66
Based on Oleksandr's excellent design idea here is my re-implementation of his package which offers much richer set of shapes.
UPDATE from October 2019
Now my function is published in the Wolfram Function Repository what means that it is available for users of Mathematica version 12.0 or higher as ResourceFunction["PolygonMarker"]. Users of previous ...
42
The following solution is hacky work-around to overcome the lack of proper pattern directives. I don't like it 100%, but still it is usable.
Solution
The idea is simple. If you see FilledCurve documentation, it supports an object with holes:
So what if you put a huge enough rectangle as an outer boundary? Then, your polygon becomes a hole and you can see ...
42
Fortunately, Wikipedia has the answer, as long as we are content to restrict ourselves to non-intersecting closed polygons. This will probably be an acceptable limitation, given that excessively complicated plot markers tend to look slightly distracting anyway.
Because we seek an aesthetic rather than rigorously well defined result, we do not need to be ...
27
UPDATE:
The previous version of my answer worked, but did not give control on the rounding radius, nor did it fully work with as a starting point for a geometric region for further calculations. Here is a version that is still based on spline curves, but it gives full control over the corner rounding radius. It also returns a FilledCurve object that in my ...
26
Here is an alternative answer. Of course, since you answered your own question, you may not need this. But I think the following is a viable alternative that may end up looking comparable, and has additional dynamic features.
Instead of ListPlot, just use BubbleChart.
data = ConstantArray[Range[5], 4] + Range[0, 6, 2];
newData = Map[MapIndexed[Join[#2, {#...
25
ragfield's solution is a good one that can easily be extended to arbitrarily complicated polygons if we create an inPolyQ function using winding numbers.
Here is a complicated polygon (the points only) from Mathematica's example data.
poly = Rescale[ExampleData[{"Statistics", "WesternUgandaBorder"}]];
This function determines whether a point is in the ...
20
The reason that these are called "pursuit polygons" is because they are formed from a dynamical system in which different agents pursue each other. Example:
In this image, one agent starts in each corner of the triangle. The agent starting in the lower right corner pursues the agent starting in the top corner, the agent in the top corner pursues the agent ...
18
Just wanted to add purely mathematical approach using complex mapping technique.
PolyMap[n_, z_] := z Hypergeometric2F1[1/n, 2/n, (n + 1)/n, z^n]
(* Integrate[1/(1 - ξ^n)^(2/n), {ξ, 0, z}] *)
g = GraphicsGrid[
Table[
ParametricPlot[
z = PolyMap[n, r (Cos[t] + I Sin[t])]; {Re[z], Im[z]},
{t, 0, 2 π}, PlotRange -> All, Axes -> False] /.
...
17
Mathematica's graphics language lacks a clipping primitive. However, you can sometimes simulate the same appearance using the sophisticated visualization functions.
RegionPlot[((0 <= x <= .5) && (0 <= y <= 1 + x)) || ((.5 < x <= 1) &&
0 <= y <= 2 - x), {x, -1, 2}, {y, -1, 2}, Mesh -> 20,
MeshFunctions -> ...
17
Since you mention that you want to use the rounded polygon in NDSolve[] as a region, you might want to look at the following construction:
With[{r = 1/5 (* rounding radius *)},
rp = DiscretizeRegion[
ImplicitRegion[RegionDistance[
Polygon[CirclePoints[{1 - 2 Sqrt[5 - 2 Sqrt[5]] r, π/10}, 5]], {x, y}] <=
r Sqrt[(5 - ...
answered Mar 31 '16 at 4:14
16
Late to the party~
A slight modification to make them more similar:
steps = AnglePath@Table[{r-0.015 r^2, 1.002*(2 Pi/5)}, {r, .1, 25, 0.1}];
ls=Thread[{Join[ConstantArray[Opacity@1,7],ConstantArray[Opacity@.7,13]],#}]&/@
Partition[Line/@Thread@{Most@steps,Rest@steps},20];
Graphics[{Red, ls}, Background -> Black]
Effect:
This solution focused on ...
16
Does this do what you are looking for?
The following function takes a parameterization surface and a region region and tries to mesh it with hexagons of radius meshsize. Afterwards, it maps surface over it and creates a mesh of extruded hexagons of thickness thickness
ClearAll[hexhex]
hexhex[surface_, thickness_, meshsize_, region_] :=
Module[{hex0, ...
15
FilledCurve will do the job because it can be styled by JoinForm:
Graphics[{
EdgeForm[{JoinForm["Round"], Thickness[0.05]}],
FilledCurve[Line /@ Partition[CirclePoints[5], 2, 2, 1]]
}, PlotRange -> 1.2]
MarcoB found that this simpler version also works (see comments):
Graphics[{
EdgeForm[{JoinForm["Round"], Thickness[0.05]}],
FilledCurve[Line@...
14
The FilledCurve approach in the accepted answer by @Yu-Sung Chang is very interesting, so I tried to make my own version of it. I defined the whole thing as a function that accepts graphics directives for the objects to be masked (backgroundDirectives - e.g., the fill pattern or image), and an arbitrary polygon as the mask. This includes Polygon with several ...
14
This might do the trick:
Manipulate[
ParametricPlot[
#1 {Cos[#2], Sin[#2]} & @@ {t, Log[i] Floor[t]},
{t, 0, 200}
, Background -> Black
, PlotStyle -> Purple
, Axes -> False
, PerformanceGoal -> "Quality"
, PlotRange -> {{-201, 201}, {-201, 201}}
],
{{i, 3.525}, 3.43, 3.6}
]
Since you enjoyed the animation aspect here ...
14
It's something like this:
steps = Table[{r, 1.001 (2 Pi/5)}, {r, 1, 25, 0.1}];
Graphics[{Red, Line@AnglePath[steps]}, Background -> Black]
13
(grabbing a lot of code from this answer)
This is just to get you started, adding in colors to the final GraphicsComplex should be pretty easy.
xyzString = ExportString[
qxyz /. {a_Integer, b___} :> {ElementData[a, "Abbreviation"], b},
"Table"];
{plot, coords, atoms} =
ImportString[
xyzString, {"XYZ", {"Graphics3D", "VertexCoordinates",
...
12
Here is a more general method for producing polygons with rounded corners. Using a bit of vector algebra and trigonometry, I came up with the following:
arcgen[{p1_, p2_, p3_}, r_, n_] :=
Module[{dc = Normalize[p1 - p2] + Normalize[p3 - p2], cc, th},
cc = p2 + r dc/EuclideanDistance[dc, Projection[dc, p1 - p2]];
th = Sign[...
answered Apr 2 '16 at 0:15
12
In case the undocumented internal function Region`Mesh`MeshCellNormals[meshregion, dimension] is of use to someone:
reg = BoundaryDiscretizeRegion[Ball[], PlotTheme -> "SmoothShading",
PrecisionGoal -> 1, MaxCellMeasure -> 0.1];
Graphics3D[
GraphicsComplex[
MeshCoordinates[reg],
{EdgeForm[], Thread[MeshCells[reg, 2], Polygon]},
...
11
Paint Poly with background colored stripes
Just for fun, I painted the lines in the background color.
The unnecessary white space is avoided by using AbsoluteOptions[g,PlotRange] from the polygon when graphed alone.
poly[n, color] makes a (solid colored) polygon with n vertices.
s[p, bc] puts stripes of a background color bc on polygon p. pr is the ...
11
EDIT: Reported to WRI as CASE:4237867 - fixed on V12. :)
Seems like a bug in my opinion. Consider the following:
With[{poly0 = Polygon[{{0, 0}, {4, 0}, {2, 2}, {4, 4}, {0, 4}}]},
With[{poly = Transpose[IdentityMatrix[2].Transpose@#] & /@ poly0},
{poly0 === poly, Area /@ {poly0, poly}}]]
{True, {12, Area[Polygon[{{0, 0}, {4, 0}, {2, 2}, {4, 4}, {...
10
This ought to do it:
GeoGraphics[{GeoStyling[RGBColor[0.1, 0.1, 0.3], Opacity[1]],
EntityClass["AdministrativeDivision", "ContinentalUSStates"][
"Polygon"]}, GeoBackground -> None]
What was needed was a different Entity (or rather in this case an EntityClass). This is a collection of US states, so I suspect it has information about their ...
10
I think it's worth reporting this to support. As a workaround you can numericize the polygon:
RegionMember[
Polygon[
N@{
{11,10},{10,11},{9,12},{9,13},{8,14},{7,15},{7,16},
{6,17},{6,18},{5,18},{4,18},{3,19},{2,18},{1,19},
{-1,19},{-2,19},{-3,19},{-4,19},{-5,18},{-4,18},
{-3,16},{-3,16},{-2,14},{-1,...
10
This seems like a bug. It might even be a bug I have reported in the past.
In particular, your polygon is slightly degenerate: it has {-3, 16} twice in a row, creating a zero-length edge. (This doesn't seem to be a sufficient condition alone to make RegionMember fail.)
You can remove such edges by the following replacement rule:
RegionMember[
Polygon[...
10
This is a packed array issue:
Area @ Polygon[Developer`ToPackedArray @ {{0, 0}, {4, 0}, {2, 2}, {4, 4}, {0, 4}}]
Area[Polygon[{{0, 0}, {4, 0}, {2, 2}, {4, 4}, {0, 4}}]]
This will be fixed in M12.
9
A shot in the dark: Reverse the orientation. Hey, it works...but I don't know why...???
Graphics3D[Polygon /@ Reverse@*pentagram /@ (vert[[#]] & /@ ind)]
A guess at what's happening. I'm not sure of the reason why things work correctly when one coordinate is the same for all vertices and do not work when the plane of the polygon is oblique. In the ...
9
H3110 customer,
thank you for contacting StackExchange coding service. We propose the following solution to your urgent problem:
pts = Table[{Cos[phi], Sin[phi]}, {phi, 0, 2 Pi - 2 Pi/5, 2 Pi/5}];
Graphics[{FaceForm[], EdgeForm[Black], Polygon[pts],
Dashed, Red,
Line[{pts[[-1]], pts[[1]] - 2*(pts[[1]] - pts[[2]])*{1, 0}, pts[[2]]}]
}]
9
xyz = Rest /@ qxyz;
Show[
ListPlot3D[xyz,
Mesh -> None,
BoxRatios -> Automatic,
Boxed -> False,
Axes -> False,
InterpolationOrder -> 1,
ColorFunction -> "Rainbow"],
NearestNeighborGraph[xyz, {All, 1.6}]]
8
Antialiasing -> False will do it, which surprisingly can be used in-line as a directive:
Graphics[{Antialiasing -> False, Polygon[{{0, 0}, {1, 0}, {0, 1}}],
Polygon[{{1, 0}, {0, 1}, {1, 1}}]}]
Because the multiple-polygon form of Polygon (presently?) renders without antialiasing by default this also works:
Graphics[{Polygon[{{{0, 0}, {1, 0}, {0, ...
8
The generated outside points are called Steiner points. You can prohibit the generation of these with
TriangulateMesh[bmr, MaxCellMeasure -> \[Infinity],
MeshQualityGoal -> 1, "SteinerPoints" -> None]
Now, other mesh elements then triangles are not possible currently. However, as Szabolcs points out in his answer you can merge connected ...
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