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copied!<p>The <a href="http://docs.python.org/2.7/library/index.html" rel="nofollow">Python Standard Library</a> is your friend:</p> <pre><code>from collections import defaultdict from itertools import combinations tagged = { 'apple': ['fruit', 'green', 'nature'], 'sometree': ['tree', 'green', 'wood', 'nature'], 'banana': ['fruit', 'nature'], 'someplant': ['green', 'wood', 'nature'], 'otherplant': ['green', 'wood', 'nature'] } tag_sets = defaultdict(set) for each, tags in tagged.items(): tag_sets[frozenset(tags)].add(each) for a, b in combinations(tag_sets, 2): if a < b: tag_sets[a].update(tag_sets[b]) elif b < a: tag_sets[b].update(tag_sets[a]) for k, v in tag_sets.items(): print '(%s): %s' % (', '.join(k), ', '.join(v)) </code></pre> <p>Result:</p> <pre><code>(wood, green, nature): sometree, someplant, otherplant (fruit, green, nature): apple (fruit, nature): banana, apple (green, wood, tree, nature): sometree </code></pre> <p>How it works: First, we transpose the mapping <em>item</em> -> <em>tag set</em> to <em>tag set</em> -> <em>set of items associated with said tag set</em>. Then we iterate over all distinct 2-sets of tag sets, represented by pairs (a, b) and check if either a or b is a proper subset of the other. If it is, we join the mapped items of the superset with the mapped items of the proper subset.</p> <p>[EDIT]</p> <p>Here is another solution:</p> <pre><code>from collections import defaultdict tagged = { 'apple': ['fruit', 'green', 'nature'], 'sometree': ['tree', 'green', 'wood', 'nature'], 'banana': ['fruit', 'nature'], 'someplant': ['green', 'wood', 'nature'], 'otherplant': ['green', 'wood', 'nature'] } intersection = set(tagged).intersection tag_sets = set() items = defaultdict(set) for item, tags in tagged.items(): tag_sets.add(frozenset(tags)) for tag in tags: items[tag].add(item) for tag_set in tag_sets: result = intersection(*(items[tag] for tag in tag_set)) print '(%s): %s' % (', '.join(tag_set), ', '.join(result)) </code></pre> <p>I don't know if it is any faster, since i don't have enough data to do a meaningful comparison.</p> <p>[/EDIT]</p>

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