在生成莫里斯序列方面，谁能提供更加Python风格的方法？

Question

我试图用python生成morris sequence。我当前的解决方案如下，但我觉得我只是用python编写了c。有人能提供一个更像Python的解决方案吗？在

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

^{}似乎非常适合！只需定义一个next_morris函数，如下所示：

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

就这些！！！看：

^{pr2}$

---

我可以用它做发电机：

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