我正在嘗試解析大量包含大量嵌套元素的 XML 文件，以收集稍后使用的特定信息。由于文件數量巨大，我試圖盡可能高效地完成此操作，以減少處理時間。我可以使用 xpath 提取所需的信息，如下所示，但似乎效率很低。特別是必須運行第二個 for 循環來使用另一個 xpath 搜索提取結果值。我可以使用更有效的方法來獲得下面所需的輸出嗎？我可以通過單個 xpath 查詢收集所需的信息嗎？所需的解析格式：Id? ? ? ? ? ? ?Object? ? Type? ? ? ? ? ? ?ResultPackages? ? ? ?total? ? ?totalPackages? ? 1200DeliveryMethod priority? packagesSent? ? ?100DeliveryMethod express? ?packagesSent? ? ?200DeliveryMethod ground? ? packagesSent? ? ?300DeliveryMethod priority? packagesReceived 100DeliveryMethod express? ?packagesReceived 200DeliveryMethod ground? ? packagesReceived 300XML 示例：<?xml version="1.0" encoding="utf-8"?>? ? <Data>? ? ? ? <Location localDn="Chicago"/>? ? ? ? <Info Id="Packages">? ? ? ? ? ? <job jobId="1"/>? ? ? ? ? ? <Type pos="1">totalPackages</Type>? ? ? ? ? ? <Value Object="total">? ? ? ? ? ? ? ? <result pos="1">1200</result>? ? ? ? ? ? </Value>? ? ? ? </Info>? ? ? ? <Info Id="DeliveryMethod">? ? ? ? ? ? <job jobId="1"/>? ? ? ? ? ? <Type pos="1">packagesSent</Type>? ? ? ? ? ? <Type pos="2">packagesReceived</Type>? ? ? ? ? ? <Value Object="priority">? ? ? ? ? ? ? ? <result pos="1">100</result>? ? ? ? ? ? ? ? <result pos="2">100</result>? ? ? ? ? ? </Value>? ? ? ? ? ? <Value Object="express">? ? ? ? ? ? ? ? <result pos="1">200</result>? ? ? ? ? ? ? ? <result pos="2">200</result>? ? ? ? ? ? </Value>? ? ? ? ? ? <Value Object="ground">? ? ? ? ? ? ? ? <result pos="1">300</result>? ? ? ? ? ? ? ? <result pos="2">300</result>? ? ? ? ? ? </Value>? ? ? ? </Info>? </Data>是否可以通過迭代來獲取所有信息tree.xpath('//*')？
2 回答
慕村225694
TA貢獻1880條經驗 獲得超4個贊
其中一項優化不會像您現在使用tree.xpath('//*')if 語句那樣遍歷所有標簽并進行檢查。這可以替換為tree.xpath('//Type')
接下來需要優化的是迭代值。Value您無需一遍又一遍地迭代（ tree.xpath('//Value')），您可以獲得標簽的所有同級Values標簽Typeelem.xpath('./following-sibling::Value')
from lxml import etree
xml_file = open('stack_sample.xml')
tree = etree.parse(xml_file)
root = tree.getroot()
for elem in tree.xpath('//Type'):
_id = elem.getparent().attrib["Id"]
_type = elem.text
_position = elem.attrib["pos"]
values = elem.xpath('./following-sibling::Value')
for value in values:
_object = value.attrib['Object']
_result = value.xpath(f'./result[@pos={_position}]/text()')[0]
print(_id, _type, _object, _result)
這將打印出：
Packages totalPackages total 1200
DeliveryMethod packagesSent priority 100
DeliveryMethod packagesSent express 200
DeliveryMethod packagesSent ground 300
DeliveryMethod packagesReceived priority 100
DeliveryMethod packagesReceived express 200
DeliveryMethod packagesReceived ground 300
編輯
這是針對特定情況的解決方案，其中我們確定resultinValue標簽的數量等于與其他解決方案Type是同級的標簽的數量Value，另外解決方案假設Type和result按相同pos屬性排序。
請記住，這是一種非常具體的解決方案，而不是通用的解決方案。
from lxml import etree
xml_file = open('stack_sample.xml')
tree = etree.parse(xml_file)
root = tree.getroot()
for elem in tree.xpath('//Type'):
_id = elem.getparent().attrib["Id"]
_type = elem.text
_objects = elem.xpath('./following-sibling::Value/@Object')
_results = elem.xpath('./following-sibling::Value/result/text()')
for _object, _result in zip(_objects, _results):
print(_id, _type, _object, _result)
輸出：
Packages totalPackages total 1200
DeliveryMethod packagesSent priority 100
DeliveryMethod packagesSent express 100
DeliveryMethod packagesSent ground 200
DeliveryMethod packagesReceived priority 100
DeliveryMethod packagesReceived express 100
DeliveryMethod packagesReceived ground 200
茅侃侃
TA貢獻1842條經驗 獲得超21個贊
//*如果您不迭代所有標簽（ ），而只是迭代，也許性能會更高<Value>：
from lxml import etree
xml_file = open('stack_sample.xml')
tree = etree.parse(xml_file)
root = tree.getroot()
for val in tree.xpath('//Value'):
t = {t.get('pos'): t.text for t in val.getparent().xpath('./Type')}
for r in val.xpath('./result'):
print(val.getparent().get('Id'), val.get('Object'), t[r.get('pos')], r.text)
印刷：
Packages total totalPackages 1200
DeliveryMethod priority packagesSent 100
DeliveryMethod priority packagesReceived 100
DeliveryMethod express packagesSent 200
DeliveryMethod express packagesReceived 200
DeliveryMethod ground packagesSent 300
DeliveryMethod ground packagesReceived 300
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