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}}
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}}
SortBy[{{1, 2, 3}, {2, 3, 1}, {3, 1, 2}, {2, 2}}, {Minus@*Last}] will ensure a stable sort.
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– Alan
Nov 5 '17 at 21:08
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.
SortBy to be considerably faster than Sort with a custom comparison function (because each custom comparison requires a Mathematica evaluation)
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– Szabolcs
Nov 5 '17 at 21:18
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