How Can I Efficiently Convert an Alternating Key-Value List to a Dictionary in Python?

Converting '[key1,val1,key2,val2]' to a Dictionary: Pythonic Methods

Problem:

In Python, we have a list a with an alternating key-value structure, where every even element represents a dictionary key and its subsequent odd element is the corresponding value. The task is to convert this list to a dictionary b, where each key-value pair is appropriately mapped.

Example:

a = ['hello', 'world', '1', '2']

Desired Output:

b = {'hello': 'world', '1': '2'}

Solution:

The most straightforward method to solve this problem is using Python's built-in dict() and zip() functions. Here's a syntactically clean solution:

b = dict(zip(a[::2], a[1::2]))

This method is efficient and achieves the desired result. However, if performance is a concern, an alternative approach is to use Python's iterator tools:

from itertools import izip
i = iter(a)
b = dict(izip(i, i))

This method is especially useful when dealing with very large lists, as it doesn't create any temporary lists in the process.

In Python 3, a dict comprehension also provides a concise solution:

b = {a[i]: a[i+1] for i in range(0, len(a), 2)}

While zip() is already lazy in Python 3, you can also use the walrus operator (:=) for a single-line solution in Python 3.8 and above:

b = dict(zip(i := iter(a), i))

Ultimately, the best choice depends on the specific requirements and Python version you're using. These methods offer clean and efficient solutions for converting lists of alternating key-value pairs into dictionaries in Python.

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