如何在使用cElementTree的iterparse方法時知道元素的父級？

Question

我想循環一個xml文件的元素併產生每個元素，除非父元素是一個特徵。

所以在僞

for event, element in cElementTree.iterparse('../test.xml'): 
     if parentOf_element != 'feature': 
     yield element 

我怎樣才能獲得該元素的父？我知道使用tree.getiterator（）函數是可能的，但我不想構建完整的樹，因爲xml文件很大。

Answer 1

如果啓用start事件，您可以通過使用堆棧跟蹤祖先節點。如果你真的想要鎮壓一個<feature>的所有後代，而不是僅僅是兒童，你可以使用一個簡單的標誌，如另一個答案中所示。您可以使用root.clear()來吹走所有已完成的元素。閱讀this。

代碼：

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)) 

輸出如下。您會注意到從lev1節點root.clear()不會吹走尚未完全解析的元素。這意味着，存儲器使用量是O（樹的深度），而不是O（樹中元素的總數量）

=== 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? 

Answer 2

你可以用lxml來做到這一點。它有getparent（）。

或者，它可能處理start和end事件並跳過feature兒童cElementTree：

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