Home Backend Development Python Tutorial How Can I Efficiently Iterate Over Overlapping Pairs (or Triples) of Elements in a Python List?

How Can I Efficiently Iterate Over Overlapping Pairs (or Triples) of Elements in a Python List?

Dec 16, 2024 pm 02:53 PM

How Can I Efficiently Iterate Over Overlapping Pairs (or Triples) of Elements in a Python List?

Iterating Over Overlapping Value Pairs in Python

When iterating over a list in Python, you may need to access both the current element and the next element simultaneously. Traditionally, this has been accomplished using code like:

for current, next in zip(the_list, the_list[1:]):
    # Do something
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However, Python 3.8 introduced a more efficient way to achieve this:

Utilizing the pairwise Function

The Python documentation provides a simple pairwise function for this purpose:

import itertools

def pairwise(iterable):
    a, b = itertools.tee(iterable)
    next(b, None)
    return zip(a, b)
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This function creates two iterators, a and b, which point to the first element of the input iterable. The iterator b is then advanced one step, resulting in a pointing to the current element and b pointing to the next element. The zip function is then used to create pairs of these elements.

For Python 2

For Python 2, you can use a similar pairwise function with the itertools.izip function instead of zip:

import itertools

def pairwise(iterable):
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.izip(a, b)
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Generalizing to Multiple Elements

The pairwise function can be generalized to produce larger windows of elements by adjusting the n parameter in the tee call. For example, to create pairs of three elements, you can use:

def threes(iterator):
    a, b, c = itertools.tee(iterator, 3)
    next(b, None)
    next(c, None)
    next(c, None)
    return zip(a, b, c)
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Caveat

It's important to note that this technique can consume a significant amount of memory if one iterator advances further than others. This can happen when the window size is large or if there are many elements in the original iterable.

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