How can I find the vertexes of a Polygon?

Question

I made a polygon with the following vertexes :

p = {{0, 0, 0}, {0, 0, 300}, {0, 300, 300}, {0, 300, 0}};(*Vertices of the polygon*)
myGraphics=Graphics3D[{Polygon[p]}, Boxed -> False, Lighting -> {Gray}];

How can I find the vertexes of myGraphics?

Answer 1

Try the following:

In[2]:= Cases[myGraphics, Polygon[pts_] -> pts, Infinity]
Out[2]= {{{0, 0, 0}, {0, 0, 300}, {0, 300, 300}, {0, 300, 0}}}

If your Graphics3D object had many polygons you would get a list comprised of lists of the vertexes of all of them.

Update

In response to István Zachar's commentL: to confine the pattern to match only 3D polygons, use

Cases[myGraphics, Polygon[pts:{{_,_,_}..}] -> pts, Infinity]

Answer 2

A function to extract graphics primitives:

ClearAll[getVrtxCoords];
getVrtxCoords[plot_, prims : {___}] := Module[{pts, pts2,
   prms = Alternatives @@ prims,
vcoords = Cases[plot, GraphicsComplex[{coords___}, __] :> {coords}, {0, Infinity}]},
pts = Cases[plot, prms[points : {({_, _} | {_, _, _}) ..}, ___] :> points, {0, 
 Infinity}];
If[vcoords === {},
If[pts === {}, {}, First@pts],
If[(pts2 =  Cases[plot, prms[{points___}, ___] :> 
   ((First[vcoords][[#]]) & /@ {points}), {0, Infinity}]) == {}, {}, First@pts2]]]

Examples:

(* OP's example : *)
p = {{0, 0, 0}, {0, 0, 300}, {0, 300, 300}, {0, 300, 0}};
myGraphics =  Graphics3D[{Polygon[p]}, Boxed -> False, Lighting -> {Gray}];
Grid[{{"image", "Points", "Lines", "Polygons"},
   Prepend[Column[N[getVrtxCoords[myGraphics, {#}]]] & /@ 
  {Point, Line, Polygon},  myGraphics]},  Spacings -> {1, 1}, Dividers -> All]

A 2D example with GraphicsComplex:

 v = Table[15 {Cos[t], Sin[t]}, {t, 0, 4 Pi, 4 Pi/5}];
grphcs2 = Graphics[GraphicsComplex[v, {Red, Thick, Polygon[{1, 2, 3, 4, 5, 6}],
 Blue, Opacity[.7], Polygon[{{1, 2, 3}, {3, 4, 5}},
  VertexColors -> {{Yellow, Yellow, Yellow}, {Blue, Blue, Blue}}],
 Opacity[1], Dashed, Thickness[.02], Brown, Line[{1, 4, 2}],
 PointSize[Large], Red, Point[{1, 2, 3, 4, 5}]}]];
Grid[{{"image", "Points", "Lines", "Polygons"},
  Prepend[Column[N[getVrtxCoords[grphcs2, {#}]]] & /@
    {Point, Line, Polygon}, grphcs2]},   Spacings -> {1, 1}, Dividers -> All]

A 3D example:

v3 = PolyhedronData["Dodecahedron", "VertexCoordinates"];
Short[i = PolyhedronData["Dodecahedron", "FaceIndices"]];
dodec = Graphics3D[{Yellow, GraphicsComplex[v3, Polygon[i]]}, ImageSize -> 300];
modifieddodec =  Graphics3D[{Opacity[.8], EdgeForm[Opacity[.3]], 
 Polygon[#, VertexColors -> Table[Hue[RandomReal[]], {25}]] & /@ 
  getVrtxCoords[dodec, {Polygon}]}, Lighting -> "Neutral",  ImageSize -> 300];
Grid[{{"image", "reconstructed", "Points", "Lines", "Polygons"}, 
   {dodec, modifieddodec,  N@getVrtxCoords[dodec, {Point}], 
    N@getVrtxCoords[dodec, {Line}], 
    Short[N@getVrtxCoords[dodec, {Polygon}], 10]}}, Spacings -> {1, 1},
 Dividers -> All, Alignment -> {Center, Top}]

Answer 3

Another version using pattern matching which can be quite useful for other purposes as well (other than in @m_goldberg's patent solution, all points will be returned in one list, the primitive structure will not be preserved):

Cases[myGraphics, {x_?NumericQ, y_?NumericQ, z_?NumericQ}, Infinity]

{{0, 0, 0}, {0, 0, 300}, {0, 300, 300}, {0, 300, 0}}

Why is that approach (esp. the pattern) useful? You can easily work on the coordinates and it works on several important Graphics3D primitives (Point, Line, Polygon,Text) out of the box regardless of the structure of your expression. Apart from the {x,y,z} tuples your expression is not changed.

myGraphics /. {x_?NumericQ, y_?NumericQ, z_?NumericQ} :> {x, 2 y-z, 3 z}

For multi-primitive syntax, GraphicsComplex etc. this approach works as well if you get a bit more fancy on the pattern side.