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First solution in Clear category for Group Equal consecutive by dshimbori
def group_equal(els):
if els == []:
return []
group = []
group.append(els[0])
result = []
for e in els[1:]:
if e in group:
group.append(e)
else:
result.append(group)
group = []
group.append(e)
if group !=[]:
result.append(group)
return result
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!")
Nov. 7, 2018
