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409423
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0
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0
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4
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2009-01-03T17:29:07.390
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2018-04-02T20:59:46.563
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2018-04-02T20:59:46.563
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409370
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2
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50
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J.F. Sebastian
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For the first question, the first thing you should do is sort the list by the second field using <a href="https://docs.python.org/3/library/operator.html#operator.itemgetter" rel="nofollow noreferrer">itemgetter</a> from the operator module: <pre><code>x = [ ['4', '21', '1', '14', '2008-10-24 15:42:58'], ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['5', '21', '3', '19', '2008-10-24 15:45:45'], ['6', '21', '1', '1somename', '2008-10-24 15:45:49'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ] from operator import itemgetter x.sort(key=itemgetter(1)) </code></pre> Then you can use itertools' <a href="https://docs.python.org/3/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">groupby</a> function: <pre><code>from itertools import groupby y = groupby(x, itemgetter(1)) </code></pre> Now y is an iterator containing tuples of (element, item iterator). It's more confusing to explain these tuples than it is to show code: <pre><code>for elt, items in groupby(x, itemgetter(1)): print(elt, items) for i in items: print(i) </code></pre> Which prints: <pre><code>21 <itertools._grouper object at 0x511a0> ['4', '21', '1', '14', '2008-10-24 15:42:58'] ['5', '21', '3', '19', '2008-10-24 15:45:45'] ['6', '21', '1', '1somename', '2008-10-24 15:45:49'] 22 <itertools._grouper object at 0x51170> ['3', '22', '4', '2somename', '2008-10-24 15:22:03'] ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] </code></pre> For the second part, you should use list comprehensions as mentioned already here: <pre><code>from pprint import pprint as pp pp([y for y in x if y[3] == '2somename']) </code></pre> Which prints: <pre><code>[['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51']] </code></pre>
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