A better XML Serializer for Python 3 - python-3.x

Question

I tried xml_marshaller as follows:
from xml_marshaller import xml_marshaller
class Person:
firstName = "John"
lastName = "Doe"
person1 = Person()
strXmlPerson = xml_marshaller.dumps(person1);
print(strXmlPerson)
The output from above is:
b'<marshal><object id="i2" module="__main__" class="Person"><tuple/><dictionary id="i3"><string>firstName</string><string>John</string><string>lastName</string><string>Doe</string></dictionary></object></marshal>'
which when formatted looks like this, which in my opinion is the ugliest XML possible:
b'<marshal>
<object id="i2" module="__main__" class="Person">
<tuple/>
<dictionary id="i3">
<string>firstName</string>
<string>John</string>
<string>lastName</string>
<string>Doe</string>
</dictionary>
</object>
</marshal>'
What is the b and quotes doing there? Means "binary" maybe? Is that really part of the data, or just a side effect of printing it to the console?
Is there any Python 3 library that will create something more closer to "human" like this:
<Person>
<firstname>John</firstname>
<lastname>Doe<lastname>
</Person>
I'm looking for something close to what .NET creates (see http://mylifeismymessage.net/xml-serializerdeserializer/.
Please don't tell me try JSON or YAML, that's not the question. I might want to run the file through XSLT for example.
Update 2 days later:
I like the Peter Hoffman answer here:
How can I convert XML into a Python object?
person1 = Person("John", "Doe")
#strXmlPerson = xml_marshaller.dumps(person1);
person = objectify.Element("Person")
strXmlPerson = lxml.etree.tostring(person1, pretty_print=True)
print(strXmlPerson)
gives error:
TypeError: Type 'Person' cannot be serialized.
In my scenario I might already have a class structure, and don't want to switch to they way they are doing it You can I serialize my "Person" class?

Answer 1

The reason the output is showing xml as a dictionary is most likely because the properties don't have a reference to the class. You should consider using self. and assigning values within a __init__ function.
class Person:
def __init__(self):
self.firstName = "John"
self.lastName = "Doe"
There are many ways to convert an object to XML. However try using the package dicttoxml. As the name suggest you'll need to convert object to a dictionary, which can be done using vars().
The full solution:
from dicttoxml import dicttoxml
class Person:
def __init__(self):
self.firstName = "John"
self.lastName = "Doe"
person = vars(Person()) # vars is pythonic way of converting to dictionary
xml = dicttoxml(person, attr_type=False, custom_root='Person') # set root node to Person
print(xml)
Output:
b'<?xml version="1.0" encoding="UTF-8" ?><Person><firstName>John</firstName><lastName>Doe</lastName></Person>'
If you want to format the XML nicely, then you can use the built in xml.dom.minidom.parseString library.
from dicttoxml import dicttoxml
from xml.dom.minidom import parseString
class Person:
def __init__(self):
self.firstName = "John"
self.lastName = "Doe"
person = vars(Person()) # vars is pythonic way of converting to dictionary
xml = dicttoxml(person, attr_type=False, custom_root='Person') # set root node to Person
print(xml)
dom = parseString(xml)
print(dom.toprettyxml())
Output:
<?xml version="1.0" ?>
<Person>
<firstName>John</firstName>
<lastName>Doe</lastName>
</Person>
Do check out the documentation https://pypi.org/project/dicttoxml/ as you can pass additional arguments to alter the output.