Python 有没有迭代__u插槽并动态赋值的方法？_Python_Slots

Python 有没有迭代__u插槽并动态赋值的方法？

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Python 有没有迭代__u插槽并动态赋值的方法？,python,slots,Python,Slots,我正在为一个基于代理的模型创建一个框架，我有一个叫做代理的类。由于模拟中将有数千个代理，我想使用_uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu，它将替换默认的uuuuuuuuuuuuuuuuuuuuuuuuuuuuu。我对代码进行了结构化，以便它从数据表中获取代理的参数，并且我希望将存储在表中的值分配给具有表头名称的属性如果以下是数据表 | agent_name | location | attr1 | at

我正在为一个基于代理的模型创建一个框架，我有一个叫做代理的类。由于模拟中将有数千个代理，我想使用_uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu，它将替换默认的uuuuuuuuuuuuuuuuuuuuuuuuuuuuu。我对代码进行了结构化，以便它从数据表中获取代理的参数，并且我希望将存储在表中的值分配给具有表头名称的属性

如果以下是数据表

| agent_name | location | attr1 | attr2 | |--------------|------------|-------------|------------| | agent smith | NY | some value | some value | | Neo | NY | some value | some value | | Morpheus | Zion | some value | some value | 我知道你可以在for循环中使用.eval方法，但我觉得这并不干净，我觉得应该有更好的方法。我想知道是否有一种方法可以遍历插槽并为每个属性赋值。

一些指针：

# illustration of what I want
header_of_table = ["agent_name", "location", "attr1", "attr2"]

class agent:
    __slots__ = header_of_table
    def __init__(self, values_in_row):
        # what I ideally want, I know this syntax doesnt work, but this is to get the idea across
        for slotted_attribute, value in zip(self.__slots__, values_in_row):
            self.slotted_attribute = value

使用

\uuuu插槽时，需要从对象继承。
Python 2


类应始终大写
正如Michael在评论中所写，您可以使用setattr
它被称为\uuuuuuuuuuuuuuuuuuuuuuuuu
而不是\uuuuuuuuuuuuuuuuuu


下面是一个基于您的代码的工作示例：
# illustration of what I want
header_of_table = ["agent_name", "location", "attr1", "attr2"]

class Agent:
    __slots__ = header_of_table
    def __init__(self, values_in_row):
        for i, slot in enumerate(self.__slots__):
            self.__setattr__(slot, values_in_row[i])

agent = Agent(["foo", "bar", "yellow", "green"])
print(agent.agent_name, agent.location, agent.attr1, agent.attr2)

>>> foo bar yellow green

编辑评论：
如果我理解正确的话，我会这样做以避免污染全球范围
slotheaders.py：
class SlotHeaders:
    __slots__ = ["agent_name", "location", "attr1", "attr2"]

agent.py：
from slotheaders import SlotHeaders

class Agent(SlotHeaders):
    def __init__(self, values_in_row):
        for i, slot in enumerate(self.__slots__):
            self.__setattr__(slot, values_in_row[i])

agent = Agent(["foo", "bar", "yellow", "green"])
print(agent.agent_name, agent.location, agent.attr1, agent.attr2)

几点建议：

使用\uuuu插槽时，需要从对象继承。
Python 2

类应始终大写
正如Michael在评论中所写，您可以使用setattr
它被称为\uuuuuuuuuuuuuuuuuuuuuuuuu
而不是\uuuuuuuuuuuuuuuuuu

下面是一个基于您的代码的工作示例：
# illustration of what I want
header_of_table = ["agent_name", "location", "attr1", "attr2"]

class Agent:
    __slots__ = header_of_table
    def __init__(self, values_in_row):
        for i, slot in enumerate(self.__slots__):
            self.__setattr__(slot, values_in_row[i])

agent = Agent(["foo", "bar", "yellow", "green"])
print(agent.agent_name, agent.location, agent.attr1, agent.attr2)

>>> foo bar yellow green

编辑评论：
如果我理解正确的话，我会这样做以避免污染全球范围
slotheaders.py：
class SlotHeaders:
    __slots__ = ["agent_name", "location", "attr1", "attr2"]

agent.py：
from slotheaders import SlotHeaders

class Agent(SlotHeaders):
    def __init__(self, values_in_row):
        for i, slot in enumerate(self.__slots__):
            self.__setattr__(slot, values_in_row[i])

agent = Agent(["foo", "bar", "yellow", "green"])
print(agent.agent_name, agent.location, agent.attr1, agent.attr2)

这里有一个“setattr”。应该读一下：它说，例如，您的类需要从对象继承。另外，它不是\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。我替换了所有这些输入错误。这里有“setattr”。应该读一下：它说，例如，您的类需要从对象继承。另外，它不是\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。我替换了所有这些输入错误。从对象
继承只是Python 2中的一件事。在Python3中，这是自动发生的，除非需要与Python2向后兼容，否则不需要指定它（现在可能没有）。我从编辑器复制了粘贴，所以我没有注意到我忘记了将类名大写，感谢您指出，我会确保更正它。@Mandera我对当前的答案很满意，但是行的头\u
在全局范围内，如果我想限制范围，那么我需要将代理类嵌套在可以存储行的头\u
的函数中。此元函数的目的仅是将行的标题分配给\uuuu插槽\uuu
。这是我唯一能想到的限制范围的方法，有没有更好的方法将变量分配给\uuuuuu slots\uuuuu
？或者我应该创建一个新问题，因为它与原始问题不同？@KukaiNakahata编辑了我的答案，你认为如何？@Mandera这是一个优雅的解决方案，比将其包装在函数中要好。谢谢，这是一个很好的答案。从object
继承只是Python 2中的一件事。在Python3中，这是自动发生的，除非需要与Python2向后兼容，否则不需要指定它（现在可能没有）。我从编辑器复制了粘贴，所以我没有注意到我忘记了将类名大写，感谢您指出，我会确保更正它。@Mandera我对当前的答案很满意，但是行的头\u
在全局范围内，如果我想限制范围，那么我需要将代理类嵌套在可以存储行的头\u
的函数中。此元函数的目的仅是将行的标题分配给\uuuu插槽\uuu
。这是我唯一能想到的限制范围的方法，有没有更好的方法将变量分配给\uuuuuu slots\uuuuu
？或者我应该创建一个新问题，因为它与原始问题不同？@KukaiNakahata编辑了我的答案，你认为如何？@Mandera这是一个优雅的解决方案，比将其包装在函数中要好。谢谢，这是一个很好的回答。