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    <p>Here is a solution using lxml. The main idea is to find all the preceding and following elements and then do a roundrobin iteration through those elements:</p> <pre><code>def find_nearest(elt): preceding = elt.xpath('preceding::*/@href')[::-1] following = elt.xpath('following::*/@href') parent = elt.xpath('parent::*/@href') for href in roundrobin(parent, preceding, following): return href </code></pre> <p>A similar solution using BeautifulSoups' (or bs4's) <a href="http://www.crummy.com/software/BeautifulSoup/bs4/doc/#next-elements-and-previous-elements" rel="nofollow">next_elements and previous_elements</a> should also be possible.</p> <hr> <pre><code>import lxml.html as LH import itertools def find_nearest(elt): preceding = elt.xpath('preceding::*/@href')[::-1] following = elt.xpath('following::*/@href') parent = elt.xpath('parent::*/@href') for href in roundrobin(parent, preceding, following): return href def roundrobin(*iterables): "roundrobin('ABC', 'D', 'EF') --&gt; A D E B F C" # http://docs.python.org/library/itertools.html#recipes # Author: George Sakkis pending = len(iterables) nexts = itertools.cycle(iter(it).next for it in iterables) while pending: try: for n in nexts: yield n() except StopIteration: pending -= 1 nexts = itertools.cycle(itertools.islice(nexts, pending)) tekst = ''' &lt;li&gt; &lt;div class="views-field-field-webrubrik-value"&gt; &lt;h3&gt; &lt;a href="/307046"&gt;Claus Hjort spiller med mrkede kort&lt;/a&gt; &lt;/h3&gt; &lt;/div&gt; &lt;div class="views-field-field-skribent-uid"&gt; &lt;div class="byline"&gt;Af: &lt;span class="authors"&gt;Dennis Kristensen&lt;/span&gt;&lt;/div&gt; &lt;/div&gt; &lt;div class="views-field-field-webteaser-value"&gt; &lt;div class="webteaser"&gt;Claus Hjort Frederiksens argumenter for at afvise trepartsforhandlinger har ikke hold i virkeligheden. Hans rinde er nok snarere at forberede det ideologiske grundlag for en Løkke Rasmussens genkomst som statsministe &lt;/div&gt; &lt;/div&gt; &lt;span class="views-field-view-node"&gt; &lt;span class="actions"&gt; &lt;a href="/307046"&gt;Ls mere&lt;/a&gt; | &lt;a href="/307046/#comments"&gt;Kommentarer (4)&lt;/a&gt; &lt;/span&gt; &lt;/span&gt; &lt;/li&gt; ''' to_find = "Rasmussen" doc = LH.fromstring(tekst) for x in doc.xpath('//*[contains(text(),{s!r})]'.format(s = to_find)): print(find_nearest(x)) </code></pre> <p>yields</p> <pre><code>/307046 </code></pre>
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    1. POFind nearest link with BeautifullSoup (python)
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    1. POFind nearest link with BeautifullSoup (python)
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    1. COThanks - so this finds the nearest element. What if there is a match in both directions 3 iterations down? Will both links be returned?
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    2. COIt will return only the first nearest element. The `roundrobin` iterator yields an element from `parent`, then `preceding`, then `following`. So if there are more than one nearest element, the one in `parent` has highest precedence, then the one in `preceding` and then `following`.
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