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First solution in Clear category for Sort Array by Element Frequency by _Chico_
#Answer to Sort Array by Element Frequency - https://py.checkio.org/en/mission/sort-array-by-element-frequency/
def frequency_sort(items):
dictionary = {}
for i in items:
if i in dictionary:
dictionary[i] += 1
else:
dictionary[i] = 1
dictionary = sorted(dictionary.items(), key=lambda k:k[1], reverse = True)
result = []
for i in dictionary:
for j in range(i[1]):
result.append(i[0])
return result
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!")
May 12, 2021
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