How to Initialize Lists with Specific Length in Python

In Python, lists are dynamic arrays that grow as you append items. However, there are scenarios, such as creating a placeholder array for algorithms, setting up a grid for a game, or pre-allocating memory for performance, where you need to initialize a list with a specific length and a default value (like 0 or None).

This guide explores the standard multiplication method, the safe approach for mutable objects (like lists of lists), and high-performance alternatives.

Method 1: Using the Multiplication Operator (Standard)

The fastest and most Pythonic way to initialize a list with a constant, immutable value (like integers, floats, strings, or None) is using the multiplication operator *.

Syntax:

[default_value] * length

Example:

length = 5

# ✅ Correct: Initialize list of integers
zeros = [0] * length
print(f"Zeros: {zeros}")

# ✅ Correct: Initialize list of None (Placeholders)
placeholders = [None] * length
print(f"Placeholders: {placeholders}")

# ✅ Correct: Initialize list of strings
strings = ["empty"] * 3
print(f"Strings: {strings}")

Output:

Zeros: [0, 0, 0, 0, 0]
Placeholders: [None, None, None, None, None]
Strings: ['empty', 'empty', 'empty']

note

This method works perfectly for immutable data types. However, if you use this to create a list of lists (e.g., a matrix), you will encounter the "Reference Trap" discussed in Section 3.

Method 2: List Comprehensions (Safe for Mutables)

If you need to initialize a list where every element is a new, distinct mutable object (like a dictionary, a set, or another list), you must use a list comprehension. This ensures that the constructor is called repeatedly for each element.

Syntax:

[Expression for _ in range(length)]

Example:

length = 3

# ✅ Correct: Creating a list of independent empty lists
matrix = [[] for _ in range(length)]

# Modifying one element does not affect the others
matrix[0].append(1)

print(f"Matrix: {matrix}")

Output:

Matrix: [[1], [], []]

The Common Pitfall: The Mutable Reference Trap

A frequent bug occurs when developers use the multiplication operator * to create multi-dimensional lists. Python copies the reference to the object, not the object itself. This means every element in the list points to the exact same object in memory.

Reproducing the Error

length = 3

# ⛔️ Incorrect: This creates 3 references to the SAME list object
matrix = [[]] * length

print(f"Initial: {matrix}")

# We try to modify only the first row
matrix[0].append(99)

# Result: ALL rows are modified because they are the same object
print(f"After modification: {matrix}")

Output:

Initial: [[], [], []]
After modification: [[99], [99], [99]]

warning

Never use [[]] * N or [{}] * N. While it looks correct initially, modifying one item will modify all of them. Always use the list comprehension method from Section 2 for mutable objects.

Method 3: Initializing Sequences with Range

If you need a list initialized with a sequence of numbers (e.g., 0, 1, 2, 3...) rather than a single repeating value, use the list() constructor with range().

length = 5

# ✅ Correct: Create a sequence from 0 to N-1
sequence = list(range(length))
print(f"Sequence: {sequence}")

# ✅ Correct: Create a sequence of even numbers
evens = list(range(0, 10, 2))
print(f"Evens: {evens}")

Output:

Sequence: [0, 1, 2, 3, 4]
Evens: [0, 2, 4, 6, 8]

Conclusion

To initialize lists with a specific length in Python:

1. Use [val] * N for immutable types like int, str, or None. It is the fastest method.
2. Use [[] for _ in range(N)] for mutable types like lists or dicts to avoid shared reference bugs.
3. Use list(range(N)) to create sequential numerical lists.