error case for cut

[Argons]

在cut类似“东方XXX”的词时，eg：东方电视台，会无法返回分词结果，但也不会报错。

[jannson]

好的,我跟踪下 :)

[zyson]

个人认为是在yaha/init.py文件的人名前缀匹配函数SurnameCutting2()中，while循环中的判断条件没有考虑到klen2大于2的情况，会造成死循环导致的。可以这么改看看
`
class SurnameCutting2(CuttingBase):
def init(self):
super(CuttingBase, self).init()
self.stage = 4
self.dict = get_dict(DICTS.SURNAME)
self.stop_dict = get_dict(DICTS.EXT_STOPWORD)

def cut_stage4(self, cuttor, sentence, graph, contex):
    start = contex['index']
    path = contex['path']
    new_path = contex['new_path']
    n = len(path)
    i = start
    while i < n:
        klen = path[i] - path[i-1]
        print 'surname', klen, sentence[path[i-1]:path[i]]
        if klen >=1 and klen <= 2 and self.dict.has_key(sentence[path[i-1]:path[i]]):
            if i < n-2:
                klen2 = path[i+1]-path[i]
                klen3 = path[i+2]-path[i+1]
                if klen2==1 and klen3==1 and not self.stop_dict.has_key(sentence[path[i+1]:path[i+2]]):
                    new_path.append(path[i+2])
                    i += 3
                    continue
                elif klen2 <= 2 and klen2 >= 1:
                    new_path.append(path[i+1])
                    i += 2
                    continue
                else:
                    new_path.append(path[i])
                    i += 1
                    continue
            elif i < n-1:
                klen2 = path[i+1]-path[i]
                if klen2 <= 2 and klen2 >= 1:
                    new_path.append(path[i+1])
                    # the code like a bug
                    # i = path[i+1]
                    i += 2
                    continue
                else:
                    new_path.append(path[i])
                    i += 1
                    continue
            else:
                return i
        else:
            # next stage to doit
            return i
    return i

`

[jannson]

好的，感谢，我改着试试。