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Get the quantity of uniq elements and the position of the first one from them solution in Clear category for Sort Array by Element Frequency by nickgryg
def frequency_sort(items):
dct = {}
lst = []
for item in items:
dct[item] = [items.count(item), items.index(item) * -1]
for item in sorted(dct, key=dct.get, reverse=True):
lst += [item] * dct[item][0]
return lst
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!")
Oct. 14, 2020
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