Should memory usage increase when using ElementTree.iterparse() when clear()ing trees?

前端 未结 3 1280
生来不讨喜
生来不讨喜 2021-01-12 16:27
import os
import xml.etree.ElementTree as et

for ev, el in et.iterparse(os.sys.stdin):
    el.clear()

Running the above on the ODP structure RDF d

相关标签:
3条回答
  • 2021-01-12 17:01

    You are clearing each element but references to them remain in the root document. So the individual elements still cannot be garbage collected. See this discussion in the ElementTree documentation.

    The solution is to clear references in the root, like so:

    # get an iterable
    context = iterparse(source, events=("start", "end"))
    
    # turn it into an iterator
    context = iter(context)
    
    # get the root element
    event, root = context.next()
    
    for event, elem in context:
        if event == "end" and elem.tag == "record":
            ... process record elements ...
            root.clear()
    

    Another thing to remember about memory usage, which may not be affecting your situation, is that once the VM allocates memory for heap storage from the system, it generally never gives that memory back. Most Java VMs work this way too. So you should not expect the size of the interpreter in top or ps to ever decrease, even if that heap memory is unused.

    0 讨论(0)
  • 2021-01-12 17:06

    I ran into the same issue. The documentation doesn't make things very clear. The issue in my case was:

    1) Calling clear does release memory for the children nodes. Documentation says that it releases all memory. Clear does not release the memory for which clear is called, because that memory belongs to the parent which allocated it. 2) Calling root.clear(), that depends on what root is. If root is the parent then it would work. Otherwise, it will not free the memory.

    The fix was to keep a reference to the parent, and when we no longer need the node, we call parent.remove(child_node). This worked and it kept the memory profile at a few KBs.

    0 讨论(0)
  • 2021-01-12 17:09

    As mentioned in the answer by Kevin Guerra, the "root.clear()" strategy in the ElementTree documentation only removes fully parsed children of the root. If those children are anchoring huge branches, it's not very helpful.

    He touched on the ideal solution, but didn't post any code, so here is an example:

    element_stack = []
    context = ET.iterparse(stream, events=('start', 'end'))
    for event, elem in context:
        if event == 'start':
            element_stack.append(elem)
        elif event == 'end':
            element_stack.pop()
            # see if elem is one of interest and do something with it here
            if element_stack:
                element_stack[-1].remove(elem)
    del context
    

    The element of interest will not have subelements; they'll have been removed as soon as their end tags were seen. This might be OK if all you need is the element's text or attributes.

    If you want to query into the element's descendants, you need to let a full branch be built for it. For this, maintain a flag, implemented as a depth counter for those elements. Only call .remove() when the depth is zero:

    element_stack = []
    interesting_element_depth = 0
    context = ET.iterparse(stream, events=('start', 'end'))
    for event, elem in context:
        if event == 'start':
            element_stack.append(elem)
            if elem.tag == 'foo':
                interesting_element_depth += 1
        elif event == 'end':
            element_stack.pop()
            if elem.tag == 'foo':
                interesting_element_depth -= 1
                # do something with elem and its descendants here
            if element_stack and not interesting_element_depth:
                element_stack[-1].remove(elem)
    del context
    
    0 讨论(0)
提交回复
热议问题