Path: csiph.com!newsfeed.hal-mli.net!feeder3.hal-mli.net!newsfeed.hal-mli.net!feeder1.hal-mli.net!newsfeed.xs4all.nl!newsfeed1.news.xs4all.nl!xs4all!newsgate.cistron.nl!newsgate.news.xs4all.nl!post.news.xs4all.nl!not-for-mail Return-Path: X-Original-To: python-list@python.org Delivered-To: python-list@mail.python.org X-Spam-Status: OK 0.001 X-Spam-Evidence: '*H*': 1.00; '*S*': 0.00; 'else:': 0.03; 'root': 0.05; 'element': 0.07; 'nested': 0.07; '-larry': 0.09; 'function,': 0.09; 'iterate': 0.09; 'parsing': 0.09; 'stack.': 0.09; 'subject:fields': 0.09; 'subject:parsing': 0.09; 'cc:addr :python-list': 0.11; 'def': 0.12; '"e"': 0.16; 'iterable,': 0.16; 'nesting': 0.16; 'node,': 0.16; 'node.': 0.16; 'nodes': 0.16; 'parent:': 0.16; 'pop()': 0.16; 'preserve': 0.16; 'redundant.': 0.16; 'subject:XML': 0.16; 'wrote:': 0.18; 'result.': 0.19; 'stefan': 0.19; 'examples': 0.20; 'seems': 0.21; 'putting': 0.22; 'cc:addr:python.org': 0.22; 'print': 0.22; 'parse': 0.24; 'tells': 0.24; 'fairly': 0.24; 'cc:2**0': 0.24; 'cc:no real name:2**0': 0.24; 'pass': 0.26; 'header:In-Reply-To:1': 0.27; 'tried': 0.27; 'function': 0.29; 'am,': 0.29; 'xml': 0.29; 'statement': 0.30; 'message-id:@mail.gmail.com': 0.30; "i'm": 0.30; 'gives': 0.31; 'code': 0.31; 'reply.': 0.31; 'depth': 0.31; 'end,': 0.31; 'node': 0.31; 'file': 0.32; 'this.': 0.32; 'thanks!': 0.32; 'up.': 0.33; 'running': 0.33; 'beginning': 0.33; 'plain': 0.33; 'could': 0.34; 'problem': 0.35; 'subject:with': 0.35; "can't": 0.35; 'common': 0.35; 'done.': 0.35; 'but': 0.35; 'received:google.com': 0.35; 'i.e.': 0.36; 'done': 0.36; 'thanks': 0.36; 'should': 0.36; 'list': 0.37; 'level': 0.37; 'nov': 0.38; 'called': 0.40; 'how': 0.40; 'remove': 0.60; 'skip:n 30': 0.60; 'then,': 0.60; 'reaching': 0.61; 'entire': 0.61; 'simply': 0.61; "you're": 0.61; 'times': 0.62; "you'll": 0.62; 'show': 0.63; 'visit': 0.64; '26,': 0.68; 'recursion,': 0.84; 'unclear': 0.84; '2013': 0.98 DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:in-reply-to:references:date:message-id:subject:from:to :cc:content-type; bh=tpk1HXm5VxEc6xKPOiNdQk2NRjRTO09rqNhmR5zlZxI=; b=dxYQiByzKqLXKp4Wc9CgAsyHvdcIUDwUYDHgDHWH63U0wxlOIZM5drmXUZg5FHmZYx SxWOs92ea0oBeMGtJ9JSbKTrPED9OJfa7iCuCYCONIjPjKWXLqkCGUrhUv4loJgDvnY5 3GSbSZgU0wSQcAFei13LB0BkYv0UJOKYAWgyPbHZr0wObWy/4HAQE++SJhdYdP/X6ovr uo/vYEzsqdIgnPBnmX5wXCTm61CfaUAm9jNy+WpOCavPG7NlLA8zf18IegzL69znKu15 SRR+PAvsJ8K9H2BT+OyYKeXwR9D5FugE4zZQNGyWgPTY13skMHaieUAavjVqXaQ5FePz r4fA== MIME-Version: 1.0 X-Received: by 10.180.185.130 with SMTP id fc2mr17677426wic.43.1385468600196; Tue, 26 Nov 2013 04:23:20 -0800 (PST) In-Reply-To: References: Date: Tue, 26 Nov 2013 07:23:20 -0500 Subject: Re: parsing nested unbounded XML fields with ElementTree From: Larry Martell To: Stefan Behnel Content-Type: text/plain; charset=UTF-8 Cc: python-list@python.org X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.15 Precedence: list List-Id: General discussion list for the Python programming language List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Newsgroups: comp.lang.python Message-ID: Lines: 64 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1385468604 news.xs4all.nl 15975 [2001:888:2000:d::a6]:48664 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:60506 On Tue, Nov 26, 2013 at 2:38 AM, Stefan Behnel wrote: > Larry.Martell...@gmail.com, 25.11.2013 23:22: >> I have an XML file that has an element called "Node". These can be nested to any depth and the depth of the nesting is not known to me. I need to parse the file and preserve the nesting. For exmaple, if the XML file had: >> >> >> >> >> >> >> >> When I'm parsing Node "E" I need to know I'm in A/B/C/D/E. Problem is I don't know how deep this can be. This is the code I have so far: >> >> nodes = [] >> >> def parseChild(c): >> if c.tag == 'Node': >> if 'Name' in c.attrib: >> nodes.append(c.attrib['Name']) >> for c1 in c: >> parseChild(c1) >> else: >> for node in nodes: >> print node, >> print c.tag >> >> for parent in tree.getiterator(): >> for child in parent: >> for x in child: >> parseChild(x) > > This seems hugely redundant. tree.getiterator() already returns a recursive > iterable, and then, for each nodes in your document, you are running > recursively over its entire subtree. Meaning that you'll visit each node as > many times as its depth in the tree. > > >> My problem is that I don't know when I'm done with a node and I should >> remove a level of nesting. I would think this is a fairly common >> situation, but I could not find any examples of parsing a file like >> this. Perhaps I'm going about it completely wrong. > > Your recursive traversal function tells you when you're done. If you drop > the getiterator() bit, reaching the end of parseChild() means that you're > done with the element and start backing up. So you can simply pass down a > list of element names that you append() at the beginning of the function > and pop() at the end, i.e. a stack. That list will then always give you the > current path from the root node. Thanks for the reply. How can I remove getiterator()? Then I won't be traversing the nodes of the tree. I can't iterate over tree. I am also unclear on where to do the pop(). I tried putting it just after the recursive call to parseChild() and I tried putting as the very last statement in parseChild() - neither one gave the desired result. Can you show me in code what you mean? Thanks! -larry > > Alternatively, if you want to use lxml.etree instead of ElementTree, you > can use it's iterwalk() function, which gives you the same thing but > without recursion, as a plain iterator. > > http://lxml.de/parsing.html#iterparse-and-iterwalk