lxml node comparison

Is there a way to check nodes equal with lxml library? For example in php DOMDocument there is isSameNode:

a->isSameNode(b) //return b开发者_如何学Pythonoolean

I need it to do something like this:

def find_special_node(special_node, xml):
    #iterating nodes in xml
    if node == special_node:
        return node

You may be able to use xpath to find the item, using attributes of "special_node" as required attributes (and values) in find(). I assume that two nodes are determined to be the same if all attributes of the nodes are the same (such as type, value, size, etc.)
Just a guess

My implementation:

import lxml.etree as etree
def isSameNode(a,b,tagsame=True):
    for attrib in a.attrib:
        if a.get(attrib) != b.get(attrib):
            print a.get(attrib),b.get(attrib)
            return False
        else:
            return False
    if a.text != b.text:
        return False
    if tagsame==True:
        if a.tag != b.tag:
            return False
    if a.prefix != b.prefix:
        return False
    if a.tail != b.tail:
        return False
    if a.values()!=b.values(): #may be redundant to the attrib matching
        return False
    if a.keys() != b.keys(): #may also be redundant to the attrib matching
        return False
    return True

def find_alike(xml,special_element,tagsame=True):
    tree = etree.fromstring(xml)
    for node in tree.iter():
        if isSameNode(node,special_element,tagsame):
            return node
    return None

Not sure if you want to check NodeA is NodeB or NodeA == NodeB

You may try:

for node in X.iter():
    if node == special_node: # maybe "node is special_node" ?
        return node

Anyway, since you already have special_node, why would you search for it?