我的数据库存储了一些标准地址行字段(地址行1,2 ..城市,国家,邮政编码),我想将它们连接成人形。我编写了下面的代码但是我不确定这对于一堆if语句是否非常有效。 (用python编写,但它是我关心的算法)
def human_readable_address(self):
'''
Return human readable address
If address1 is empyty, return None
'''
addr = ""
if(self.address1):
addr += self.address1 + ", "
else:
return None
if(self.address2):
addr += self.address2 + ", "
if(self.city):
addr += self.city + ", "
if(self.postal_code):
addr += self.postal_code + ", "
if(self.country):
addr += self.country + ", "
return addr
你们有什么感想?有没有更好的办法?
您可以使用str.join()
而不是使用字符串连接(每次创建一个新字符串):
def human_readable_address(self):
'''
Return human readable address
If address1 is empty, return None
'''
if not self.address1:
return None
parts = list(filter(None, [self.address1, self.address2, self.city,
self.postal_code, self.country]))
return ', '.join(parts)
您可以使用列表解析,加入字符串,以您需要的格式获取地址:
address_1 = "first"
address_2 = False
address_3 = "third"
def make_address(*args):
return "".join(["%s,\n" % x for x in args if x])[:-2]
address = make_address(address_1, address_2, address_3)
print(address)