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First solution in Clear category for Sort Array by Element Frequency by vlad.bezden
"""
Sort the given iterable so that its elements end up in the decreasing frequency order,
that is, the number of times they appear in elements.
If two elements have the same frequency,
they should end up in the same order as the first appearance in the iterable.
Input: Iterable
Output: Iterable
Precondition: elements can be integers or strings
"""
from typing import List, Any, Iterator, Union
from collections import Counter
InputData = List[Any]
OutputData = Union[InputData, Iterator[Any]]
def frequency_sort(items: InputData) -> OutputData:
c = Counter(items)
result = sorted(c.elements(), key=lambda k: c[k], reverse=True)
return result
if __name__ == "__main__":
assert list(frequency_sort([4, 6, 2, 2, 6, 4, 4, 4])) == [4, 4, 4, 4, 6, 6, 2, 2]
assert list(frequency_sort([4, 6, 2, 2, 2, 6, 4, 4, 4])) == [
4,
4,
4,
4,
2,
2,
2,
6,
6,
]
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("DONE!")
Feb. 16, 2019
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