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itertools groupby solution in Speedy category for Group Equal consecutive by caldeius
from itertools import groupby
def group_equal(els):
# your code here
return ([list(g) for _,g in groupby(els)])
if __name__ == '__main__':
print("Example:")
print(group_equal([1, 1, 4, 4, 4, "hello", "hello", 4]))
# These "asserts" are used for self-checking and not for an auto-testing
assert group_equal([1, 1, 4, 4, 4, "hello", "hello", 4]) == [[1,1],[4,4,4],["hello","hello"],[4]]
assert group_equal([1, 2, 3, 4]) == [[1], [2], [3], [4]]
assert group_equal([1]) == [[1]]
assert group_equal([]) == []
print("Coding complete? Click 'Check' to earn cool rewards!")
Jan. 22, 2020
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