从我的 XML 文件中提取信息并为其分配一个向量

Question

我想用 python 解析我的计算机上的一些 XML 文件并从每个文件中提取一些信息这是我的其中之一的 xml 文件：（如果您想要文本在这里： https://github.com/peldszus/arg-microtexts/blob/master/corpus/en/micro_b002.xml）作为第一级，我已经完成了第一级：myList = []                #read the whole text from for root, dirs, files in os.walk(path):    for file in files:        if file.endswith('.xml'):            with open(os.path.join(root, file), encoding="UTF-8") as content:                tree = ET.parse(content)                myList.append(tree)在 myList 中，我有一些 XMl 文件 <xml.etree.ElementTree.ElementTree at 0x1f0fb1f8430>现在对于根“边缘”，它们没有 type="seg" <edge id="c1" src="a1" trg="a3" type="sup"/>  <edge id="c2" src="a2" trg="a3" type="sup"/>  <edge id="c4" src="a4" trg="a3" type="reb"/>  <edge id="c5" src="a5" trg="c4" type="und"/>我想提取标签“src”，我想提取标签=Src，  src="a1"    src="a2"    src="a4"   src="a5" 然后我想分配的数字不在src中，因为这句话称为前提，例如这里...我想说“a3”是所谓的“前提”（因为它不是标签src）例如这里(0,0,1,0,0) 应该是我的过程的结果，因为 a3 没有被应用，我将第三个数组设置为 1，其余的设置为零一般来说，我想提取信息以注释我的文本，该文本已使用 xml 进行了一些注释

Answer 1

您的问题中并非所有内容都清楚...

以下是数据提取部分

import xml.etree.ElementTree as ET

xml = '''<?xml version='1.0' encoding='UTF-8'?>

<arggraph id="micro_b002" topic_id="higher_dog_poo_fines" stance="pro">

  <edu id="e1"><![CDATA[One can hardly move in Friedrichshain or Neukölln these days without permanently scanning the ground for dog dirt.]]></edu>

  <edu id="e2"><![CDATA[And when bad luck does strike and you step into one of the many 'land mines' you have to painstakingly scrape the remains off your soles.]]></edu>

  <edu id="e3"><![CDATA[Higher fines are therefore the right measure against negligent, lazy or simply thoughtless dog owners.]]></edu>

  <edu id="e4"><![CDATA[Of course, first they'd actually need to be caught in the act by public order officers,]]></edu>

  <edu id="e5"><![CDATA[but once they have to dig into their pockets, their laziness will sure vanish!]]></edu>

  <adu id="a1" type="pro"/>

  <adu id="a2" type="pro"/>

  <adu id="a3" type="pro"/>

  <adu id="a4" type="opp"/>

  <adu id="a5" type="pro"/>

  <edge id="c6" src="e1" trg="a1" type="seg"/>

  <edge id="c7" src="e2" trg="a2" type="seg"/>

  <edge id="c8" src="e3" trg="a3" type="seg"/>

  <edge id="c9" src="e4" trg="a4" type="seg"/>

  <edge id="c10" src="e5" trg="a5" type="seg"/>

  <edge id="c1" src="a1" trg="a3" type="sup"/>

  <edge id="c2" src="a2" trg="a3" type="sup"/>

  <edge id="c4" src="a4" trg="a3" type="reb"/>

  <edge id="c5" src="a5" trg="c4" type="und"/>

</arggraph>'''

root = ET.fromstring(xml)

interesting_edges_src = [e.attrib['src'] for e in root.findall('.//edge') if e.attrib['type'] != 'seg' ]

print(interesting_edges_src)

输出

['a1', 'a2', 'a4', 'a5']

Answer 2

这里可以被认为是某种接近最终答案的答案

myList = []  

myEdgesList=[]

#read the whole text from 

for root, dirs, files in os.walk(path):

    for file in files:

        if file.endswith('.xml'):

            with open(os.path.join(root, file), encoding="UTF-8") as content:

                tree = ET.parse(content)

                myList.append(tree)

                

for k in myList:

    Edge= [e.attrib['src'] for e in k.findall('.//edge') if e.attrib['type'] != 'seg' ]

    myEdgesList.append(Edge)

这提供

['a1', 'a2', 'a4', 'a5'] 对于上面的示例以及所有其他示例的列表

[['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a4', 'a5'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a1', 'a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3', 'a4', 'a5'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a1', 'a2', 'a3'],

 ['a2', 'a3', 'a4', 'a5'],

.

.

.

只剩下将此列表转换为

(0,0,0,0,1) <----- ['a1', 'a2', 'a3', 'a4']

#as a5 is missing 

(0,0,1,0,0) <------  ['a1', 'a2', 'a4', 'a5']

#as a3 is misisng 

.

.

.

(0,0,1)    <-------   ['a2', 'a3']

#as a1 is missing 

等等

如果您有任何想法请告诉我，我也在努力

Answer 3

对于下一个问题

myEdgtlistmap=[]

for lst in myEdgesList:

    tp=[]

    for el in lst:

        if el=="a1":

            tp.append(1)

        if el=="a2":

            tp.append(2)

        if el=="a3":

            tp.append(3)

        if el=="a4":

            tp.append(4)

        if el=="a5":

            tp.append(5)

        if el=="a6":

            tp.append(6)

    myEdgtlistmap.append(tp)

label=[]

for le in myEdgtlistmap:

    b=[1]*(len(le)+1)

    for v in le: 

        b[v-1]=0

    label.append(b)

y=[l for lab in label for l in lab ]