python使用re匹配字符串中重复出现的字母

Question

（1）现在想要将字符串中连续出现的同个字母去重，如"abbbcccbba" -> "abcba"，使用re模块的话如何优雅的完成这件事情？

---

（2）完成需求(1)后，能否顺便统计连续出现的个数，如"abbbcccbba" -> "a1b3c3b2a1"?

---

Answer 1

(1)

>>> import re
>>> p = re.compile(ur"([a-zA-Z])(\1+)")
>>> s = "abbbcccbba"
>>> p.sub(ur"\1",s)
'abcba'
>>> 

(2)

>>> import re
>>> p = re.compile(ur"([a-zA-Z])(\1*)")
>>> s = "abbbcccbba"
>>> p.sub(lambda m: m.group(1)+str(1+len(m.group(2))), s)
'a1b3c3b2a1'

Answer 2

（1）

>>> import re
>>> p=re.compile(ur"(\w)(\1+)")
>>> s="abbbcccbba"
>>> p.sub(ur"\1",s)
'abcba'
>>> 

（2）

import re


def count(s):
    p = re.compile(ur"(\w)(\1+)")
    keys = list(p.sub(ur"\1", s))
    words = list(s)
    result = []
    # print keys, words
    for k in keys:
        n = 0
        # print words
        while len(words) > n and k == words[n]:
            n = n + 1
        words = words[n:]
        result.append((k, n))
        # print result
    return result

if __name__ == '__main__':
    s = "abbbcccbba"
    result = count(s)
    print ''.join(["%s%s" % x for x in result])