如果源文件是一个xml文件，就像这个示例一样

<pa:Process xmlns:pa="http://sssss">
        <pa:firsttag>SAMPLE</pa:firsttag>
    </pa:Process>

您可以尝试下面的代码

from lxml import etree, objectify
metadata = 'C:\\Users\\PROCS.xml' # this is sample xml file the contents are shown above
parser = etree.XMLParser(remove_blank_text=True) # this line removes the  name space from the xml in this sample the name space is --> http://sssss
tree = etree.parse(metadata, parser) # this line parses the xml file which is PROCS.xml
root = tree.getroot() # we get the root of xml which is process and iterate using a for loop
for elem in root.getiterator():
    if not hasattr(elem.tag, 'find'): continue  # (1)
    i = elem.tag.find('}')
    if i >= 0:
        elem.tag = elem.tag[i+1:]

dict={}  # a python dictionary is declared
for elem in tree.iter(): #iterating through the xml tree using a for loop
    if elem.tag =="firsttag": # if the tag name matches the name that is equated then the text in the tag is stored into the dictionary
        dict["FIRST_TAG"]=str(elem.text)
        print(dict)

输出将是

{'FIRST_TAG': 'SAMPLE'}

你可以使用BeautifulSoup:

from bs4 import BeautifulSoup

x="""<foo>
   <bar>
      <type foobar="1"/>
      <type foobar="2"/>
   </bar>
</foo>"""

y=BeautifulSoup(x)
>>> y.foo.bar.type["foobar"]
u'1'

>>> y.foo.bar.findAll("type")
[<type foobar="1"></type>, <type foobar="2"></type>]

>>> y.foo.bar.findAll("type")[0]["foobar"]
u'1'
>>> y.foo.bar.findAll("type")[1]["foobar"]
u'2'

为了增加另一种可能性，可以使用untangle，因为它是一个简单的xml-to-python-object库。这里有一个例子:

安装:

pip install untangle

用法:

你的XML文件(有一点变化):

<foo>
   <bar name="bar_name">
      <type foobar="1"/>
   </bar>
</foo>

使用untangle访问属性:

import untangle

obj = untangle.parse('/path_to_xml_file/file.xml')

print obj.foo.bar['name']
print obj.foo.bar.type['foobar']

输出将是:

bar_name
1

更多关于untangle的信息可以在“untangle”中找到。

此外，如果您感兴趣，可以在“Python和XML”中找到使用XML和Python的工具列表。您还将看到前面的答案中提到的最常见的问题。

如果源文件是一个xml文件，就像这个示例一样

<pa:Process xmlns:pa="http://sssss">
        <pa:firsttag>SAMPLE</pa:firsttag>
    </pa:Process>

您可以尝试下面的代码

from lxml import etree, objectify
metadata = 'C:\\Users\\PROCS.xml' # this is sample xml file the contents are shown above
parser = etree.XMLParser(remove_blank_text=True) # this line removes the  name space from the xml in this sample the name space is --> http://sssss
tree = etree.parse(metadata, parser) # this line parses the xml file which is PROCS.xml
root = tree.getroot() # we get the root of xml which is process and iterate using a for loop
for elem in root.getiterator():
    if not hasattr(elem.tag, 'find'): continue  # (1)
    i = elem.tag.find('}')
    if i >= 0:
        elem.tag = elem.tag[i+1:]

dict={}  # a python dictionary is declared
for elem in tree.iter(): #iterating through the xml tree using a for loop
    if elem.tag =="firsttag": # if the tag name matches the name that is equated then the text in the tag is stored into the dictionary
        dict["FIRST_TAG"]=str(elem.text)
        print(dict)

输出将是

{'FIRST_TAG': 'SAMPLE'}

这里有一个使用cElementTree的非常简单但有效的代码。

try:
    import cElementTree as ET
except ImportError:
  try:
    # Python 2.5 need to import a different module
    import xml.etree.cElementTree as ET
  except ImportError:
    exit_err("Failed to import cElementTree from any known place")      

def find_in_tree(tree, node):
    found = tree.find(node)
    if found == None:
        print "No %s in file" % node
        found = []
    return found  

# Parse a xml file (specify the path)
def_file = "xml_file_name.xml"
try:
    dom = ET.parse(open(def_file, "r"))
    root = dom.getroot()
except:
    exit_err("Unable to open and parse input definition file: " + def_file)

# Parse to find the child nodes list of node 'myNode'
fwdefs = find_in_tree(root,"myNode")

这是来自“python xml解析”。

我很受伤，没有人建议熊猫。Pandas有一个read_xml()函数，它非常适合这种扁平的xml结构。

import pandas as pd

xml = """<foo>
   <bar>
      <type foobar="1"/>
      <type foobar="2"/>
   </bar>
</foo>"""

df = pd.read_xml(xml, xpath=".//type")
print(df)

输出:

   foobar
0       1
1       2