First speedy solution for Sort Array by Element Frequency by ollegattor - python coding challenges

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First solution in Speedy category for Sort Array by Element Frequency by ollegattor

def frequency_sort(data: list) -> list:
   data=sorted(data,key=lambda s1:data.index(s1))
   return sorted(data,key=lambda s2:data.count(s2) ,reverse=True)


if __name__ == '__main__':
    print("Example:")
    print(frequency_sort([4, 6, 2, 2, 6, 4, 4, 4]))

    # These "asserts" are used for self-checking and not for an auto-testing
    assert list(frequency_sort([4, 6, 2, 2, 6, 4, 4, 4])) == [4, 4, 4, 4, 6, 6, 2, 2]
    assert list(frequency_sort(['bob', 'bob', 'carl', 'alex', 'bob'])) == ['bob', 'bob', 'bob', 'carl', 'alex']
    assert list(frequency_sort([17, 99, 42])) == [17, 99, 42]
    assert list(frequency_sort([])) == []
    assert list(frequency_sort([1])) == [1]
    print("Coding complete? Click 'Check' to earn cool rewards!")

Feb. 20, 2020

Comments:

Splitter on March 7, 2020, 4:59 p.m.
I like the solution, but maybe another variable name would improve the code readability, e.g.: ```pyton def frequencysort(data: list) -> list: byequality = sorted(data, key=lambda s1: data.index(s1)) return sorted(byequality, key=lambda s2: byequality.count(s2), reverse=True) ``` Correct me if am wrong.

ollegattor on March 7, 2020, 6:17 p.m.
Thank you for your comment, you are right!)))

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