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    <p>Here is my approach using <code>Pick</code></p> <pre><code>maxBy[list_, n_] := With[{s = list[[All, n]]}, Pick[list, s, Max[s]]] maxBy[{{4, 3}, {5, 10}, {-2, 1}, {3, 7}}, 2] (* output: {{5, 10}} *) </code></pre> <p>This version works on any number of elements per sublist provided <code>n</code> is less than or equal to the length of the shortest sublist. </p> <p>Timings for this version on my machine</p> <pre><code>list2 = RandomInteger[{-10^7, 10^7}, {10^6, 2}]; list3 = RandomInteger[{-10^7, 10^7}, {10^6, 3}]; list9 = RandomInteger[{-10^7, 10^7}, {10^6, 9}]; maxBy[list2, 2]; // Timing maxBy[list3, 2]; // Timing maxBy[list9, 2]; // Timing (* output: {0.030341, Null} {0.030912, Null} {0.033313, Null} *) </code></pre> <p>Compared to David's code</p> <pre><code>maxBy[list2, 2]; // Timing maxBy[list3, 2]; // Timing maxBy[list9, 2]; // Timing (* ouput: {0.186175, Null} {0.184989, Null} {0.262018, Null} *) </code></pre> <p>Yoda's code</p> <pre><code>maxBy[list2, 2]; // Timing maxBy[list3, 2]; // Timing maxBy[list9, 2]; // Timing (* ouput: {0.944016, Null} {0.83094, Null} {0.874126, Null} *) </code></pre> <p>And belisarius' code</p> <pre><code>Reverse /@ Take[#, Ordering[#, -1]] &amp;@(Reverse /@ #) &amp;@list2; // Timing Take[#, Ordering[Last /@ #, -1]] &amp;@list2; // Timing #[[Ordering[#, -1, Last@#2 &gt; Last@#1 &amp;]]] &amp;@list2; // Timing #[[First@Position[#, Max@#] &amp;@(Last /@ #)]] &amp;@list2; // Timing (* output: {0.211016, Null} {0.099253, Null} {2.03415, Null} {0.266934, Null} *) </code></pre>
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    1. POWhat's the best way to select the maximum one from a list of lists by their last element?
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    1. USHeike
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    1. COStrange. The difference in our code was far less significant on my machine (2.8 GHz MacBook Intel Core i7 with 8 GHz, 1067 MHz, DDR3) Your code ran in .10, .12, and .21 on my machine. My code ran in .16, .18, and .54 using the same lists each time.
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    2. CO@David: Strange indeed. My computer is slower than yours (2.4 GHz Macbook Intel Core 2 Duo) so in theory it should have run faster. I just tried in a fresh session of Mathematica just in case but my timings are virtually the same as before.
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    3. COSuch things can happen. Regardless, your code is fast, and clean, it seems. Welcome to Stack Overflow.
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