Sort a list in a descending order

Question

Say we have a list:

{{1/2, -(Sqrt[3]/2)}, {1, 0}, {1/2, Sqrt[3]/2}, {-(1/2), Sqrt[3]/2}, {-1, 0}, {-(1/2), -(Sqrt[3]/2)}}

now we want to sort that list by looking at the second component of each sub list (call it y coordinate for ease), i.e. sort it s.t. y is in the descending order. So we'd get

{{1/2, Sqrt[3]/2},{-1/2, Sqrt[3]/2},{1,0},{-1,0},{1/2,-(Sqrt[3]/2)},{-1/2,-(Sqrt[3]/2)}}

And yeah, I should mention that if there are more than one sub lists with equal y, we would additionally (sub)sort them by descending x (first component in each of the sub lists).

I tried to use SortBy, in different forms, but I don't seem to be able to figure it out by myself. I'd appreciate any help.

Likely a duplicate of https://mathematica.stackexchange.com/q/2729/12 — Szabolcs, Feb 13 '19 at 13:59
Thanks for that link @Szabolcs. Reverse[SortBy[list, N[Last]]] seems to be doing the job. — amator2357, Feb 13 '19 at 14:12
Not sure why and how in the case of those sub lists that have equal y component it puts those with the greater x first... — amator2357, Feb 13 '19 at 14:18
Just realized that Reverse[SortBy[list, N[Last]]] doesn't actually work. — amator2357, Feb 13 '19 at 14:24
You did not say what you tried originally. N[Last] is not a function, so it's not appropriate here. Use N@*Last or N[Last[#]]&. — Szabolcs, Feb 13 '19 at 14:26
@amator, if you've figured it out from the comments, you can answer your own question. Helps the silent majority of lurkers develop their knowledge too. — MikeY, Feb 13 '19 at 14:51

score 1 · Answer 1 · answered Feb 13 '19 at 15:02

1

As suggested in the comments, the following works as desired:

Reverse[SortBy[list, N[Last[#]]&]].

answered Feb 13 '19 at 15:02

amator2357

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If you use SortBy, you might as well put a minus sign in front of the sorting function so you don't have to reverse the list after sorting it. – Sjoerd Smit Feb 13 '19 at 16:13
Nice, wasn't aware of that, thank you @SjoerdSmit ! – amator2357 Feb 13 '19 at 16:14

score 1 · Answer 2 · answered Jan 30 '24 at 00:37

list = 
  {{1/2, -(Sqrt[3]/2)}, {1, 0}, {1/2, Sqrt[3]/2}, 
  {-(1/2), Sqrt[3]/2}, {-1, 0}, {-(1/2), -(Sqrt[3]/2)}};

Using ReverseSortBy (new in 12.0)

ReverseSortBy[N @* Last] @ list

{{1/2, Sqrt[3]/2}, {-1/2, Sqrt[3]/2}, {1, 0}, {-1, 0}, {1/2, -1/2 Sqrt[3]}, {-1/2, -1/2 Sqrt[3]}}

score 1 · Answer 3 · answered Jan 30 '24 at 04:03

l = 
  {{1/2, -(Sqrt[3]/2)}, {1, 0}, {1/2, Sqrt[3]/2}, 
  {-(1/2), Sqrt[3]/2}, {-1, 0}, {-(1/2), -(Sqrt[3]/2)}};

The ordering using OrderingBy:

o = Flatten[RotateRight[Sort /@ Partition[Reverse@OrderingBy[#, Abs], 2]]] &@l;

Apply the ordering:

l[[o]]

{{-(1/2), Sqrt[3]/2}, {-(1/2), -(Sqrt[3]/2)}, {1, 0}, {-1, 0}, {1/2, -(Sqrt[3]/2)}, {1/2, Sqrt[3]/2}}

Sort a list in a descending order

3 Answers3