 'b', 'c', 'd', 'e', 'f']

Activity 9.5.1.

What is printed by the following statements?

alist = [3, 67, "cat", [56, 57, "dog"], [ ], 3.14, False]
print(alist[4:])

[ [ ], 3.14, False]
Yes, the slice starts at index 4 and goes up to and including the last item.
[ [ ], 3.14]
By leaving out the upper bound on the slice, we go up to and including the last item.
[ [56, 57, "dog"], [ ], 3.14, False]
Index values start at 0.

Activity 9.5.2.

What is printed by the following statements?

alist = [3, 67, "cat", [56, 57, "dog"], [ ], 3.14, False]
print(alist[:])

[]
An empty slice like this would not produce an empty list. Think about how the slice indexes.
[3]
This would be correct if the slice was [:1]
[3, 67, "cat", [56, 57, "dog"], [ ], 3.14, False]
Omitting both indexes in a slice will create a copy of the whole list.

Since lists are mutable, it is often useful to make a copy before performing operations that fold, spindle, or mutilate lists.

A slice operator on the left side of an assignment can update multiple elements:

Activity 9.5.3.

What is printed by the following statements?

L = [0.34, '6', 'SI106', 'Python', -2]
print(len(L[1:-1]))

2
The list begins with the second item of L and includes everything up to but not including the last item.
3
Yes, there are 3 items in this slice.
4
The list begins with the second item of L and includes everything up to but not including the last item.
5
The list begins with the second item of L and includes everything up to but not including the last item.

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