itemgetter and groupby speedy solution for Group Equal consecutive by PythonWithPI - python coding challenges

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itemgetter and groupby solution in Speedy category for Group Equal consecutive by PythonWithPI

def group_equal(els):
    from itertools import groupby
    from operator import itemgetter
    return list(map(list,map(itemgetter(1),groupby(els))))

Feb. 17, 2019

Comments:

jcfernandez on June 16, 2019, 4:44 p.m.
Great solution !!! I like the idea of using as much builtin python code as possible. Using a `map(itemgetter(1), iterator)` function for getting only the consecutive groups is a great idea. The same result can be accomplished with ... <pre class='brush: python'>def group_equal(els): from itertools import groupby # Return a list of consecutive groups return [ # Make each consecutive groups into a list list(g) for k, g in groupby(els) ] </pre> This solution uses list comprehension to archive the same result.

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