解析xml - join在同一目录中按文件名将xml文件拆分成1个pandas数据帧

Question

我在一个文件夹中有4-10个xml文件，这些文件是从一个大的单个xml文件中分解出来的。幸运的是，解析xml很容易，因为我可以使用xmltodict包。因此，我可以使用单个xml文件做任何我需要做的事情。我将其转换为pandas数据帧以满足分析需求。但是，我需要将4个xml文件组合成一个pandas数据帧。假设没有数据/索引问题，这些文件肯定按顺序正确命名为00001.xml, 00002.xml, 00003.xml, 00004.xml。

import xmltodict
import numpy as np
import pandas as pd
from collections import Counter

with open('00001.xml') as fd:
    doc = xmltodict.parse(fd.read())

def panda_maker (xml_dict):
    channel_list = xml_dict ['logs']['log']['logData']['mnemonicList'].split(",")
    logData_list = [i.split(",") for i in xml_dict ['logs']['log']['logData']['data']]
    logData_list.insert(0, xml_dict ['logs']['log']['logData']['unitList'].split(","))
    return pd.DataFrame(np.array(logData_list).reshape(len(logData_list),len(channel_list)), columns = channel_list)

logData_frame_01 = panda_maker(doc)
logData_frame_01.head() #all good

如何将logData_frame_01 + _02 + _03 + _04巧妙地组合到一个数据帧中？上面程序中的任何进一步的抽象技巧也是非常受欢迎的。

Answer 1

尝试：

doc = []
for i in range(1,5):
    with open('0000{}.xml'.format(i)) as fd:
        doc.append(xmltodict.parse(fd.read()))

def panda_maker (xml_dict):
    logData_list = []
    for xmlval in xml_dict:
        channel_list = xmlval['logs']['log']['logData']['mnemonicList'].split(",")
        temp = [i.split(",") for i in xml_dict ['logs']['log']['logData']['data']]
        temp.insert(0, xml_dict ['logs']['log']['logData']['unitList'].split(","))
        logData_list.extend(temp)
    return pd.DataFrame(np.array(logData_list).reshape(len(logData_list),len(channel_list)), columns = channel_list)

logData_frame_01 = panda_maker(doc)
logData_frame_01.head() #all good