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First-revised solution in Clear category for Group Equal consecutive by Sillte
def group_equal(els):
if not els:
return els
splits = [0] + [index + 1 for index, (a, b) in enumerate(zip(els, els[1:])) if a != b]
return [els[s:e] for s, e in zip(splits, splits[1:] + [len(els)])]
Dec. 16, 2018
