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First solution in Clear category for Group Equal consecutive by erewhon
def group_equal(els):
if els == []:
return els
separate = []
for i in range(len(els)-1):
if els[i] != els[i+1]:
separate.append(i)
start = 0
result = []
for s in separate :
result.append(els[start:s+1])
start = s+1
result.append(els[start:])
return result
Jan. 7, 2020
