我试图使用PyYAML序列化一个使用SampleClass
实例作为键的字典。它序列化好,但是当我尝试用yaml.load()
加载它时,它引发了一个异常:
AttributeError:'SampleClass'对象没有属性'name'
怎么解决这个问题? SampleClass
看起来像这样:
import uuid
class SampleClass:
def __init__(self, name = "<NO NAME>"):
self.objects = []
self.name = name
self.id = uuid.uuid1()
def __eq__(self, other):
if isinstance(other, SampleClass):
return self.name == other.name and \
self.objects == other.objects and \
self.id == other.id
else:
return False
def __hash__(self):
return hash((str(self.name), str(self.id)))
PyYAML有点过时,它只支持YAML 1.1,它在2009年被YAML 1.2取代。另请注意,虽然PyYAML可以解析YAML映射中的复杂键(例如,键序列或映射本身),但是有效的键在YAML,它无法在Python中构造它们,实际上无法加载它们。
使用ruamel.yaml
(免责声明:我是该软件包的作者),您可以简单地执行:
import sys
import uuid
import ruamel.yaml
from ruamel.yaml.compat import StringIO
class SampleClass:
def __init__(self, name = "<NO NAME>"):
self.objects = []
self.name = name
self.id = uuid.uuid1()
def __eq__(self, other):
if isinstance(other, SampleClass):
return self.name == other.name and \
self.objects == other.objects and \
self.id == other.id
else:
return False
def __hash__(self):
return hash((str(self.name), str(self.id)))
def __repr__(self):
return "SampleClass({})".format(self. name)
data = {SampleClass("abc"): 1, SampleClass("xyz"): 42}
yaml = ruamel.yaml.YAML(typ="unsafe")
buf = StringIO()
yaml.dump(data, buf)
x = yaml.load(buf.getvalue())
print(x)
这使:
{SampleClass(abc): 1, SampleClass(xyz): 42}
但我建议为to_yaml
提供from_yaml
和SampleClass
例程并注册课程(doc)。这允许您取消不安全的加载(这是BTY默认的PyYAML)。