python 正则寻找连续的字符串

Tinylamb

1. s='99669999996669999996699666699666999966699666699'

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现在想匹配到

1. ['99', '66', '999999', '666', '999999', '66', '99', '6666', '99', '666', '9999', '666', '99', '6666', '99']

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也就是所有的连续字符。
这个是我的代码，感觉麻烦了

1. import re
   
2. s='99669999996669999996699666699666999966699666699'
   
3. p=re.compile(r'(.)(\1*)')
   
4. l=p.findall(s)  #l=[('9', '9'), ('6', '6'), ('9', '99999'), ('6', '66'), ('9', '99999'), ('6', '6'), ('9', '9'), ('6', '666'), ('9', '9'), ('6', '66'), ('9', '999'), ('6', '66'), ('9', '9'), ('6', '666'), ('9', '9')]  匹配了开头，跟接下来连续的字符
   
5. 
   
6. li=[]  #现在将每个开头与连续的部分合并
   
7. for i in l:
   
8.     i=''.join(i)
   
9.     li.append(i)
   
10. 
    
11. li   #li就是结果

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抛砖引玉，应该有一个正则就搞定的方法

lkk2003rty

一定要用正则么？
t =  l = '99669999996669999996699666699666999966699666699'[0]
for i in '99669999996669999996699666699666999966699666699':
    if cmp(l,i) != 0:
        ret.append(t)
        print t
        t = l = i
        continue
    t += i