如何将嵌套类转换为 json 或 dict？

发布于 2025-01-10 09:12:29 字数 432 浏览 0 评论 0原文

我有如下的类：


class Base:
  def __init__(self):
    self.var1 = “variable”

class Child:
  def __init__(self):
    self.base = Base()
    self.var2 = False

class RootChild:
  def __init__(self):
    self.child = Child()
    self.var3 = “variable”
    self.base = Base()

我想保存“RootChild”的一个实例，其所有字段都以 json 表示。关于此任务有很多主题，但我无法理解它们的差异和用例。这个问题的大多数 Pythonic 解决方案是什么？

原文

I have class of classes as below:


class Base:
  def __init__(self):
    self.var1 = “variable”

class Child:
  def __init__(self):
    self.base = Base()
    self.var2 = False

class RootChild:
  def __init__(self):
    self.child = Child()
    self.var3 = “variable”
    self.base = Base()

I want to save an instance of ‘RootChild’ with all fields of it represented in json. There are a lot of topics about this task, but I could not understand their differences and use cases. What are most pythonic solutions to this problem?

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自演自醉 2025-01-17 09:12:29

您的里程会有所不同 - 从使用“序列化器”策略：一层机制知道如何从您的类映射到 Json 并返回，
这可能需要大量工作（手动描述应该序列化/反序列化的每个字段） - https://docs.pylonsproject.org/projects/colander/en/latest/

或者您可以构建一个自定义 JSON 编码器，该编码器仅检查它是否是已知类，然后返回一个包含所有字段的字典（可以通过内省自动提取）-（检查示例以扩展编码器：https://docs.python.org/3/library/json.html ）

无论如何，都有很多权衡和不同级别的工作 -
如果您打算在具有相同类的 Python 应用程序中读回 JSON，您可以使用“jsonpickle” - 它将向您的 JSON 添加额外的元数据，但将确保所有对象的往返完全按原样：< a href="https://jsonpickle.github.io/" rel="nofollow noreferrer">https://jsonpickle.github.io/

不使用任何其他库，并假设您只需要导出类，自定义编码器策略可以有点简单：

import json
class SimpleObject(json.JSONEncoder):
    def default(self, obj):
        if hasattr(obj, "__dict__"):
            return {key:value for key, value in obj.__dict__.items() if not key.startswith("_")}
        return super().default(obj)
a = RootChild()

在交互环境上：

In [52]: a = RootChild()
In [53]: print(json.dumps(a, cls=SimpleObject, indent=4))
{
    "child": {
        "base": {
            "var1": "variable"
        },
        "var2": false
    },
    "var3": "variable",
    "base": {
        "var1": "variable"
    }
}

Your mileage will vary - from using a "serializer" strategy: a layer of mechanisms that will know how to map from your classes to Json and back,
which can be a lot of work (to the point of hand-describing each field that should be serialized/unserialized) - https://docs.pylonsproject.org/projects/colander/en/latest/

Or you could build a custom JSON encoder that will just check if it is a known class and then return a dict with all its fields (which can be extracted automatically by introspection) - (check the example to extend the encoder here: https://docs.python.org/3/library/json.html )

Any way there are lots of trade-offs and differing level of work -
if you plan to read the JSON back in Python apps which have the same classes, you could use "jsonpickle" - it will add extra meta-data to your JSON, but will ensure round tripping of all your objects exactly as they are: https://jsonpickle.github.io/

Without using any other lib, and assuming you just need to export your classes, the custom encoder strategy can be somewhat simple:

import json
class SimpleObject(json.JSONEncoder):
    def default(self, obj):
        if hasattr(obj, "__dict__"):
            return {key:value for key, value in obj.__dict__.items() if not key.startswith("_")}
        return super().default(obj)
a = RootChild()

And on the interactive environment:

In [52]: a = RootChild()
In [53]: print(json.dumps(a, cls=SimpleObject, indent=4))
{
    "child": {
        "base": {
            "var1": "variable"
        },
        "var2": false
    },
    "var3": "variable",
    "base": {
        "var1": "variable"
    }
}

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