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copied!<p>Here's an approach that relies on <code>Transpose</code>:</p> <pre><code>maxBy = #1[[Position[t = Transpose[#1][[#2]], Max[t]][[All, 1]]]] &; </code></pre> <p>For example: list = {{4, 3}, {5, 10}, {20, 1}, {3, 7}};</p> <pre><code>maxBy[list, 1] (* {{20, 1}} *) maxBy[list, 2] (* {{5, 10}} *) </code></pre> <p>It can handle more than two elements per sublist, provided that the sublists are all the same length.</p> <pre><code>r:=RandomInteger[{-10^5,10^5}]; list3=Table[{r,r,r},{j,10^2}]; (* 3 numbers in each sublist *) list9=Table[{r,r,r,r,r,r,r,r,r},{j,10^2}]; (* 9 numbers *) maxBy[list3, 2] (* Find max in position 2 of list3 *) (* {{-93332, 99582, 4324}} *) maxBy[list9, 5] (* Find max in position 5 of list9 *) (* {{7680, 85508, 51915, -58282, 94679, 50968, -12664, 75246, -82903}} *) </code></pre> <p>Of course, the results will vary according to the random numbers you have generated.</p> <p><strong>Edit</strong></p> <p>Here's some timing data for large lists. <code>SortBy</code> is clearly slower. but doesn't seem as influenced by the number of elements in each sublist. First, my <code>maxBy</code> code followed by <code>SortBy</code>:</p> <p><img src="https://i.stack.imgur.com/pia4A.png" alt="Timing comparison"></p> <p>Using the same list2, here's some timing data for Yoda's code. Although his routine is also called maxBy, it is his that produced the output that follows:</p> <p><img src="https://i.stack.imgur.com/heCLz.png" alt="Yoda"></p> <p>Again, with the same list2, some data for Belisarius' code:</p> <p><img src="https://i.stack.imgur.com/0LCAK.png" alt="Belisarius"></p> <p>His second suggestion is the fastest of all tested.</p>

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