我想用 python 解析我的計算機上的一些 XML 文件并從每個文件中提取一些信息這是我的其中之一的 xml 文件：（如果您想要文本在這里： https://github.com/peldszus/arg-microtexts/blob/master/corpus/en/micro_b002.xml）作為第一級，我已經完成了第一級：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>現在對于根“邊緣”，它們沒有 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"/>我想提取標簽“src”，我想提取標簽=Src， src="a1" src="a2" src="a4" src="a5" 然后我想分配的數字不在src中，因為這句話稱為前提，例如這里...我想說“a3”是所謂的“前提”（因為它不是標簽src）例如這里(0,0,1,0,0) 應該是我的過程的結果，因為 a3 沒有被應用，我將第三個數組設置為 1，其余的設置為零一般來說，我想提取信息以注釋我的文本，該文本已使用 xml 進行了一些注釋
3 回答
德瑪西亞99
TA貢獻1770條經驗 獲得超3個贊
您的問題中并非所有內容都清楚...
以下是數據提取部分
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 Neuk?lln 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)
輸出
['a1', 'a2', 'a4', 'a5']
手掌心
TA貢獻1942條經驗 獲得超3個贊
這里可以被認為是某種接近最終答案的答案
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)
這提供
['a1', 'a2', 'a4', 'a5'] 對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'],
.
.
.
只剩下將此列表轉換為
(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
如果您有任何想法請告訴我，我也在努力
牧羊人nacy
TA貢獻1862條經驗 獲得超7個贊
對于下一個問題
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 ]
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