scala , python哪个用来开发spark更好

 如果唯一的要求是application之前的permission ，则可以通过在包含Element位置0插入permissions （使用insert方法）来确保。 
 
import StringIO
from xml.etree import ElementTree

xml_str = "<root><something></something><application></application></root>"
permission_str = "<permission></permission>"

tree = ElementTree.parse(StringIO.StringIO(xml_str))
perm_element = ElementTree.fromstring(permission_str)

# insert our permission element as the first child
tree.getroot().insert(0, perm_element)

print ElementTree.tostring(tree.getroot())
 
 收益率： 
 
<root><permission /><something /><application /></root>
 
 编辑 ： 
 
 如果您需要更精细的控制，您有几个选择。 可能最好的选择是安装和使用lxml ，它具有insertBefore功能。 
 
 如果您更喜欢坚持xml.etree ，这里有一个更精确的插入新元素的方法： 
 
import StringIO
from xml.etree import ElementTree


def get_index(root, element):
    for idx, child in enumerate(root):
        if element == child:
            return idx
    else:
        raise ValueError("No '%s' tag found in '%s' children" %
                         (element.tag, root.tag))

xml_str = """<root><something></something><application></application></root>"""
tree = ElementTree.parse(StringIO.StringIO(xml_str))

# print our tree pre-insertion
print ElementTree.tostring(tree.getroot())

permission_str = "<permission></permission>"
perm_element = ElementTree.fromstring(permission_str)
app_element = tree.getroot().find('./application')
app_index = get_index(tree.getroot(), app_element)
tree.getroot().insert(app_index, perm_element)

# print our tree post-insertion
print ElementTree.tostring(tree.getroot())
 
 在运行时打印： 
 
<root><something /><application /></root>
<root><something /><permission /><application /></root>
 
 这有一些限制 - 如果有几个相同的元素，它会在找到的第一个元素之前插入新元素（因此，如果你有两个具有相同属性的root应用程序子元素，并希望在第一个元素之后但在第二个，你运气不好。 也就是说，我相信它能完成你所需要的。 

If the only requirement is that permission be before application, you can ensure that by inserting permissions at position 0 of the containing Element (using the insert method).
 
import StringIO
from xml.etree import ElementTree

xml_str = "<root><something></something><application></application></root>"
permission_str = "<permission></permission>"

tree = ElementTree.parse(StringIO.StringIO(xml_str))
perm_element = ElementTree.fromstring(permission_str)

# insert our permission element as the first child
tree.getroot().insert(0, perm_element)

print ElementTree.tostring(tree.getroot())
 
yields:
 
<root><permission /><something /><application /></root>
 
EDIT:
 
You have a couple of options if you need finer grained control. Probably the best option is to install and use lxml, which does have an insertBefore capability.
 
If you prefer to stick to xml.etree, here's a more precise method of inserting new elements:
 
import StringIO
from xml.etree import ElementTree


def get_index(root, element):
    for idx, child in enumerate(root):
        if element == child:
            return idx
    else:
        raise ValueError("No '%s' tag found in '%s' children" %
                         (element.tag, root.tag))

xml_str = """<root><something></something><application></application></root>"""
tree = ElementTree.parse(StringIO.StringIO(xml_str))

# print our tree pre-insertion
print ElementTree.tostring(tree.getroot())

permission_str = "<permission></permission>"
perm_element = ElementTree.fromstring(permission_str)
app_element = tree.getroot().find('./application')
app_index = get_index(tree.getroot(), app_element)
tree.getroot().insert(app_index, perm_element)

# print our tree post-insertion
print ElementTree.tostring(tree.getroot())
 
which, when run, prints:
 
<root><something /><application /></root>
<root><something /><permission /><application /></root>
 
This has limitations - if there are several identical elements, it will insert the new element before the first one it finds (so if you had two application children of root with identical attributes and wanted to place a new tag after the first one but before the second one, you'd be out of luck). That said, I believe it accomplishes what you need it to.