python - 任何人都可以提供一种更 pythonic 的方式来生成莫里斯序列吗？

Question

我正在尝试生成 morris sequence在 python 。我目前的解决方案如下，但我觉得我只是用 python 编写了 c。谁能提供更 pythonic 的解决方案？

def morris(x):
    a = ['1', '11']
    yield a[0]
    yield a[1]
    while len(a) <= x:
        s = ''
        count = 1
        al = a[-1]
        for i in range(0,len(al)):
            if i+1 < len(al) and al[i] == al[i+1]:
                count += 1
            else:
                s += '%s%s' % (count, al[i])
                count = 1
        a.append(s)
        yield s
a = [i for i in morris(30)]

Answer 1

itertools.groupby似乎非常适合!定义一个 next_morris 函数如下:

def next_morris(number):
    return ''.join('%s%s' % (len(list(group)), digit)
                   for digit, group in itertools.groupby(str(number)))

就是这样!!!看:

print next_morris(1)
11
print next_morris(111221)
312211

---

我可以用它来制作发电机:

def morris_generator(maxlen, start=1):
    num = str(start)
    while len(num) < maxlen:
        yield int(num)
        num = next_morris(num)

用法:

for n in morris_generator(10):
    print n

结果:

1
11
21
1211
111221
312211
13112221