How Does Python\'s `zip([iter(s)]*n)` Function Efficiently Chunk Lists?

Unveiling the Inner Workings of zip([iter(s)]n): A Python Chunking Technique

To efficiently split a list into chunks of specified size, Python offers an ingenious solution: zip([iter(s)]n). Understanding how this cryptic code operates can empower developers to confidently employ it for their data-processing needs.

At its core, zip() combines multiple iterables into a single object, producing tuples containing corresponding elements from each input. However, the operator in [[iter(s)]*n] requires further scrutiny.

iter() yields an iterator for the given sequence, s. By enclosing iter(s) within square brackets, we create a list of iterators, each representing a distinct iteration over the same sequence.

[x] * n produces a list of length n, where each element is x. In this case, we have a list containing n instances of iter(s).

Finally, *arg unpacks this list, allowing us to pass each iterator to zip() separately. Thus, zip() obtains an item from each iterator, resulting in tuples with n elements extracted from s.

To paint a clearer picture, consider the example:

s = [1,2,3,4,5,6,7,8,9]
n = 3

list(zip(*[iter(s)]*n)) # returns [(1,2,3),(4,5,6),(7,8,9)]

Breaking down the code:

1. iter([1,2,3,4,5,6,7,8,9]) provides an iterator for s.
2. [[iter([1,2,3,4,5,6,7,8,9])] 3] generates a list containing three copies of the iterator.
3. zip(*[]) unfolds the list, presenting zip() with three iterators.
4. Each iteration of zip() retrieves an element from each iterator, forming tuples as a result.

This technique proves indispensable when dividing a list into uniform-sized chunks for efficient processing, like pagination or batching.

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