How can ignore the member Trivial._ignore when serializing this object?
import yaml
class Trivial(yaml.YAMLObject):
yaml_tag = u'!Trivial'
def __init__(self):
self.a = 1
self.b = 2
self._ignore = 3
t = Trivial()
print(yaml.dump(t))
prints
!Trivial
_ignore: 3
a: 1
b: 2
def my_yaml_dump(yaml_obj):
my_ob = deepcopy(yaml_obj)
for item in dir(my_ob):
if item.startswith("_") and not item.startswith("__"):
del my_ob.__dict__[item]
return yaml.dump(my_ob)
something like this would ignore anything with a single (leading) underscore
I think this is what you want
class Trivial(yaml.YAMLObject):
yaml_tag = u'!Trivial'
def __init__(self):
self.a = 1
self.b = 2
self._ignore = 3
@classmethod
def to_yaml(cls, dumper, data):
# ...
my_ob = deepcopy(data)
for item in dir(my_ob):
if item.startswith("_") and not item.startswith("__"):
del my_ob.__dict__[item]
return dumper.represent_yaml_object(cls.yaml_tag, my_ob, cls,
flow_style=cls.yaml_flow_style)
although a better method (stylistically)
class SecretYamlObject(yaml.YAMLObject):
hidden_fields = []
@classmethod
def to_yaml(cls,dumper,data):
new_data = deepcopy(data)
for item in cls.hidden_fields:
del new_data.__dict__[item]
return dumper.represent_yaml_object(cls.yaml_tag, new_data, cls,
flow_style=cls.yaml_flow_style)
class Trivial(SecretYamlObject):
hidden_fields = ["_ignore"]
yaml_tag = u'!Trivial'
def __init__(self):
self.a = 1
self.b = 2
self._ignore = 3
print yaml.dump(Trivial())
this is adhering to pythons mantra of explicit is better than implicit
