Grouping symmetric points in 2D domain

Question

I have the following list of points

L = {
    {-0.980187,-1.07694}, {-0.980187,1.07694},
    {-0.788396,-1.81236},{-0.788396,1.81236},
    {-0.437793,-1.07694},{-0.437793,1.07694},
    {-0.157935,-1.81236},{-0.157935,1.81236},
    {0.157935,-1.81236},{0.157935,1.81236},
    {0.437793,-1.07694},{0.437793,1.07694},
    {0.788396,-1.81236},{0.788396,1.81236},
    {0.980187,-1.07694},{0.980187,1.07694}
}

Which are distributed in four symmetric sets each having the points $(x,y), (-x,y),(x,-y),(-x,-y)$

(see image below).

I'd like to group the points based in their symmetry in a new list L1 in the following way

$L1 = \{l1,l2,l3,l4\},$ where $l_i = \{\{x_i, y_i\},\{-x_i, y_i\},\{x_i, -y_i\},\{-x_i, -y_i\}\}, i=1,2,3,4$

Answer 1

I would use GatherBy[data,Abs]

ClearAll[L, l];

L = GatherBy[data, Abs];

l[n_] := L[[n]]

ListPlot[ L ]

Answer 2

Just another way:

Map[Extract[L, #] &, Reverse@Partition[List /@ OrderingBy[L, Abs], {4}]] // ListPlot

Answer 3

l1 = Select[L, #[[1]] > 0 && #[[2]] > 0 &]
l2 = Select[L, #[[1]] < 0 && #[[2]] > 0 &]
l3 = Select[L, #[[1]] > 0 && #[[2]] < 0 &]
l4 = Select[L, #[[1]] < 0 && #[[2]] < 0 &]

Visualization

ListPlot[{l1, l2, l3, l4}, PlotRange -> {{-1.1, 1.1}, {-2.1, 2.1}}]

Answer 4

list = {
    {-0.980187,-1.07694}, {-0.980187,1.07694},
    {-0.788396,-1.81236}, {-0.788396,1.81236},
    {-0.437793,-1.07694}, {-0.437793,1.07694},
    {-0.157935,-1.81236}, {-0.157935,1.81236},
    {0.157935,-1.81236}, {0.157935,1.81236},
    {0.437793,-1.07694}, {0.437793,1.07694},
    {0.788396,-1.81236}, {0.788396,1.81236},
    {0.980187,-1.07694}, {0.980187,1.07694}};

Using SequenceCases and SortBy

ListPlot @ Reverse @ SequenceCases[SortBy[list, Abs], {_, _, _, _}]