How to convert list to dictionary in Python?

Converting a list to a dictionary in Python is a common operation, especially when you have paired data elements. This article demonstrates different methods to transform a list where odd position elements become keys and even position elements become values.

Understanding the Conversion

Given a list like [1, 'Delhi', 2, 'Kolkata', 3, 'Bangalore', 4, 'Noida'], we want to create a dictionary where:

• Elements at index 0, 2, 4... become keys
• Elements at index 1, 3, 5... become values

The result would be: {1: 'Delhi', 2: 'Kolkata', 3: 'Bangalore', 4: 'Noida'}

Using Loop Iteration

This method iterates through the list with a step of 2, pairing consecutive elements ?

def convert_list_to_dict(data):
    result = {}
    for i in range(0, len(data), 2):
        result[data[i]] = data[i+1]
    return result

cities = [1, 'Delhi', 2, 'Kolkata', 3, 'Bangalore', 4, 'Noida']
print(convert_list_to_dict(cities))

{1: 'Delhi', 2: 'Kolkata', 3: 'Bangalore', 4: 'Noida'}

Using zip() with Iterator

The zip() function with an iterator provides a more elegant solution by pairing adjacent elements ?

def convert_list_to_dict(data):
    iterator = iter(data)
    return dict(zip(iterator, iterator))

cities = [1, 'Delhi', 2, 'Kolkata', 3, 'Bangalore', 4, 'Noida']
print(convert_list_to_dict(cities))

{1: 'Delhi', 2: 'Kolkata', 3: 'Bangalore', 4: 'Noida'}

Using Dictionary Comprehension

A concise one-liner approach using dictionary comprehension ?

cities = [1, 'Delhi', 2, 'Kolkata', 3, 'Bangalore', 4, 'Noida']

result = {cities[i]: cities[i+1] for i in range(0, len(cities), 2)}
print(result)

{1: 'Delhi', 2: 'Kolkata', 3: 'Bangalore', 4: 'Noida'}

Comparison

Conclusion

Use zip() with iterator for the most efficient solution, dictionary comprehension for concise code, or loop iteration when you need maximum clarity and control over the conversion process.