[Tut] How to Create a Python List of Size n?

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How to Create a Python List of Size n?

To create a list of

Code:

n

placeholder elements, multiply the list of a single placeholder element with

Code:

n

. For example, use

Code:

[None] * 5

to create a list

Code:

[None, None, None, None, None]

with five elements

Code:

None

. You can then overwrite some elements with index assignments. In the example,

Code:

lst[2] = 42

would result in the changed list

Code:

[None, None, 42, None, None]

.



Let’s play with an interactive code shell before you’ll dive into the detailed solution!

https://trinket.io/embed/python/8b63f214b1

Exercise: Initialize the list with

Code:

n=20

placeholder elements

Code:

-1

and run the code.

Next, you’ll learn about the more formal problem and dive into the step-by-step solution.

Problem: Given an integer

Code:

n

. How to initialize a list with

Code:

n

placeholder elements?

Code:

# n=0 --> [] # n=1 --> [None] # n=5 --> [None, None, None, None, None]

Solution: Use the list concatenation operation

Code:

*

.

Code:

n = 5 lst = [None] * n print(lst) # [None, None, None, None, None]

You can modify the element

Code:

n

as you like. In subsequent operations, you can overwrite all placeholder

Code:

None

list elements using simple index assignment operations:

Code:

lst[0] = 'Alice' lst[1] = 0 lst[2] = 42 lst[3] = 12 lst[4] = 'hello' print(lst) # ['Alice', 0, 42, 12, 'hello']

However, there’s a small problem if you want to create a list with mutable objects (such as alist of lists):

Code:

lst = [[]] * n print(lst) # [[], [], [], [], []] lst[2].append(42) print(lst) # [[42], [42], [42], [42], [42]]

Changing one list element changes all list elements because all list elements refer to the same list object in memory:

https://pythontutor.com/iframe-embed.htm...nces=false

The solution is to use list comprehension (see my detailed blog tutorial on list comprehension for a complete guide):

Code:

lst = [[] for _ in range(n)] print(lst) # [[], [], [], [], []] lst[2].append(42) print(lst) # [[], [], [42], [], []]

In the following visualization, you can see how each element now refers to an independent list object in memory:

https://pythontutor.com/iframe-embed.htm...nces=false

Exercise: Run the visualization and convince yourself that only one element is modified! Why is this the case?

Where to Go From Here?

Enough theory, let’s get some practice!

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