我想循环遍历xml文件的元素并产生每个元素,除非父元素是一个特征。
所以在伪代码中
for event, element in cElementTree.iterparse('../test.xml'):
if parentOf_element != 'feature':
yield element
如何获取元素的父元素?我知道有可能使用tree.getiterator()函数,但我不想构建完整的树,因为xml文件很大。
答案 0 :(得分:2)
您可以使用lxml执行此操作。它有getparent()。
或者,可以处理start
和end
个事件,并使用feature
跳过cElementTree
个孩子:
from xml.etree import cElementTree as etree
in_feature_tag = False
for event, element in etree.iterparse('test.xml', events=('start', 'end')):
if element.tag == 'feauture':
in_feature_tag = event == 'start'
if event == 'end' and not in_feature_tag:
yield element
答案 1 :(得分:1)
如果启用start
事件,则可以使用堆栈跟踪祖先节点。如果你真的想要压制<feature>
的所有后代,而不仅仅是孩子,你可以使用一个简单的标志,如另一个答案所示。
您可以使用root.clear()
吹走所有已完成的元素。阅读this。
代码:
import xml.etree.cElementTree as et
# Produces identical answers with import lxml.etree as et
import cStringIO
def normtext(t):
return repr("" if t is None else t.strip())
def dump(el):
print el.tag, normtext(el.text), normtext(el.tail), el.attrib
def my_filtered_elements(source, skip_parent_tag="feature"):
# get an iterable
context = et.iterparse(source, events=("start", "end"))
# turn it into an iterator
context = iter(context)
# get the root element
event, root = context.next()
tag_stack = [None, root.tag]
for event, elem in context:
# print event, elem.tag, tag_stack
if event == "start":
tag_stack.append(elem.tag)
else:
assert event == "end"
my_tag = tag_stack.pop()
assert my_tag == elem.tag
parent_tag = tag_stack[-1]
if parent_tag is not None and parent_tag != skip_parent_tag:
dump(elem)
# yield elem
root.clear()
def other_filtered_elements(source, skip_parent_tag="feature"):
in_feature_tag = False
for event, element in et.iterparse(source, events=('start', 'end')):
if element.tag == skip_parent_tag:
in_feature_tag = event == 'start'
if event == 'end' and not in_feature_tag:
dump(element)
test_input = """
<top>
<lev1 guff="1111">
<lev2>aaaaa</lev2>
<lev2>bbbbb</lev2>
</lev1>
<feature>
feat text 1
<fchild>fcfcfcfc
<fgchild>ggggg</fgchild>
</fchild>
feat text 2
</feature>
<lev1 guff="2222">
<lev2>ccccc</lev2>c-tail
<lev2>ddddd</lev2>d-tail
<notext1></notext1>e-tail
<notext2 />f-tail
</lev1>g-tail
</top>
"""
print "=== me ==="
my_filtered_elements(cStringIO.StringIO(test_input))
print "=== other ==="
other_filtered_elements(cStringIO.StringIO(test_input))
输出如下。您会从lev1
节点注意到root.clear()
没有吹走尚未完全解析的元素。这意味着使用的内存量是O(树的深度),而不是O(树中元素的总数)
=== me ===
lev2 'aaaaa' '' {}
lev2 'bbbbb' '' {}
lev1 '' '' {'guff': '1111'}
fgchild 'ggggg' '' {} <<<=== do you want this?
feature 'feat text 1' '' {}
lev2 'ccccc' 'c-tail' {}
lev2 'ddddd' 'd-tail' {}
notext1 '' 'e-tail' {}
notext2 '' 'f-tail' {}
lev1 '' 'g-tail' {'guff': '2222'}
=== other ===
lev2 'aaaaa' '' {}
lev2 'bbbbb' '' {}
lev1 '' '' {'guff': '1111'}
feature 'feat text 1' '' {}
lev2 'ccccc' 'c-tail' {}
lev2 'ddddd' 'd-tail' {}
notext1 '' 'e-tail' {}
notext2 '' 'f-tail' {}
lev1 '' 'g-tail' {'guff': '2222'}
top '' '' {} <<<=== do you want this?