CODEWITHSUNDEEP
pythonxml-parsing
guettli
guettli

Reputation: 27969

xml.etree.ElementTree getElementByID()?

How to get the equivalent of getElementByID() with the Python library xml.etree.ElementTree?

There seems to be a method called parseid() but my tree is already parsed. I don't want to parse it again.

Upvotes: 3

Views: 4476

Answers (2)

user1556435
user1556435

Reputation: 1056

The accepted answer works indeed, but performance can be quite abysmal as - my guess is, I didn't verify this, perhaps also related to the complexity of xpath - the tree is traversed on every to findall(), which may or may not be a concern for your use case.

Probably parseid() is indeed what you want if performance is a concern. If you want to obtain such an id mapping on an existing tree, you can also easily perform the traversal once manually.

class getElementById():

    def __init__(self, tree):
        self.di = {}

        def v(node):
            i = node.attrib.get("id")
            if i is not None:
                self.di[i] = node

            for child in node:
                v(child)

        v(tree.getroot())


    def __call__(self, k):
        return self.di[k]

Upvotes: 4

guettli
guettli

Reputation: 27969

I found it myself:

tree.findall('''.//*[@id='fooID']''')[0]

Better or other solutions are still welcome. :-)

Upvotes: 3

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