您几乎肯定要使用itertools.groupby：
l = [0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,2,2,2]answer = []for key, iter in itertools.groupby(l): answer.append((key, len(list(iter))))# answer is [(0, 6), (1, 6), (0, 4), (2, 3)]
如果要提高内存效率，但又增加了复杂性，则可以添加长度函数：
def length(l): if hasattr(l, '__len__'): return len(l) else: i = 0 for _ in l: i += 1 return il = [0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,2,2,2]answer = []for key, iter in itertools.groupby(l): answer.append((key, length(iter)))# answer is [(0, 6), (1, 6), (0, 4), (2, 3)]
请注意，尽管我尚未对length（）函数进行基准测试，但很有可能会使您变慢。
