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If the returned order does not matter, you can try using <code>groupby</code> from <code>itertools</code> to group the items by their first element (after sorting by the first element), and then pulling out the maximum value with the <code>max</code> function (also, it should be noted that this returns a new list as opposed to modifying in place): <pre><code>In [1]: from itertools import groupby In [2]: l = [[['a'],[24],214,1] ,[['b'],[24],312,1] ,[['a'],[24],3124,1] , [['c'],[24],34,1]] In [3]: result = [] In [4]: for k, g in groupby(sorted(l, key=lambda x: x[0]), key=lambda x: x[0]): ...: result.append(max(g, key=lambda m: m[2])) ...: ...: In [5]: result Out[5]: [[['a'], [24], 3124, 1], [['b'], [24], 312, 1], [['c'], [24], 34, 1]] </code></pre> Extending this a bit, if you want to maintain original order, you can modify <code>l</code> by including only those items the are in <code>results</code>, which will maintain order: <pre><code>In [6]: l = [i for i in l if i in result] In [7]: l Out[7]: [[['b'], [24], 312, 1], [['a'], [24], 3124, 1], [['c'], [24], 34, 1]] </code></pre> And to combine that into a true abomination of a one-liner, you could (but probably shouldn't :) ) do this: <pre><code>In [10]: l = [[['a'],[24],214,1] ,[['b'],[24],312,1] ,[['a'],[24],3124,1] , [['c'],[24],34,1]] In [11]: [i for i in l if i in [max(g, key=lambda m: m[2]) for k, g in groupby(sorted(l, key=lambda x: x[0]), key=lambda x: x[0])]] Out[11]: [[['b'], [24], 312, 1], [['a'], [24], 3124, 1], [['c'], [24], 34, 1]] </code></pre>
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1. COcode golf : `[max(v, key=lambda item:item[2]) for _, v in groupby(sorted(l), lambda item: item[0])]`
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2. CO@Boud Haha, I was honestly just about to type that with the disclaimer 'In an attempt to confuse anyone reading your code, you could try ...'. I will leave the glory to you!
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