Re: lxml, comparing nodes

[code_berzerker]

On Jul 23, 6:29*pm, Stefan Behnel <[e-mail]:

Your requirements for a single Element are simple enough to write it in three
to five lines of Python code (depending on your definition of equality).
Checking this equality recursively is another two to three lines. Not complex
enough to be considered a wheel in the first place.

Forgive my ignorance as I am new to both Python and lxml ;)

[Fredrik Lundh]

code_berzerker wrote:

>Your requirements for a single Element are simple enough to write it in three
to five lines of Python code (depending on your definition of equality).
Checking this equality recursively is another two to three lines. Not complex
enough to be considered a wheel in the first place.

Forgive my ignorance as I am new to both Python and lxml ;)

off the top of my head (untested):

>>def equal(a, b):

.... if a.tag != b.tag or a.attrib != b.attrib:
.... return False
.... if a.text != b.text or a.tail != b.tail:
.... return False
.... if len(a) != len(b):
.... return False
.... if any(not equal(a, b) for a, b in zip(a, b)):
.... return False
.... return True

this should work for arbitrary ET implementations (lxmk, xml.etree, ET,
etc). tweak as necessary.

</F>