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我正在尝试编写一个解析算法来有效地从 xml 文档中提取数据。我目前正在基于元素和子元素滚动文档，但想改用 iterparse。一个问题是我有一个元素列表，当找到这些元素时，我想从中提取子数据，但似乎使用 iterparse 我的选项是基于一个元素名称进行过滤，或者获取每个元素。

示例 xml：

<?xml version="1.0" encoding="UTF-8"?>
<data_object xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
 <source id="0">
  <name>Office Issues</name>
  <datetime>2012-01-13T16:09:15</datetime>
  <data_id>7</data_id>
 </source>
 <event id="125">
  <date>2012-11-06</date>
  <state_id>7</state_id>
 </event>
 <state id="7">
  <name>Washington</name>
 </state>
 <locality id="2">
  <name>Olympia</name>
  <state_id>7</state_id>
  <type>City</type>
 </locality>
 <locality id="3">
  <name>Town</name>
  <state_id>7</state_id>
  <type>Town</type>
 </locality>
</data_object>

代码示例：

from lxml import etree

fname = "test.xml"
ELEMENT_LIST = ["source", "event", "state", "locality"]

with open(fname) as xml_doc:
    context = etree.iterparse(xml_doc, events=("start", "end"))

    context = iter(context)

    event, root = context.next()

    base = False
    b_name = ""

    for event, elem in context:
        if event == "start" and elem.tag in ELEMENT_LIST:
            base = True
            bname = elem.tag
            children = elem.getchildren()
            child_list = []
            for child in children:
                child_list.append(child.tag)
            print bname + ":" + str(child_list)
        elif event == "end" and elem.tag in ELEMENT_LIST:
            base = False
            root.clear()
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3 回答 3

1

由于iterparse您无法将解析限制为某些类型的标签，因此您只能使用一个标签（通过传递参数tag）来执行此操作。但是，很容易手动完成您想要实现的目标。在以下代码段中：

from lxml import etree

fname = "test.xml"
ELEMENT_LIST = ["source", "event", "state", "locality"]

with open(fname) as xml_doc:
    context = etree.iterparse(xml_doc, events=("start", "end"))

    for event, elem in context:
        if event == "start" and elem.tag in ELEMENT_LIST:
            print "this elem is interesting, do some processing: %s: [%s]" % (elem.tag, ", ".join(child.tag for child in elem))
        elem.clear()

您将搜索限制在有趣的标签上。重要的部分iterparse是elem.clear()在项目过时时清除内存。这就是它具有内存效率的原因，请参阅http://lxml.de/parsing.html#modifying-the-tree

于 2012-05-07T19:28:00.577 回答
0

我会改用XPath。它比你自己阅读文档要优雅得多，而且我认为肯定更有效率。

于 2012-05-07T20:32:03.080 回答
0

利用tag='{http://www.sitemaps.org/schemas/sitemap/0.9}url'

有正确答案的类似问题https://stackoverflow.com/a/7019273/1346222

#!/usr/bin/python
# coding: utf-8
""" Parsing xml file. Basic example """
from StringIO import StringIO
from lxml import etree
import urllib2

sitemap = urllib2.urlopen(
    'http://google.com/sitemap.xml',
    timeout=10
).read()


NS = {
    'x': 'http://www.sitemaps.org/schemas/sitemap/0.9',
    'x2': 'http://www.google.com/schemas/sitemap-mobile/1.0'
}


res = []

urls = etree.iterparse(StringIO(sitemap), tag='{http://www.sitemaps.org/schemas/sitemap/0.9}url')

for event, url in urls:
    t = []
    t = url.xpath('.//x:loc/text() | .//x:priority/text()', namespaces=NS)
    t.append(url.xpath('boolean(.//x2:mobile)', namespaces=NS))
    res.append(t)
于 2013-03-10T20:29:49.923 回答