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First solution in Clear category for Frequency Sorting by wo.tomasz
def frequency_sorting(numbers):
en = [(i, numbers.count(i)) for i in numbers]
en.sort()
en.sort(reverse=True, key=lambda k: k[1])
numbers = [n[0] for n in en]
return numbers
if __name__ == '__main__':
print("Example:")
print(frequency_sorting([1, 2, 3, 4, 5]))
#These "asserts" using only for self-checking and not necessary for auto-testing
assert frequency_sorting([1, 2, 3, 4, 5]) == [1, 2, 3, 4, 5], "Already sorted"
assert frequency_sorting([3, 4, 11, 13, 11, 4, 4, 7, 3]) == [4, 4, 4, 3, 3, 11, 11, 7, 13], "Not sorted"
assert frequency_sorting([99, 99, 55, 55, 21, 21, 10, 10]) == [10, 10, 21, 21, 55, 55, 99, 99], "Reversed"
print("Coding complete? Click 'Check' to earn cool rewards!")
Dec. 13, 2020
