Question

如何打破此Python行以保持在PEP 8的79个字符的限制内？

config["network"]["connection"]["client_properties"]["service"] = config["network"]["connection"]["client_properties"]["service"].format(service=service)

Answer 1

考虑到Python可与引用一起使用，您可以执行以下操作：

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Answer 2

使用\：

config["network"]["connection"]["client_properties"]["service"] = \
config["network"]["connection"]["client_properties"]["service"].format(
service=service)

Answer 3

使用black，自以为是的可复制代码格式化程序：

config["network"]["connection"]["client_properties"][
    "service"
] = config["network"]["connection"]["client_properties"][
    "service"
].format(
    service=service
)

Answer 4

方括号允许隐式连续行。例如，

config["network"
]["connection"
]["client_properties"
]["service"] = config["network"]["connection"]["client_properties"]["service"].format(
service=service)

那就是说，我不认为每个括号应该在哪一行上达成共识。（就我个人而言，我从未找到过看起来特别“正确”的任何选择。）

更好的解决方案可能是引入一个临时变量。

d = config["network"]["connection"]["client_properties"]
d["service"] = d["service"].format(service=service)

Answer 5

您也可以使用变量来更好地阅读：

client_service = config["network"]["connection"]["client_properties"]["service"]
client_service = client_service.format(service=service)

# If you are using the value later in your code keeping it in an variable may
# increase readability
...
# else you can put it back
config["network"]["connection"]["client_properties"]["service"] = client_service

Answer 6

其他人试图将其分解为两行，但您的问题是它仍然很冗长。您可以将config["network"]["connection"]["client_properties"]["service"]分配给它自己的变量。

config_service = config["network"]["connection"]["client_properties"]["service"]
config_service = config_service.format(service=service)

# If you really need it at the original variable
config["network"]["connection"]["client_properties"]["service"] = config_service

如何打破这条Python线？

6 个答案: