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Python challenge: can you solve this in a readable and efficient way?

Kevin Naidoo on February 13, 2024

I have a list of search terms and their associated "metadata": search_tags = [ "iPhone 14", "metadata", "Samsung Galaxy S3", "me...
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Fabrizio • 
search_tags_dict = dict(zip(search_tags[::2], search_tags[1::2]))
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Kevin Naidoo •

Nice! Very compact.

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Robert Groves • 
search_dict = {}
for i in range(0, len(search_tags), 2):
    key = search_tags[i]
    search_dict[key] = search_tags[i + 1]

# ...or the same using a list comprehension...

search_dict = {
    search_tags[i]: search_tags[i + 1]
    for i in range(0, len(search_tags), 2)
}
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Kevin Naidoo •

Nice! This is indeed correct.

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Ryan Lee • • Edited 
from itertools import pairwise
search_tags_dict = dict(pairwise(search_tags))
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Paul McGuire • 
search_iter = iter(search_tags)
search_tags_dict = dict(zip(search_iter, search_iter))
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Kevin Naidoo •

Great! Another good way of doing it.

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fialex137 • • Edited

This is my solution!!!

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