nukl nukl - 1 month ago 12
Python Question

Efficient way to iterate throught xml elements

i have a xml like this:


I need to iterate through all
tags, but i don't know how many of them are in document. So i use
to handle that:

from lxml import etree

doc = etree.fromstring(xml)

atags = doc.xpath('//a')
for a in atags:
btags = a.xpath('b')
for b in btags:
print b

It works, but i have pretty big files, and
shows me that
is very expensive to use.

I wonder, maybe there is there more efficient way to iterate through indefinitely number of xml-elements?


XPath should be fast. You can reduce the number of XPath calls to one:

doc = etree.fromstring(xml)
btags = doc.xpath('//a/b')
for b in btags:
    print b.text

If that is not fast enough, you could try Liza Daly's fast_iter. This has the advantage of not requiring that the entire XML be processed with etree.fromstring first, and parent nodes are thrown away after the children have been visited. Both of these things help reduce the memory requirements. Below is a modified version of fast_iter which is more aggressive about removing other elements that are no longer needed.

def fast_iter(context, func, *args, **kwargs):
    fast_iter is useful if you need to free memory while iterating through a
    very large XML file.
    Based on Liza Daly's fast_iter
    See also
    for event, elem in context:
        func(elem, *args, **kwargs)
        # It's safe to call clear() here because no descendants will be
        # accessed
        # Also eliminate now-empty references from the root node to elem
        for ancestor in elem.xpath('ancestor-or-self::*'):
            while ancestor.getprevious() is not None:
                del ancestor.getparent()[0]
    del context

def process_element(elt):

context=etree.iterparse(io.BytesIO(xml), events=('end',), tag='b')
fast_iter(context, process_element)

Liza Daly's article on parsing large XML files may prove useful reading to you too. According to the article, lxml with fast_iter can be faster than cElementTree's iterparse. (See Table 1).