你几乎肯定要使用 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）] 
 

  def length （l）：
 if hasattr（l，'__len__'）：
 return len（l）
 else：
i = 0 
 for _ in l：
i + = 1 
 return i 

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，length（iter））） 

＃answer is [（0，6），（1,6），（0,4），（2,3）] 
 $请注意，虽然我没有基准的length（）函数，它很可能会减慢你。 
I have a list as follows:
l = [0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,2,2,2]
I want to determine the length of a sequence of equal items, i.e for the given list I want the output to be:
[(0, 6), (1, 6), (0, 4), (2, 3)]
(or a similar format). 

I thought about using a defaultdict but it counts the occurrences of each item and accumulates it for the entire list, since I cannot have more than one key '0'.

Right now, my solution looks like this:
out = []
cnt = 0

last_x = l[0]  
for x in l:
    if x == last_x:
        cnt += 1
    else:
        out.append((last_x, cnt))
        cnt = 1
    last_x = x
out.append((last_x, cnt))

print out
I am wondering if there is a more pythonic way of doing this.
You almost surely want to use 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)]
If you want to make it more memory efficient, yet add more complexity, you can add a length function:
def length(l):
    if hasattr(l, '__len__'):
        return len(l)
    else:
        i = 0
        for _ in l:
            i += 1
        return i

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, length(iter)))

# answer is [(0, 6), (1, 6), (0, 4), (2, 3)]
Note though that I have not benchmarked the length() function, and it's quite possible it will slow you down.