将Python对象转换为JSON输出 [英] Turn Python object into JSON output

问题描述

Python的新手，试图定义一个非常简单的类，该类将包含一些值，然后将其输出为JSON表示法.

New to Python, trying to define a very simple class that will hold a few values, and then get it output into JSON notation.

import json

class Multiple:
    def __init__(self, basis):
            self.double = basis * 2
            self.triple = basis * 3
            self.quadruple = basis * 4


m = Multiple(100)
json.dumps(m)

我希望看到类似的东西

{
  "double":"200",
  "triple":"300",
  "quadruple":"400"
}

但收到错误:"TypeError:< 主要.位于0x0149A3F0>的多个对象不是JSON可序列化的"

but instead get an error: "TypeError: <main.Multiple object at 0x0149A3F0> is not JSON serializable"

推荐答案

您可以序列化m的__dict__属性:

In [214]: json.dumps(m.__dict__)
Out[214]: '{"quadruple": 400, "double": 200, "triple": 300}'

---

您可以也调用

---

You can also call vars:

In [216]: json.dumps(vars(m))
Out[216]: '{"quadruple": 400, "double": 200, "triple": 300}'

使用什么以及为什么: 使用`__dict__`或`vars()`?

What to use and why: Use `__dict__` or `vars()`?

对于更复杂的类，请考虑使用 jsonpickle .

For more complicated classes, consider the use of jsonpickle.

jsonpickle是用于序列化和反序列化的Python库 复杂的Python对象与JSON之间的交互.标准的Python 用于将Python编码为JSON的库，例如stdlib的json， simplejson和demjson，只能处理具有 等效的直接JSON(例如dict s，list s，str ings，int s等). jsonpickle建立在这些库的顶部，并允许更复杂 数据结构要序列化为JSON.

jsonpickle is a Python library for serialization and deserialization of complex Python objects to and from JSON. The standard Python libraries for encoding Python into JSON, such as the stdlib’s json, simplejson, and demjson, can only handle Python primitives that have a direct JSON equivalent (e.g. dicts, lists, strings, ints, etc.). jsonpickle builds on top of these libraries and allows more complex data structures to be serialized to JSON.

强调我的.

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