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我正在尝试从 4 个不同的列表中获取以下输出，我需要调用 address_machine 方法并从列表中压缩项目并使用 for 循环遍历新列表，然后将它们打印在新行上 --> 预期输出以下：

["T Cruise","D Francis","C White"]
["2 West St","65 Deadend Cls","15 Magdalen Rd"]
["Canterbury", "Reading", "Oxford"]
["CT8 23RD", "RG4 1FG", "OX4 3AS"]

我的方法如下所示：

def address_machine(name, street_address, town, postcode):
    address = "{0},{1},{2},{3}".format(name, street_address, town, postcode)
    return address

print([address_machine(name, street_address, town, postcode) for name, street_address, town, postcode in
       zip(["T Cruise", "D Francis", "C White"], ["2 West St", "65 Deadend Cls", "15 Magdalen Rd"],
           ["Canterbury", "Reading", "Oxford"], ["CT8 23RD", "RG4 1FG", "OX4 3AS"])])

这是一个问题，因为我得到的输出看起来像这样，并且全部在一行上：

['T Cruise,2 West St,Canterbury,CT8 23RD', 'D Francis,65 Deadend Cls,Reading,RG4 1FG', 'C White,15 Magdalen Rd,Oxford,OX4 3AS']

我需要如何编写这个 Python adress_machine 函数来获得预期的输出？

4

2 回答 2

1

你可以使用正常for的循环

for name, street_address, town, postcode in zip(["T Cruise", "D Francis", "C White"], ["2 West St", "65 Deadend Cls", "15 Magdalen Rd"],
       ["Canterbury", "Reading", "Oxford"], ["CT8 23RD", "RG4 1FG", "OX4 3AS"]):

     print([name, street_address, town, postcode])

或首先创建包含所有元素的列表，然后for-loop 来显示它

data = [[name, street_address, town, postcode] for name, street_address, town, postcode in zip(["T Cruise", "D Francis", "C White"], ["2 West St", "65 Deadend Cls", "15 Magdalen Rd"],
       ["Canterbury", "Reading", "Oxford"], ["CT8 23RD", "RG4 1FG", "OX4 3AS"])]

for item in data:
    print(item)

如果您期望列表的结果，不明白为什么要使用address_machine()将列表转换为字符串的函数。

于 2021-01-03T22:27:10.600 回答
1

将您的 address_machine 函数更改为：

def address_machine(name, street_address, town, postcode):
    return "{0},{1},{2},{3}\n".format(name, street_address, town, postcode)

您必须返回一个额外的换行符。并且直接返回而不是首先定义一个变量使代码更短！

于 2021-01-04T01:28:08.733 回答