Group a list based on multiples of sublists

Question

I have a list of lists like this one, where the sublists vary in length.

list = {{0, 1, 2}, {0, -2, 4}, {0, 3, 6}, {1, 2, 3}, {0, 0}, {2, 4, 6}, {-2, -4, -6}, {1, 2, -3}, {1, -2, 3}, {-6, -8, -10}, {0.3, 0.4, 0.5}, {0, 0}};

I want to rearrange the list so that each group has all the multiples of the sublists grouped together. The above example should give this reorganized list:

{{0, 1, 2}, {0, 3, 6}}
{{0, -2, 4}}
{{1, 2, 3}, {2, 4, 6}, {-2, -4, -6}}
{{0, 0}, {0, 0}}
{{1, 2, -3}}
{{1, -2, 3}}
{{-6, -8, -10}, {0.3, 0.4, 0.5}}

I have read this post, but it’s mainly suitable for positive integer multiples and equal-length sublists. This method fails with my example.

GatherBy[list, #/Max[1, GCD @@ #] &]

I prefer the solution based on GatherBy instead of Gather, because GatherBy is much more efficient.

Answer 1

Updated

list = {{0, 1, 2}, {0, -2, 4}, {0, 3, 6}, {1, 2, 3}, {0, 0}, {2, 4, 
    6}, {-2, -4, -6}, {1, 2, -3}, {1, -2, 3}, {-6, -8, -10}, {0.3, 
    0.4, 0.5}, {0, 0}};
f[w_] := Sort@{Normalize[Rationalize@w], -Normalize[Rationalize@w]}
GatherBy[list, f]

{{{0, 1, 2}, {0, 3, 6}}, {{0, -2, 4}}, {{1, 2, 3}, {2, 4, 6}, {-2, -4, -6}}, {{0, 0}, {0, 0}}, {{1, 2, -3}}, {{1, -2, 3}}, {{-6, -8, -10}, {0.3, 0.4, 0.5}}}

The trick is, for a vector $(x,y,z)$, we map it into two vectors $\{ (x,y,z),-(x,y,z) \}$, and Sort the two vectors in this set.

So the two vectors $(a,b,c)$ and $(-a,-b,-c)$ are map into the same order set $$\{(a,b,c),(-a,-b,-c)\}$$

so they regard as the same object!

For example,

u = {1, -2, 3};
v = {-1, 2, -3};
Sort[{u, -u}] === Sort[{v, -v}]
(* True *)

Original

Not so effect. Here we define a normalize function.

Function[w, (Sign@First@w)*Normalize[Rationalize@w, Sqrt[#.#] &]]

for example, if w={x,y,z},we first Rationalize w,and then calculate $$ \frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}}$$

after this,we make the $\frac{x}{\sqrt{x^2+y^2+z^2}}$ is positive by multiple the vector the sign of x.

list = {{1, 2, 3}, {0, 0}, {2, 4, 6}, {-2, -4, -6}, {1, 
    2, -3}, {1, -2, 3}, {-6, -8, -10}, {0.3, 0.4, 0.5}, {0, 0}};
GatherBy[list, 
 Function[w, (Sign@First@w)*Normalize[Rationalize@w, Sqrt[#.#] &]]]

{{{1, 2, 3}, {2, 4, 6}, {-2, -4, -6}}, {{0, 0}, {0, 0}}, {{1, 2, -3}}, {{1, -2, 3}}, {{-6, -8, -10}, {0.3, 0.4, 0.5}}}

Other idea

Maybe MatrixRank is another way.

Answer 2

ClearAll[proj]
proj = KroneckerProduct[#, #] &[Normalize @ Rationalize @ #] &;

GatherBy[list, proj] // Column

ClearAll[fit]
fit = Fit[{Table[0, Length@#], #}, Array[x, Length[#]-1], Array[x, Length[#]-1]]&;

GatherBy[list, fit] // Column

Answer 3

If we does not insist on using GatherBy ,then there other way can be use to get the same result by using Gather.

Gather is easy to handle.

Method I

list = {{0, 1, 2}, {0, -2, 4}, {0, 3, 6}, {1, 2, 3}, {0, 0}, {2, 4, 
    6}, {-2, -4, -6}, {1, 2, -3}, {1, -2, 3}, {-6, -8, -10}, {0.3, 
    0.4, 0.5}, {0, 0}, {Sqrt[3], Sqrt[3]}, {1, 1}, {0, 0, 0}, {1, 1, 
    1}, {2., 2., 2.}, {π Sqrt[3], 2 π Sqrt[3], 
    3 π Sqrt[3]}};
Gather[list, 
 RegionEqual[AffineSpace[{#1, -#1}], AffineSpace[{#2, -#2}]] &]

Method II

list = {{0, 1, 2}, {0, -2, 4}, {0, 3, 6}, {1, 2, 3}, {0, 0}, {2, 4, 
    6}, {-2, -4, -6}, {1, 2, -3}, {1, -2, 3}, {-6, -8, -10}, {0.3, 
    0.4, 0.5}, {0, 0}, {Sqrt[3], Sqrt[3]}, {1, 1}, {0, 0, 0}, {1, 1, 
    1}, {2., 2., 2.}, {π Sqrt[3], 2 π Sqrt[3], 
    3 π Sqrt[3]}};
Gather[list, 
 Length[#1] == Length[#2] && Norm[#1] == Norm[#2] == 0 || 
   Length[#1] == Length[#2] && Norm[#1]*Norm[#2] != 0 && 
    MatrixRank[{#1, #2}] == 1 &]

{{{0, 1, 2}, {0, 3, 6}}, {{0, -2, 4}}, {{1, 2, 3}, {2, 4, 6}, {-2, -4, -6}, {Sqrt[3] π, 2 Sqrt[3] π, 3 Sqrt[3] π}}, {{0, 0}, {0, 0}}, {{1, 2, -3}}, {{1, -2, 3}}, {{-6, -8, -10}, {0.3, 0.4, 0.5}}, {{Sqrt[3], Sqrt[3]}, {1, 1}}, {{0, 0, 0}}, {{1, 1, 1}, {2., 2., 2.}}}