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我想定义一个可以在目录中的每个 xml 文件上实现的函数，以便解析它并从数据框中的标签中获取内容。


from xml.etree import ElementTree

def func(path, filename):

    for filename in os.listdir(path):
        with open(os.path.join(path, filename)) as file:
        # Read each line in the file, readlines() returns a list of lines
            content = file.readlines()
        # Combine the lines in the list into a string
            content = "".join(content)
            bs_content = bs(content, "lxml")

            headline = bs_content.find_all("headline")
            eventtitle = bs_content.find_all("eventtitle")
            city = bs_content.find_all("city")
            companyname = bs_content.find_all("companyname")
            companyticker = bs_content.find_all("companyticker")
            startdate = bs_content.find_all("startdate")
            eventstory = bs_content.find_all("eventstory")

            data = []
            for i in range(0,len(companyname)):
                rows = [companyname[i].get_text(),headline[i].get_text(),
                       city[i].get_text(),eventtitle[i].get_text(),
                       companyticker[i].get_text(),startdate[i].get_text(),
                      eventstory[i].get_text()]
                data.append(rows)
 
    df = pd.DataFrame(data,columns = ['companyname','headline',
                                  'city','eventtitle','companyticker',
                                  'startdate','eventstory'], dtype = float)

当我调用一个函数时，我收到此错误。不幸的是，每个现有的解决方案都不适用于我。

func('./Calls/', '1000015_T.xml')

---------------------------------------------------------------------------
UnicodeDecodeError                        Traceback (most recent call last)
Input In [58], in <module>
----> 1 func('./Calls/', '1000015_T.xml')

Input In [57], in func(path, filename)
      7 for filename in os.listdir(path):
      8     with open(os.path.join(path, filename)) as file:
      9     # Read each line in the file, readlines() returns a list of lines
---> 10         content = file.readlines()
     11     # Combine the lines in the list into a string
     12         content = "".join(content)

File /Library/Frameworks/Python.framework/Versions/3.10/lib/python3.10/codecs.py:322, in BufferedIncrementalDecoder.decode(self, input, final)
    319 def decode(self, input, final=False):
    320     # decode input (taking the buffer into account)
    321     data = self.buffer + input
--> 322     (result, consumed) = self._buffer_decode(data, self.errors, final)
    323     # keep undecoded input until the next call
    324     self.buffer = data[consumed:]

UnicodeDecodeError: 'utf-8' codec can't decode byte 0x80 in position 3131: invalid start byte

也许您还可以帮助我进行代码优化。我的任务是获取 2k xml 文件的内容，到目前为止，我决定定义一个函数，然后使用 pandarallel：parallel_apply(func)

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1 回答 1

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输入文件不是 UTF-8，它可能是其他代码页。

确定正确的编码是什么，并相应地更改您的程序。

于 2022-01-23T02:14:56.133 回答