Python：确定列表中相等项的序列长度

Question

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, ie 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.

Answer 1

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.

Answer 2

Mike's answer is good, but the itertools._grouper returned by groupby will never have a __len__ method so there is no point testing for it

I use sum(1 for _ in i) to get the length of the itertools._grouper

>>> import itertools as it
>>> L = [0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,2,2,2]
>>> [(k, sum(1 for _ in i)) for k, i in it.groupby(L)]
[(0, 6), (1, 6), (0, 4), (2, 3)]