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I am looking at the two functions SortBy and ReverseSort.

SortBy[{{1, 2, 3}, {2, 3, 1}, {3, 1, 2}, {2, 2}}, Last]

{{2, 3, 1}, {2, 2}, {3, 1, 2}, {1, 2, 3}}

How do I sort this by the last element, but in the reverse order? I want to get this:

{{1, 2, 3}, {3, 1, 2}, {2, 2}, {2, 3, 1}}

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    $\begingroup$ may be I am missing your question. Is this what you want? Reverse@SortBy[{{1,2,3},{2,3,1},{3,1,2},{2,2}},Last] which gives {{1,2,3},{3,1,2},{2,2},{2,3,1}} $\endgroup$ – Nasser Nov 5 '17 at 18:21
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If you don't care about the ordering of {3, 1, 2} and {2, 2}, do this:

SortBy[{{1, 2, 3}, {2, 3, 1}, {3, 1, 2}, {2, 2}}, Minus @* Last]
   {{1, 2, 3}, {2, 2}, {3, 1, 2}, {2, 3, 1}}

Otherwise, add a second sorting criterion; e.g.

SortBy[{{1, 2, 3}, {2, 3, 1}, {3, 1, 2}, {2, 2}}, {Minus @* Last, Minus @* Length}]
   {{1, 2, 3}, {3, 1, 2}, {2, 2}, {2, 3, 1}}

or just wrap the sorting criterion in a list (per Alan):

SortBy[{{1, 2, 3}, {2, 3, 1}, {3, 1, 2}, {2, 2}}, {Minus @* Last}]
   {{1, 2, 3}, {3, 1, 2}, {2, 2}, {2, 3, 1}}
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  • $\begingroup$ I think SortBy[{{1, 2, 3}, {2, 3, 1}, {3, 1, 2}, {2, 2}}, {Minus@*Last}] will ensure a stable sort. $\endgroup$ – Alan Nov 5 '17 at 21:08
  • $\begingroup$ Ah, right. Let me add that note. $\endgroup$ – J. M.'s technical difficulties♦ Nov 5 '17 at 21:10
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Using:

l = {{2, 3, 1}, {2, 2}, {3, 1, 2}, {1, 2, 3}};

One option, just using Sort:

Sort[l, Last[#2] < Last[#1] &]

Another option, using SortBy and Reverse:

Reverse@SortBy[l, Last]

I'm not completely sure about ReverseSort (my version of Mathematica does not have it), but I imagine something like this would be at least a start (not too sure about "tie" cases, or how concerned you are with them):

ReverseSort[l, Last[#1] < Last[#2] &]

Not an expert on this, so anyone should feel free to correct me if I'm wrong.

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    $\begingroup$ I'd expect SortBy to be considerably faster than Sort with a custom comparison function (because each custom comparison requires a Mathematica evaluation) $\endgroup$ – Szabolcs Nov 5 '17 at 21:18

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