我通过 Flickr 的 API 提取图像数据,打印出来的是几千个 xml 对象,如下所示:
<photo accuracy="15" context="0" dateupload="1398279194" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="320" id="13986079375" isfamily="0" isfriend="0" ispublic="1" latitude="41.828482" license="0" longitude="-87.624506" owner="100231432@N02" pathalias="perspectivesschools" place_id="cF8n.mJTWrhYf0uBEw" secret="f46eef0b1d" server="7308" title="Sean Gallagher, Pulitzer Photojournalist visits MSA" url_n="https://farm8.staticflickr.com/7308/13986079375_f46eef0b1d_n.jpg" width_n="213" woeid="28297331" />
<photo accuracy="12" context="0" dateupload="1394558054" farm="4" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086071753" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="265103ac38" server="3040" title="" url_n="https://farm4.staticflickr.com/3040/13086071753_265103ac38_n.jpg" width_n="320" woeid="13978" />
<photo accuracy="12" context="0" dateupload="1394558019" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086343854" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="a6858f84d2" server="7451" title="" url_n="https://farm8.staticflickr.com/7451/13086343854_a6858f84d2_n.jpg" width_n="320" woeid="13978" />
现在我想在一次运行中提取属性“lat”和“long”的数据。另一个是属性“url_n”的数据。我怎样才能在Python中做到这一点?我没有解析 xml 数据的经验,不知道从哪里开始。
非常感谢!
最佳答案
使用lxml
虽然Python中有多个与XML相关的包,包括。 stdlib 之一,我更喜欢使用 lxml
,如
它提供了我需要的一切(良好的 XPath 支持、模式验证等),我更喜欢保留数字
我使用的软件包数量很少。
对于 Flickr 中的 xml 文档,解决方案可能如下所示
脚本flickr.py
from lxml import etree
xmllines = """
<photo accuracy="15" context="0" dateupload="1398279194" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="320" id="13986079375" isfamily="0" isfriend="0" ispublic="1" latitude="41.828482" license="0" longitude="-87.624506" owner="100231432@N02" pathalias="perspectivesschools" place_id="cF8n.mJTWrhYf0uBEw" secret="f46eef0b1d" server="7308" title="Sean Gallagher, Pulitzer Photojournalist visits MSA" url_n="https://farm8.staticflickr.com/7308/13986079375_f46eef0b1d_n.jpg" width_n="213" woeid="28297331" />
<photo accuracy="12" context="0" dateupload="1394558054" farm="4" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086071753" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="265103ac38" server="3040" title="" url_n="https://farm4.staticflickr.com/3040/13086071753_265103ac38_n.jpg" width_n="320" woeid="13978" />
<photo accuracy="12" context="0" dateupload="1394558019" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086343854" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="a6858f84d2" server="7451" title="" url_n="https://farm8.staticflickr.com/7451/13086343854_a6858f84d2_n.jpg" width_n="320" woeid="13978" />
"""
for line in xmllines.strip().splitlines():
doc = etree.fromstring(line)
urls = doc.xpath("/photo/@url_n")
if urls:
url = urls[0]
print url
else:
print "---no attribute url_n was found---"
它将输出:
$ python flickr.py
https://farm8.staticflickr.com/7308/13986079375_f46eef0b1d_n.jpg
https://farm4.staticflickr.com/3040/13086071753_265103ac38_n.jpg
https://farm8.staticflickr.com/7451/13086343854_a6858f84d2_n.jpg
关于python - 使用 Python 进行基本 xml 解析,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24851593/