Python How to Convert Dictionary Keys to List

You can convert dictionary keys to a list in Python by using list(your_dict.keys()), which returns a list of all keys.

Examples

Input{'a': 1, 'b': 2, 'c': 3}

Output['a', 'b', 'c']

Input{1: 'apple', 2: 'banana'}

Output[1, 2]

Input{}

Output[]

How to Think About It

To convert dictionary keys to a list, think of the dictionary as a collection of pairs. You want to take only the first part of each pair, which are the keys, and gather them into a new list. Python provides a simple way to get all keys, then you just turn that collection into a list.

Algorithm

1

Get the dictionary input.

2

Extract all keys from the dictionary.

3

Convert the keys collection into a list.

4

Return or print the list of keys.

Code

python

my_dict = {'a': 1, 'b': 2, 'c': 3}
keys_list = list(my_dict.keys())
print(keys_list)

Output

['a', 'b', 'c']

Dry Run

Let's trace {'a': 1, 'b': 2, 'c': 3} through the code

1

Start with dictionary

my_dict = {'a': 1, 'b': 2, 'c': 3}

2

Extract keys

my_dict.keys() returns dict_keys(['a', 'b', 'c'])

3

Convert to list

list(dict_keys(['a', 'b', 'c'])) becomes ['a', 'b', 'c']

Step	Operation	Result
1	Dictionary	{'a': 1, 'b': 2, 'c': 3}
2	Get keys	dict_keys(['a', 'b', 'c'])
3	Convert to list	['a', 'b', 'c']

Why This Works

Step 1: Dictionary keys method

The keys() method returns a view of all keys in the dictionary.

Step 2: Convert keys view to list

Using list() on the keys view creates a new list containing all keys.

Alternative Approaches

Using list comprehension

python

my_dict = {'a': 1, 'b': 2, 'c': 3}
keys_list = [key for key in my_dict]
print(keys_list)

This manually collects keys into a list; slightly longer but clear.

Using unpacking operator

python

my_dict = {'a': 1, 'b': 2, 'c': 3}
keys_list = [*my_dict]
print(keys_list)

Unpacks keys directly into a list; concise and modern syntax.

Complexity: O(n) time, O(n) space

Time Complexity

Extracting keys and converting to a list requires visiting each key once, so it takes linear time relative to the number of keys.

Space Complexity

A new list is created to hold all keys, so space grows linearly with the number of keys.

Which Approach is Fastest?

Using list(dict.keys()) or unpacking [*dict] are both efficient and clear; list comprehension is slightly more verbose but similar in speed.

Approach	Time	Space	Best For
list(dict.keys())	O(n)	O(n)	Clarity and standard usage
List comprehension	O(n)	O(n)	Explicit iteration, good for adding conditions
Unpacking operator [*dict]	O(n)	O(n)	Concise and modern syntax