python-将层次结构(树状)的XML读取到pandas数据框中,以保留层次结构
作者：互联网
我有一个XML文档,其中包含分层的树状结构,请参见下面的示例.
该文档包含多个< Message>标签(为方便起见,我仅复制了其中之一).
每个< Message>本身具有一些关联的数据(ID,状态,优先级).
此外,每个< Message>可以包含一个或多个< Street>子级又具有一些相关数据(< name>,< length>).
此外,每个< Street>可以具有一个或多个< Link>子级又具有自己的相关数据(< id>,<方向>).
XML文档示例：
<?xml version="1.0" encoding="ISO-8859-1"?>
<Root xmlns="someNamespace">
<Messages>
<Message id='12345'>
<status>Active</status>
<priority>Low</priority>
<Area>
<Streets>
<Street>
<name>King Street</name>
<length>Short</length>
<Link>
<id>75838745</id>
<direction>North</direction>
</Link>
<Link>
<id>168745</id>
<direction>South</direction>
</Link>
<Link>
<id>975416</id>
<direction>North</direction>
</Link>
</Street>
<Street>
<name>Queen Street</name>
<length>Long</length>
<Link>
<id>366248</id>
<direction>West</direction>
</Link>
<Link>
<id>745812</id>
<direction>East</direction>
</Link>
</Street>
</Streets>
</Area>
</Message>
</Messages>
</Root>
用Python解析XML并将相关数据存储在变量中不是问题-例如,我可以使用lxml库并读取整个文档,然后执行一些xpath表达式以获取相关字段,或者逐行读取iterparse方法.
但是,我想将数据放入熊猫数据框中,同时保留其中的层次结构.目标是查询单个消息(例如,通过布尔表达式(例如,如果status == Active,然后获取消息及其所有街道及其街道的链接)),并获取属于特定消息(其街道及其街道)的所有数据’链接).如何最好地做到这一点？
我尝试了不同的方法,但是都遇到了问题.
如果我为每个包含信息的XML行创建一个数据框行,然后在[MessageID,StreetName,LinkID]上设置一个MultiIndex,则会得到其中包含很多NaN的索引(通常不建议这样做),因为MessageID不知道其子级街道和链接呢.此外,我不知道如何通过布尔条件选择某些子数据集,而不是仅获取一些没有子项的单行.
在[MessageID,StreetName,LinkID]上执行GroupBy时,我不知道如何从熊猫GroupBy对象取回(可能是MultiIndex)数据帧,因为这里没有要聚合的内容(没有均值/标准差/总和/无论如何,值应保持不变).
有什么建议可以有效地处理吗？
解决方法:
我终于设法解决了如上所述的问题,这就是方法.
我扩展了上面给出的XML文档,使其包含两个消息而不是一个.这就是有效的Python字符串的样子(当然也可以从文件中加载它)：
xmlDocument = '''<?xml version="1.0" encoding="ISO-8859-1"?> \
<Root> \
<Messages> \
<Message id='12345'> \
<status>Active</status> \
<priority>Low</priority> \
<Area> \
<Streets> \
<Street> \
<name>King Street</name> \
<length>Short</length> \
<Link> \
<id>75838745</id> \
<direction>North</direction> \
</Link> \
<Link> \
<id>168745</id> \
<direction>South</direction> \
</Link> \
<Link> \
<id>975416</id> \
<direction>North</direction> \
</Link> \
</Street> \
<Street> \
<name>Queen Street</name> \
<length>Long</length> \
<Link> \
<id>366248</id> \
<direction>West</direction> \
</Link> \
<Link> \
<id>745812</id> \
<direction>East</direction> \
</Link> \
</Street> \
</Streets> \
</Area> \
</Message> \
<Message id='54321'> \
<status>Inactive</status> \
<priority>High</priority> \
<Area> \
<Streets> \
<Street> \
<name>Princess Street</name> \
<length>Mid</length> \
<Link> \
<id>744154</id> \
<direction>West</direction> \
</Link> \
<Link> \
<id>632214</id> \
<direction>South</direction> \
</Link> \
<Link> \
<id>654785</id> \
<direction>East</direction> \
</Link> \
</Street> \
<Street> \
<name>Prince Street</name> \
<length>Very Long</length> \
<Link> \
<id>1022444</id> \
<direction>North</direction> \
</Link> \
<Link> \
<id>4474558</id> \
<direction>South</direction> \
</Link> \
</Street> \
</Streets> \
</Area> \
</Message> \
</Messages> \
</Root>'''
为了将分层XML结构解析为一个扁平的pandas数据帧,我使用了Python的ElementTree iterparse方法,该方法提供了类似于SAX的接口,可以逐行遍历XML文档,并在特定XML标签开始或结束时触发事件.
对于每个解析的XML行,给定的信息都存储在字典中.使用三个字典,每个字典以某种方式属于一起(消息,街道,链接),然后将其存储在其自己的数据帧行中.当收集到这样一行的所有信息时,该字典将附加到以适当顺序存储所有行的列表中.
这就是XML解析的样子(有关更多说明,请参见内联注释)：
# imports
import xml.etree.ElementTree as ET
import pandas as pd
# initialize parsing from Bytes buffer
from io import BytesIO
xmlDocument = BytesIO(xmlDocument.encode('utf-8'))
# initialize dictionaries storing the information to each type of row
messageRow, streetRow, linkRow = {}, {}, {}
# initialize list that stores the single dataframe rows
listOfRows = []
# read the xml file line by line and throw signal when specific tags start or end
for event, element in ET.iterparse(xmlDocument, events=('start', 'end')):
##########
# get all information on the current message and store in the appropriate dictionary
##########
# get current message's id attribute
if event == 'start' and element.tag == 'Message':
messageRow = {} # re-initialize the dictionary for the current row
messageRow['messageId'] = element.get('id')
# get current message's status
if event == 'end' and element.tag == 'status':
messageRow['status'] = element.text
# get current message's priority
if event == 'end' and element.tag == 'priority':
messageRow['priority'] = element.text
# when no more information on the current message is expected, append it to the list of rows
if event == 'end' and element.tag == 'priority':
listOfRows.append(messageRow)
##########
# get all information on the current street and store in row dictionary
##########
if event == 'end' and element.tag == 'name':
streetRow = {} # re-initialize the dictionary for the current street row
streetRow['streetName'] = element.text
if event == 'end' and element.tag == 'length':
streetRow['streetLength'] = element.text
# when no more information on the current street is expected, append it to the list of rows
if event == 'end' and element.tag == 'length':
# link the street to the message it belongs to, then append
streetRow['messageId'] = messageRow['messageId']
listOfRows.append(streetRow)
##########
# get all information on the current link and store in row dictionary
##########
if event == 'end' and element.tag == 'id':
linkRow = {} # re-initialize the dictionary for the current link row
linkRow['linkId'] = element.text
if event == 'end' and element.tag == 'direction':
linkRow['direction'] = element.text
# when no more information on the current link is expected, append it to the list of rows
if event == 'end' and element.tag == 'direction':
# link the link to the message it belongs to, then append
linkRow['messageId'] = messageRow['messageId']
listOfRows.append(linkRow)
现在,listOfRows是词典的列表,每个词典在其中存储要放入一个数据帧行中的信息.可以使用此列表作为数据源创建数据框
# create dataframe from list of rows and pass column order (would be random otherwise)
df = pd.DataFrame.from_records(listOfRows, columns=['messageId', 'status', 'priority', 'streetName', 'streetLength', 'linkId', 'direction'])
print(df)
并给出“原始”数据框：
messageId status priority streetName streetLength linkId \
0 12345 Active Low NaN NaN NaN
1 12345 NaN NaN King Street Short NaN
2 12345 NaN NaN NaN NaN 75838745
3 12345 NaN NaN NaN NaN 168745
4 12345 NaN NaN NaN NaN 975416
5 12345 NaN NaN Queen Street Long NaN
6 12345 NaN NaN NaN NaN 366248
7 12345 NaN NaN NaN NaN 745812
8 54321 Inactive High NaN NaN NaN
9 54321 NaN NaN Princess Street Mid NaN
10 54321 NaN NaN NaN NaN 744154
11 54321 NaN NaN NaN NaN 632214
12 54321 NaN NaN NaN NaN 654785
13 54321 NaN NaN Prince Street Very Long NaN
14 54321 NaN NaN NaN NaN 1022444
15 54321 NaN NaN NaN NaN 4474558
direction
0 NaN
1 NaN
2 North
3 South
4 North
5 NaN
6 West
7 East
8 NaN
9 NaN
10 West
11 South
12 East
13 NaN
14 North
15 South
现在,我们可以将感兴趣的列(messageId,streetName,linkId)设置为该数据帧上的MultiIndex：
# set the columns of interest as MultiIndex
df = df.set_index(['messageId', 'streetName', 'linkId'])
print(df)
这使：
status priority streetLength direction
messageId streetName linkId
12345 NaN NaN Active Low NaN NaN
King Street NaN NaN NaN Short NaN
NaN 75838745 NaN NaN NaN North
168745 NaN NaN NaN South
975416 NaN NaN NaN North
Queen Street NaN NaN NaN Long NaN
NaN 366248 NaN NaN NaN West
745812 NaN NaN NaN East
54321 NaN NaN Inactive High NaN NaN
Princess Street NaN NaN NaN Mid NaN
NaN 744154 NaN NaN NaN West
632214 NaN NaN NaN South
654785 NaN NaN NaN East
Prince Street NaN NaN NaN Very Long NaN
NaN 1022444 NaN NaN NaN North
4474558 NaN NaN NaN South
尽管通常不应该考虑将NaN包含在索引中,但对于此用例,我认为它没有任何问题.
最后,为了获得按其messageId(包括其所有“子级”街道和链接)访问单个消息的预期效果,必须将MultiIndexed数据帧按最外部的索引级别进行分组：
# group by the most outer index
groups = df.groupby(level='messageId')
现在,您可以使用以下方法遍历所有消息(并对其进行任何处理)
# iterate over all groups
for key, group in groups:
print('key: ' + key)
print('group:')
print(group)
print('\n')
哪个返回
key: 12345
group:
status priority streetLength direction
messageId streetName linkId
12345 NaN NaN Active Low NaN NaN
King Street NaN NaN NaN Short NaN
NaN 75838745 NaN NaN NaN North
168745 NaN NaN NaN South
975416 NaN NaN NaN North
Queen Street NaN NaN NaN Long NaN
NaN 366248 NaN NaN NaN West
745812 NaN NaN NaN East
key: 54321
group:
status priority streetLength direction
messageId streetName linkId
54321 NaN NaN Inactive High NaN NaN
Princess Street NaN NaN NaN Mid NaN
NaN 744154 NaN NaN NaN West
632214 NaN NaN NaN South
654785 NaN NaN NaN East
Prince Street NaN NaN NaN Very Long NaN
NaN 1022444 NaN NaN NaN North
4474558 NaN NaN NaN South
或者您可以通过messageId访问特定的消息,返回包含messageId以及其所有专用街道和链接的行：
# get groups by key
print('specific group only:')
print(groups.get_group('54321'))
给
specific group only:
status priority streetLength direction
messageId streetName linkId
54321 NaN NaN Inactive High NaN NaN
Princess Street NaN NaN NaN Mid NaN
NaN 744154 NaN NaN NaN West
632214 NaN NaN NaN South
654785 NaN NaN NaN East
Prince Street NaN NaN NaN Very Long NaN
NaN 1022444 NaN NaN NaN North
4474558 NaN NaN NaN South
希望这对某人有所帮助.
标签：pandas,tree,hierarchical-data,xml,python 来源： https://codeday.me/bug/20191120/2047894.html