使用 Python 解析 USPTO 批量 XML 文件

Question

import xml.etree.ElementTree as ET
import csv
import re
import codecs
import io


xml = open('ipa110106.xml')
line_num=0
f = open('workfile.xml', 'w')

for  line in xml:
   line_num+=1
   if line_num == 1:
       print (line)

   if '<?xml version="1.0" encoding="UTF-8"?>' in line and line_num !=1:
       count =count+1
       line = line.replace('<?xml version="1.0" encoding="UTF-8"?>', '')
   if '<!DOCTYPE us-patent-application SYSTEM "us-patent-application-v42-2006-08-23.dtd" [ ]>' in line:
       line = line.replace('<!DOCTYPE us-patent-application SYSTEM "us-patent-application-v42-2006-08-23.dtd" [ ]>', '')  
       count2+=1
   if "!DOCTYPE" in line:
    line=line.replace('<!DOCTYPE sequence-cwu SYSTEM "us-sequence-listing.dtd" [ ]>','')  
   f.write(line)  
f.close()

with open("workfile.xml") as f:
 xml = f.read()
 tree = ET.fromstring(re.sub(r"(<\?xml[^>]+\?>)", r"\1<root>", xml) + "</root>")
 root= tree.getroot()

结果：

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

0
Traceback (most recent call last):

  File "<ipython-input-164-4d6fc9ea9aac>", line 1, in <module>
    runfile('C:/Users/Harshit/Downloads/ipa110106 (1)/parsing_test5.py', wdir='C:/Users/Harshit/Downloads/ipa110106 (1)')

  File "C:\Users\Harshit\Anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 866, in runfile
    execfile(filename, namespace)

  File "C:\Users\Harshit\Anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 102, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)

  File "C:/Users/Harshit/Downloads/ipa110106 (1)/parsing_test5.py", line 41, in <module>
    root= tree.getroot()

AttributeError: 'xml.etree.ElementTree.Element' object has no attribute 'getroot'

我正在尝试解析 USPTO XML 文件以提取相关信息。这些文件是多个 XML 文件的串联，并遵循本论坛中给出的标准建议，我删除了多个实例：<?xml version="1.0" encoding="UTF-8"?> 和 <!DOCTYPE us-patent-application SYSTEM "us-patent-application-v42-2006-08-23.dtd" [ ]> 因为它们也导致了错误：

ParseError: not well-formed (invalid token): line 2, column 2.

最后，在从 XML 中删除这些麻烦的元素后，我创建了一个合成父根来将此文件转换为适当的 XML 格式。但是，当我试图解析这个文件并访问它的根时，我遇到了一个错误。我在帖子中附上了代码。

import xml.etree.ElementTree as ET
import csv
import re
import codecs
import io


xml = open('ipa110106.xml')
line_num=0
f = open('workfile.xml', 'w')

for  line in xml:
   line_num+=1
   if line_num == 1:
       print (line)

   if '<?xml version="1.0" encoding="UTF-8"?>' in line and line_num !=1:
       count =count+1
       line = line.replace('<?xml version="1.0" encoding="UTF-8"?>', '')
   if '<!DOCTYPE us-patent-application SYSTEM "us-patent-application-v42-2006-08-23.dtd" [ ]>' in line:
       line = line.replace('<!DOCTYPE us-patent-application SYSTEM "us-patent-application-v42-2006-08-23.dtd" [ ]>', '')  
       count2+=1
   if "!DOCTYPE" in line:
    line=line.replace('<!DOCTYPE sequence-cwu SYSTEM "us-sequence-listing.dtd" [ ]>','')  
   f.write(line)  
f.close()

with open("workfile.xml") as f:
 xml = f.read()
 tree = ET.fromstring(re.sub(r"(<\?xml[^>]+\?>)", r"\1<root>", xml) + "</root>")
 root= tree.getroot()

另外，XML 文件很大，我只能分享它的链接-enter link description here

XML（类似）文件的小样本：

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE us-patent-application SYSTEM "us-patent-application-v42-2006-08-23.dtd" [ ]>
<us-patent-application lang="EN" dtd-version="v4.2 2006-08-23" file="US20110000001A1-20110106.XML" status="PRODUCTION" id="us-patent-application" country="US" date-produced="20101222" date-publ="20110106">
<us-bibliographic-data-application lang="EN" country="US">
<publication-reference>
<document-id>
<country>US</country>
<doc-number>20110000001</doc-number>
<kind>A1</kind>
<date>20110106</date>
</document-id>
</publication-reference>
<application-reference appl-type="utility">
<document-id>
<country>US</country>
<doc-number>12838840</doc-number>
<date>20100719</date>
</document-id>
</application-reference>
<us-application-series-code>12</us-application-series-code>
<priority-claims>
<priority-claim sequence="01" kind="national">
<country>IL</country>
<doc-number>189088</doc-number>
<date>20080128</date>
</priority-claim>
</priority-claims>
<classifications-ipcr>
<classification-ipcr>

Answer 1

如果您在 XML 声明中拆分它们并分别处理每个发布，则当前 PTO XML 文件是有效的 XML。我希望尝试一次处理它们以使用大量内存。无论哪种方式，都不需要您进行的替换。

我的解决方案是创建一个拥有 zipfile 的类（对于可能不知道的其他人，数据是一个 zip 文件，其中包含一个包含连接的 XML 文件的文件）并具有依次生成每个 XML 文件的函数。然后我使用ET.XML() 来处理这些文件。