策略模式与函数式编程应用

策略模式 | 单一职责原则（Single Responsibility Principle, SRP）：islenone和islentwo分别根据特定条件返回电话号码
函数式编程： ‘’ if pd.isna(self.note1) else len(re.findall(r’\d+', self.note1))
重复代码： 当前代码重复还是太高，消除了重复的赋值操作和类似的条件分支。

class hmtq:def __init__(self, phone1, phone2, note1, note3, phonegg):self.phone1 = phone1self.phone2 = phone2self.note1 = note1self.note3 = note3self.phonegg = phoneggself.dict_len = {}def islen(self):self.dict_len['b'] = '' if pd.isna(self.note1) else len(re.findall(r'\d+', self.note1))self.dict_len['c'] =  '' if pd.isna(self.note3) else len(re.findall(r'\d+', self.note3))self.dict_len['a'] = False if pd.isna(self.phonegg)  else self.phoneggreturn self.dict_lendef islenone(self, indata, phone):return phone if int(re.findall(r'\d+', indata)[0]) == 1 else ''def islentwo(self, indata, phone, phone2):return phone if int(re.findall(r'\d+', indata)[0]) == 1 else \(phone2 if int(re.findall(r'\d+', indata)[1]) == 1 else '')def hm(self):self.islen()if self.dict_len['a']==True:a = self.phoneggelif self.dict_len['b'] == 1:a = self.islenone(self.note1, self.phone1)elif self.dict_len['b'] == 2:a = self.islentwo(self.note1, self.phone1, self.phone2)else:a=''if a !='':passelif  self.dict_len['c'] == 1:a = self.islenone(self.note3, self.phone1)elif  self.dict_len['c'] == 2:a = self.islentwo(self.note3, self.phone1, self.phone2)else:a = ''return a

重复代码改进，hm方法使用。hm主处理方法，根据条件选择电话号码，也用上三元运算。
把重复的代码能抽取出来，self.dict_len[‘c’] 和self.dict_len[‘b’] 用到多次了。

class hmtqupdate:def __init__(self, phone1, phone2, note1, note3, phonegg):"""初始化hmtq类实例，设置电话号码和笔记信息。参数:- phone1, phone2: 电话号码1和2- note1, note3: 笔记1和3，可能包含数字- phonegg: 特殊电话号码标记"""self.phone1 = phone1self.phone2 = phone2self.note1 = note1self.note3 = note3self.phonegg = phoneggself.dict_len = self.islen()  # 初始化时直接计算dict_lendef islen(self):"""计算note中数字的数量，并处理phonegg的逻辑。"""self.dict_len = {'b': len(re.findall(r'\d+', self.note1)) if not pd.isna(self.note1) else '','c': len(re.findall(r'\d+', self.note3)) if not pd.isna(self.note3) else '','a': not pd.isna(self.phonegg)}return self.dict_lendef islenone(self, indata, phone):"""如果indata中的第一个数字为1，则返回phone。"""return phone if re.findall(r'\d+', indata) and int(re.findall(r'\d+', indata)[0]) == 1 else ''def islentwo(self, indata, phone, phone2):"""根据indata中的数字决定返回哪个电话号码。"""digits = re.findall(r'\d+', indata)if digits and int(digits[0]) == 1:return phoneif len(digits) > 1 and int(digits[1]) == 1:return phone2return ''def hm(self):"""主处理方法，根据条件选择电话号码。"""# 直接尝试获取'a'源的电话号码a = self.phonegg if self.dict_len['a'] else ''# 如果'a'源未成功获取，尝试'b'源if not a and self.dict_len['b']:a = self.islenone(self.note1, self.phone1) if self.dict_len['b'] == 1 else \self.islentwo(self.note1, self.phone1, self.phone2)# 最后尝试'c'源if not a and self.dict_len['c']:a = self.islenone(self.note3, self.phone1) if self.dict_len['c'] == 1 else \self.islentwo(self.note3, self.phone1, self.phone2)return a