有效地连接地址线

问题描述 投票:1回答:2

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?

python algorithm concatenation street-address
2个回答
2
投票

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)

0
投票

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)

如:https://repl.it/repls/LiveJuniorLanservers

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