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Impara Indexing and Slicing | Section
Working with Numbers in Python: Integers, Floats, and Type Conversion - 1769704232138

Indexing and Slicing

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Strings are sequences: each character has a position (an index). Python uses zero-based indexing, so the first character is at index 0. You can take single characters with indexing and ranges of characters with slicing.

Indexing

Use square brackets with a single position.

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s = "python" print(s[0]) # 'p' (first character) print(s[5]) # 'n' (sixth character)

Negative indices count from the end.

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s = "python" print(s[-1]) # 'n' (last character) print(s[-2]) # 'o' (second from the end)

Indexing must hit an existing position, otherwise you get IndexError.

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s = "python" print(s[10]) # IndexError: string index out of range

Also, strings are immutable, so you can read s[i] but not assign to it.

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s = "python" s[0] = 'P' # TypeError: 'str' object does not support item assignment

Slicing

A slice uses start:stop:step and returns a new string. stop is exclusive (it's not included).

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s = "python" print(s[1:4]) # 'yth' (indices 1,2,3) print(s[:4]) # 'pyth' (start defaults to 0) print(s[3:]) # 'hon' (stop defaults to len(s)) print(s[::2]) # 'pto' (every 2nd character)

Slices are forgiving: going past the ends just trims to valid bounds (no error).

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s = "python" print(s[0:100]) # 'python'

Negative Indices and Reversing

You can mix negative indices in slices, and a negative step walks backward.

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s = "python" print(s[-3:]) # 'hon' (last three) print(s[::-1]) # 'nohtyp' (reverse)
Note
Note

step cannot be 0. Leaving out step implies 1. Leaving out start or stop means "from the beginning" / "to the end".

1. What value will this code output?

2. What value will this code output?

3. Which statement raises an error for u = "hello"?

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What value will this code output?

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question mark

What value will this code output?

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question mark

Which statement raises an error for u = "hello"?

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Sezione 1. Capitolo 14

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Sezione 1. Capitolo 14
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