Python 抓取时保存网页中的图像/表格

我需要从这个网站上抓取一张图片： 例如，对于

堆栈溢出，towardsdatascience

URL

stackoverflow.com
towardsdatascience.com

我不知道如何将表格/图像上的信息包含在

<div class="sparkline" style="width: 1225px;"><div id="wm-graph-anchor"><div id="wm-ipp-sparkline" title="Explore captures for this URL" style="height: 77px;"><canvas class="sparkline-canvas" width="1225" height="75" alt="sparklines"></canvas></div></div><div id="year-labels"><span class="sparkline-year-label">1996</span><span class="sparkline-year-label">1997</span><span class="sparkline-year-label">1998</span><span class="sparkline-year-label">1999</span><span class="sparkline-year-label">2000</span><span class="sparkline-year-label">2001</span><span class="sparkline-year-label">2002</span><span class="sparkline-year-label">2003</span><span class="sparkline-year-label">2004</span><span class="sparkline-year-label">2005</span><span class="sparkline-year-label">2006</span><span class="sparkline-year-label">2007</span><span class="sparkline-year-label">2008</span><span class="sparkline-year-label">2009</span><span class="sparkline-year-label">2010</span><span class="sparkline-year-label">2011</span><span class="sparkline-year-label">2012</span><span class="sparkline-year-label">2013</span><span class="sparkline-year-label">2014</span><span class="sparkline-year-label">2015</span><span class="sparkline-year-label">2016</span><span class="sparkline-year-label">2017</span><span class="sparkline-year-label">2018</span><span class="sparkline-year-label">2019</span><span class="sparkline-year-label selected-year">2020</span></div></div>

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你能给我一些关于如何获取这些图像的信息吗？

如果你在for循环中有每个图像URL，你可以使用python库

urllib下载图像。request

函数

urretrive

：

首先在脚本开始时使用

import os
from urllib.parse import urlparse
import urllib.request

然后使用

for url in df_url['URL']:
  urllib.request.urlretrieve(url,os.path.basename(urlparse(url).path))

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如果您不想使用URL basename保存，请不要进行前两次导入。

谢谢。您知道导入iv class=“sparkline”的标签是否正确吗？我不知道哪一个对应于那个图像。更新：它不工作，因为我得到一个ValueError:

ValueError:未知url类型：

。url类型正确，但如果使用url lib.request，则会出现此错误

for url in df_url['URL']:
  urllib.request.urlretrieve(url,os.path.basename(urlparse(url).path))