为什么解析的字典相等而腌制的字典不相等？

Question

我正在开发一个聚合配置文件解析工具，希望它能支持.json、.yaml和.toml文件。所以，我做了下一个测试：

example.json 配置文件如下：

{
  "DEFAULT":
  {
    "ServerAliveInterval": 45,
    "Compression": true,
    "CompressionLevel": 9,
    "ForwardX11": true
  },
  "bitbucket.org":
    {
      "User": "hg"
    },
  "topsecret.server.com":
    {
      "Port": 50022,
      "ForwardX11": false
    },
  "special":
    {
      "path":"C:\\Users",
      "escaped1":"\n\t",
      "escaped2":"\\n\\t"
    }  
}

example.yaml 配置文件如下：

DEFAULT:
  ServerAliveInterval: 45
  Compression: yes
  CompressionLevel: 9
  ForwardX11: yes
bitbucket.org:
  User: hg
topsecret.server.com:
  Port: 50022
  ForwardX11: no
special:
  path: C:\Users
  escaped1: "\n\t"
  escaped2: \n\t

example.toml 配置文件如下：

[DEFAULT]
ServerAliveInterval = 45
Compression = true
CompressionLevel = 9
ForwardX11 = true
['bitbucket.org']
User = 'hg'
['topsecret.server.com']
Port = 50022
ForwardX11 = false
[special]
path = 'C:\Users'
escaped1 = "\n\t"
escaped2 = '\n\t'

然后，带有输出的测试代码如下：

import pickle,json,yaml
# TOML, see https://github.com/hukkin/tomli
try:
    import tomllib
except ModuleNotFoundError:
    import tomli as tomllib

path = "example.json"
with open(path) as file:
    config1 = json.load(file)
    assert isinstance(config1,dict)
    pickled1 = pickle.dumps(config1)

path = "example.yaml"
with open(path, 'r', encoding='utf-8') as file:
    config2 = yaml.safe_load(file)
    assert isinstance(config2,dict)
    pickled2 = pickle.dumps(config2)

path = "example.toml"
with open(path, 'rb') as file:
    config3 = tomllib.load(file)
    assert isinstance(config3,dict)
    pickled3 = pickle.dumps(config3)

print(config1==config2) # True
print(config2==config3) # True
print(pickled1==pickled2) # False
print(pickled2==pickled3) # True

所以，我的问题是，既然解析出来的obj都是dict，而且这些dict是相等的，为什么他们的pickled代码不一样，即为什么从@987654333解析出来的dict的pickled代码@与其他两个不同？

提前致谢。

Answer 1

您是否评估了正在创建的数据类型？可能混合了 str 和 int dtypes。祝你好运！