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How can I replace the pixels of a binarized image with a custom image in Mathematica?
I figured that once I have a matrix M of 0 or 1 entries depending on a pixel being white or black (which I can obtain by using Binarize and manipulating the output a bit), I can use Graphics[] to place a custom image with square boundary in a grid wherever there's a 1 and a flat background when there's a 0, but I'm not exactly sure how to do this.
Thank you in advance :)
Here's one way:
mat = RandomInteger[1, {10, 10}];
Graphics[MapIndexed[If[#1 == 1, Disk, Circle][#2, 0.4] &, mat, {2}]]
I like to use various versions of MapIndexed for this. Instead of Disk or Circle you can use any other graphics object. Just make a function that will take a position as its argument and will produce that object.
If M is your matrix containing 0's and 1's and image0/image1 are the images you want to display:
image0 = Graphics[{Red, Disk[]}, ImageSize -> 10];
image1 = Graphics[{Blue, Rectangle[]}, ImageSize -> 10];
M = {{0, 1, 0}, {1, 1, 1}, {1, 0, 0}};
You can just do this:
GraphicsGrid[M /. {0 -> image0, 1 -> image1}]
or, if you want the 0's to be blank:
GraphicsGrid[M /. {0 -> "", 1 -> image1}]
I have written code which draws the Sierpinski fractal. It is really slow since it uses recursion. Do any of you know how I could write the same code without recursion in order for it to be quicker? Here is my code:
midpoint[p1_, p2_] := Mean[{p1, p2}]
trianglesurface[A_, B_, C_] := Graphics[Polygon[{A, B, C}]]
sierpinski[A_, B_, C_, 0] := trianglesurface[A, B, C]
sierpinski[A_, B_, C_, n_Integer] :=
Show[
sierpinski[A, midpoint[A, B], midpoint[C, A], n - 1],
sierpinski[B, midpoint[A, B], midpoint[B, C], n - 1],
sierpinski[C, midpoint[C, A], midpoint[C, B], n - 1]
]
edit:
I have written it with the Chaos Game approach in case someone is interested. Thank you for your great answers!
Here is the code:
random[A_, B_, C_] := Module[{a, result},
a = RandomInteger[2];
Which[a == 0, result = A,
a == 1, result = B,
a == 2, result = C]]
Chaos[A_List, B_List, C_List, S_List, n_Integer] :=
Module[{list},
list = NestList[Mean[{random[A, B, C], #}] &,
Mean[{random[A, B, C], S}], n];
ListPlot[list, Axes -> False, PlotStyle -> PointSize[0.001]]]
This uses Scale and Translate in combination with Nest to create the list of triangles.
Manipulate[
Graphics[{Nest[
Translate[Scale[#, 1/2, {0, 0}], pts/2] &, {Polygon[pts]}, depth]},
PlotRange -> {{0, 1}, {0, 1}}, PlotRangePadding -> .2],
{{pts, {{0, 0}, {1, 0}, {1/2, 1}}}, Locator},
{{depth, 4}, Range[7]}]
If you would like a high-quality approximation of the Sierpinski triangle, you can use an approach called the chaos game. The idea is as follows - pick three points that you wish to define as the vertices of the Sierpinski triangle and choose one of those points randomly. Then, repeat the following procedure as long as you'd like:
Choose a random vertex of the trangle.
Move from the current point to the halfway point between its current location and that vertex of the triangle.
Plot a pixel at that point.
As you can see at this animation, this procedure will eventually trace out a high-resolution version of the triangle. If you'd like, you can multithread it to have multiple processes plotting pixels at once, which will end up drawing the triangle more quickly.
Alternatively, if you just want to translate your recursive code into iterative code, one option would be to use a worklist approach. Maintain a stack (or queue) that contains a collection of records, each of which holds the vertices of the triangle and the number n. Initially put into this worklist the vertices of the main triangle and the fractal depth. Then:
While the worklist is not empty:
Remove the first element from the worklist.
If its n value is not zero:
Draw the triangle connecting the midpoints of the triangle.
For each subtriangle, add that triangle with n-value n - 1 to the worklist.
This essentially simulates the recursion iteratively.
Hope this helps!
You may try
l = {{{{0, 1}, {1, 0}, {0, 0}}, 8}};
g = {};
While [l != {},
k = l[[1, 1]];
n = l[[1, 2]];
l = Rest[l];
If[n != 0,
AppendTo[g, k];
(AppendTo[l, {{#1, Mean[{#1, #2}], Mean[{#1, #3}]}, n - 1}] & ## #) & /#
NestList[RotateLeft, k, 2]
]]
Show#Graphics[{EdgeForm[Thin], Pink,Polygon#g}]
And then replace the AppendTo by something more efficient. See for example https://mathematica.stackexchange.com/questions/845/internalbag-inside-compile
Edit
Faster:
f[1] = {{{0, 1}, {1, 0}, {0, 0}}, 8};
i = 1;
g = {};
While[i != 0,
k = f[i][[1]];
n = f[i][[2]];
i--;
If[n != 0,
g = Join[g, k];
{f[i + 1], f[i + 2], f[i + 3]} =
({{#1, Mean[{#1, #2}], Mean[{#1, #3}]}, n - 1} & ## #) & /#
NestList[RotateLeft, k, 2];
i = i + 3
]]
Show#Graphics[{EdgeForm[Thin], Pink, Polygon#g}]
Since the triangle-based functions have already been well covered, here is a raster based approach.
This iteratively constructs pascal's triangle, then takes modulo 2 and plots the result.
NestList[{0, ##} + {##, 0} & ## # &, {1}, 511] ~Mod~ 2 // ArrayPlot
Clear["`*"];
sierpinski[{a_, b_, c_}] :=
With[{ab = (a + b)/2, bc = (b + c)/2, ca = (a + c)/2},
{{a, ab, ca}, {ab, b, bc}, {ca, bc, c}}];
pts = {{0, 0}, {1, 0}, {1/2, Sqrt[3]/2}} // N;
n = 5;
d = Nest[Join ## sierpinski /# # &, {pts}, n]; // AbsoluteTiming
Graphics[{EdgeForm#Black, Polygon#d}]
(*sierpinski=Map[Mean, Tuples[#,2]~Partition~3 ,{2}]&;*)
Here is a 3D version,https://mathematica.stackexchange.com/questions/22256/how-can-i-compile-this-function
ListPlot#NestList[(# + RandomChoice[{{0, 0}, {2, 0}, {1, 2}}])/2 &,
N#{0, 0}, 10^4]
With[{data =
NestList[(# + RandomChoice#{{0, 0}, {1, 0}, {.5, .8}})/2 &,
N#{0, 0}, 10^4]},
Graphics[Point[data,
VertexColors -> ({1, #[[1]], #[[2]]} & /# Rescale#data)]]
]
With[{v = {{0, 0, 0.6}, {-0.3, -0.5, -0.2}, {-0.3, 0.5, -0.2}, {0.6,
0, -0.2}}},
ListPointPlot3D[
NestList[(# + RandomChoice[v])/2 &, N#{0, 0, 0}, 10^4],
BoxRatios -> 1, ColorFunction -> "Pastel"]
]
I am wondering if anyone can help me to plot the Cantor dust on the plane in Mathematica. This is linked to the Cantor set.
Thanks a lot.
EDIT
I actually wanted to have something like this:
Here's a naive and probably not very optimized way of reproducing the graphics for the ternary Cantor set construction:
cantorRule = Line[{{a_, n_}, {b_, n_}}] :>
With[{d = b - a, np = n - .1},
{Line[{{a, np}, {a + d/3, np}}], Line[{{b - d/3, np}, {b, np}}]}]
Graphics[{CapForm["Butt"], Thickness[.05],
Flatten#NestList[#/.cantorRule&, Line[{{0., 0}, {1., 0}}], 6]}]
To make Cantor dust using the same replacement rules, we take the result at a particular level, e.g. 4:
dust4=Flatten#Nest[#/.cantorRule&,Line[{{0.,0},{1.,0}}],4]/.Line[{{a_,_},{b_,_}}]:>{a,b}
and take tuples of it
dust4 = Transpose /# Tuples[dust4, 2];
Then we just plot the rectangles
Graphics[Rectangle ### dust4]
Edit: Cantor dust + squares
Changed specs -> New, but similar, solution (still not optimized).
Set n to be a positive integer and choice any subset of 1,...,n then
n = 3; choice = {1, 3};
CanDChoice = c:CanD[__]/;Length[c]===n :> CanD[c[[choice]]];
splitRange = {a_, b_} :> With[{d = (b - a + 0.)/n},
CanD##NestList[# + d &, {a, a + d}, n - 1]];
cantLevToRect[lev_]:=Rectangle###(Transpose/#Tuples[{lev}/.CanD->Sequence,2])
dust = NestList[# /. CanDChoice /. splitRange &, {0, 1}, 4] // Rest;
Graphics[{FaceForm[LightGray], EdgeForm[Black],
Table[cantLevToRect[lev], {lev, Most#dust}],
FaceForm[Black], cantLevToRect[Last#dust /. CanDChoice]}]
Here's the graphics for
n = 7; choice = {1, 2, 4, 6, 7};
dust = NestList[# /. CanDChoice /. splitRange &, {0, 1}, 2] // Rest;
and everything else the same:
Once can use the following approach. Define cantor function:
cantorF[r:(0|1)] = r;
cantorF[r_Rational /; 0 < r < 1] :=
Module[{digs, scale}, {digs, scale} = RealDigits[r, 3];
If[! FreeQ[digs, 1],
digs = Append[TakeWhile[Most[digs]~Join~Last[digs], # != 1 &], 1];];
FromDigits[{digs, scale}, 2]]
Then form the dust by computing differences of F[n/3^k]-F[(n+1/2)/3^k]:
With[{k = 4},
Outer[Times, #, #] &[
Table[(cantorF[(n + 1/2)/3^k] - cantorF[(n)/3^k]), {n, 0,
3^k - 1}]]] // ArrayPlot
I like recursive functions, so
cantor[size_, n_][pt_] :=
With[{s = size/3, ct = cantor[size/3, n - 1]},
{ct[pt], ct[pt + {2 s, 0}], ct[pt + {0, 2 s}], ct[pt + {2 s, 2 s}]}
]
cantor[size_, 0][pt_] := Rectangle[pt, pt + {size, size}]
drawCantor[n_] := Graphics[cantor[1, n][{0, 0}]]
drawCantor[5]
Explanation: size is the edge length of the square the set fits into. pt is the {x,y} coordinates of it lower left corner.
Does Mathematica support hidden line removal for wire frame images? If this isn't the case, has anybody here ever come across a way to do it? Lets start with this:
Plot3D[Sin[x+y^2], {x, -3, 3}, {y, -2, 2}, Boxed -> False]
To create a wire frame we can do:
Plot3D[Sin[x+y^2], {x, -3, 3}, {y, -2, 2}, Boxed -> False, PlotStyle -> None]
One thing we can do to achieve the effect is to color the all the surfaces white. This however, is undesirable. The reason is because if we export this hidden line wire frame model to pdf we will have all of those white polygons that Mathematica uses to render the image. I want to be able to obtain a wire frame with hidden line removal in pdf and/or eps format.
UPDATE:
I have posted a solution to this problem. The problem is that the code runs very slow. In its current state it is unable to generate the wireframe for the image in this question. Feel free to play with my code. I added a link to it at the end of my post. You can also find the code in this link
Here I present a solution. First I will show how to use the function that generates the wire frame, then I will proceed to explain in detail the rest of the functions that compose the algorithm.
wireFrame
wireFrame[g_] := Module[{figInfo, opt, pts},
{figInfo, opt} = G3ToG2Info[g];
pts = getHiddenLines[figInfo];
Graphics[Map[setPoints[#] &, getFrame[figInfo, pts]], opt]
]
The input of this function is a Graphics3D object preferably with no axes.
fig = ListPlot3D[
{{0, -1, 0}, {0, 1, 0}, {-1, 0, 1}, {1, 0, 1}, {-1, 1, 1}},
Mesh -> {10, 10},
Boxed -> False,
Axes -> False,
ViewPoint -> {2, -2, 1},
ViewVertical -> {0, 0, 1},
MeshStyle -> Directive[RGBColor[0, 0.5, 0, 0.5]],
BoundaryStyle -> Directive[RGBColor[1, 0.5, 0, 0.5]]
]
Now we apply the function wireFrame.
wireFrame[fig]
As you can see wireFrame obtained most of the lines and its colors. There is a green line that was not included in the wireframe. This is most likely due to my threshold settings.
Before I proceed to explain the details of the functions G3ToG2Info, getHiddenLines, getFrame and setPoints I will show you why wire frames with hidden line removal can be useful.
The image shown above is a screenshot of a pdf file generated by using the technique described in rasters in 3D graphics combined with the wire frame generated here. This can be advantageous in various ways. There is no need to keep the information for the triangles to show a colorful surface. Instead we show a raster image of the surface. All of the lines are very smooth, with the exception of the boundaries of the raster plot not covered by lines. We also have a reduction of file size. In this case the pdf file size reduced from 1.9mb to 78kb using the combination of the raster plot and the wire frame. It takes less time to display in the pdf viewer and the image quality is great.
Mathematica does a pretty good job at exporting 3D images to pdf files. When we import the pdf files we obtain a Graphics object composed of line segments and triangles. In some cases this objects overlap and thus we have hidden lines. To make a wire frame model with no surfaces we first need to remove this overlap and then remove the polygons. I will start by describing how to obtain the information from a Graphics3D image.
G3ToG2Info
getPoints[obj_] := Switch[Head[obj],
Polygon, obj[[1]],
JoinedCurve, obj[[2]][[1]],
RGBColor, {Table[obj[[i]], {i, 1, 3}]}
];
setPoints[obj_] := Switch[Length#obj,
3, Polygon[obj],
2, Line[obj],
1, RGBColor[obj[[1]]]
];
G3ToG2Info[g_] := Module[{obj, opt},
obj = ImportString[ExportString[g, "PDF", Background -> None], "PDF"][[1]];
opt = Options[obj];
obj = Flatten[First[obj /. Style[expr_, opts___] :> {opts, expr}], 2];
obj = Cases[obj, _Polygon | _JoinedCurve | _RGBColor, Infinity];
obj = Map[getPoints[#] &, obj];
{obj, opt}
]
This code is for Mathematica 8 in version 7 you would replace JoinedCurve in the function getPoints by Line. The function getPoints assumes that you are giving a primitive Graphics object. It will see what type of object it recieves and then extract the information it needs from it. If it is a polygon it gets a list of 3 points, for a line it obtains a list of 2 points and if it is a color then it gets a list of a single list containing 3 points. This has been done like this in order to maintain consistency with the lists.
The function setPoints does the reverse of getPoints. You input a list of points and it will determine if it should return a polygon, a line or a color.
To obtain a list of triangles, lines and colors we use G3ToG2Info. This function will use
ExportString and ImportString to obtain a Graphics object from the Graphics3D version. This info is store in obj. There is some clean up that we need to perform, first we get the options of the obj. This part is necessary because it may contain the PlotRange of the image. Then we obtain all the Polygon, JoinedCurve and RGBColor objects as described in obtaining graphics primitives and directives. Finally we apply the function getPoints on all of these objects to get a list of triangles, lines and colors. This part covers the line {figInfo, opt} = G3ToG2Info[g].
getHiddenLines
We want to be able to know what part of a line will not be displayed. To do this we need to know point of intersection between two line segments. The algorithm I'm using to find the intersection can be found here.
lineInt[L_, M_, EPS_: 10^-6] := Module[
{x21, y21, x43, y43, x13, y13, numL, numM, den},
{x21, y21} = L[[2]] - L[[1]];
{x43, y43} = M[[2]] - M[[1]];
{x13, y13} = L[[1]] - M[[1]];
den = y43*x21 - x43*y21;
If[den*den < EPS, Return[-Infinity]];
numL = (x43*y13 - y43*x13)/den;
numM = (x21*y13 - y21*x13)/den;
If[numM < 0 || numM > 1, Return[-Infinity], Return[numL]];
]
lineInt assumes that the line L and M do not coincide. It will return -Infinity if the lines are parallel or if the line containing the segment L does not cross the line segment M. If the line containing L intersects the line segment M then it returns a scalar. Suppose this scalar is u, then the point of intersection is L[[1]] + u (L[[2]]-L[[1]]). Notice that it is perfectly fine for u to be any real number. You can play with this manipulate function to test how lineInt works.
Manipulate[
Grid[{{
Graphics[{
Line[{p1, p2}, VertexColors -> {Red, Red}],
Line[{p3, p4}]
},
PlotRange -> 3, Axes -> True],
lineInt[{p1, p2}, {p3, p4}]
}}],
{{p1, {-1, 1}}, Locator, Appearance -> "L1"},
{{p2, {2, 1}}, Locator, Appearance -> "L2"},
{{p3, {1, -1}}, Locator, Appearance -> "M1"},
{{p4, {1, 2}}, Locator, Appearance -> "M2"}
]
Now that we know how to far we have to travel from L[[1]] to the line segment M we can find out what portion of a line segment lies within a triangle.
lineInTri[L_, T_] := Module[{res},
If[Length#DeleteDuplicates[Flatten[{T, L}, 1], SquaredEuclideanDistance[#1, #2] < 10^-6 &] == 3, Return[{}]];
res = Sort[Map[lineInt[L, #] &, {{T[[1]], T[[2]]}, {T[[2]], T[[3]]}, {T[[3]], T[[1]]} }]];
If[res[[3]] == Infinity || res == {-Infinity, -Infinity, -Infinity}, Return[{}]];
res = DeleteDuplicates[Cases[res, _Real | _Integer | _Rational], Chop[#1 - #2] == 0 &];
If[Length#res == 1, Return[{}]];
If[(Chop[res[[1]]] == 0 && res[[2]] > 1) || (Chop[res[[2]] - 1] == 0 && res[[1]] < 0), Return[{0, 1}]];
If[(Chop[res[[2]]] == 0 && res[[1]] < 0) || (Chop[res[[1]] - 1] == 0 && res[[2]] > 1), Return[{}]];
res = {Max[res[[1]], 0], Min[res[[2]], 1]};
If[res[[1]] > 1 || res[[1]] < 0 || res[[2]] > 1 || res[[2]] < 0, Return[{}], Return[res]];
]
This function returns the the portion of the line L that needs to be deleted. For instance, if it returns {.5, 1} this means that you will delete 50 percent of the line, starting from half the segment to the ending point of the segment. If L = {A, B} and the function returns {u, v} then this means that the line segment {A+(B-A)u, A+(B-A)v} is the section of the line that its contained in the triangle T.
When implementing lineInTri you need to be careful that the line L is not one of the edges of T, if this is the case then the line does not lie inside the triangle. This is where rounding erros can be bad. When Mathematica exports the image sometimes a line lies on the edge of the triangle but these coordinates differ by some amount. It is up to us to decide how close the line lies on the edge, otherwise the function will see that the line lies almost completely inside the triangle. This is the reason of the first line in the function. To see if a line lies on an edge of a triangle we can list all the points of the triangle and the line, and delete all the duplicates. You need to specify what a duplicate is in this case. In the end, if we end up with a list of 3 points this means that a line lies on an edge. The next part is a little complicated. What we do is check for the intersection of the line L with each edge of the triangle T and store this the results in a list. Next we sort the list and find out what section, if any, of the line lies in the triangle. Try to make sense out of it by playing with this, some of the tests include checking if an endpoint of the line is a vertex of the triangle, if the line is completely inside the triangle, partly inside or completely outside.
Manipulate[
Grid[{{
Graphics[{
RGBColor[0, .5, 0, .5], Polygon[{p3, p4, p5}],
Line[{p1, p2}, VertexColors -> {Red, Red}]
},
PlotRange -> 3, Axes -> True],
lineInTri[{p1, p2}, {p3, p4, p5}]
}}],
{{p1, {-1, -2}}, Locator, Appearance -> "L1"},
{{p2, {0, 0}}, Locator, Appearance -> "L2"},
{{p3, {-2, -2}}, Locator, Appearance -> "T1"},
{{p4, {2, -2}}, Locator, Appearance -> "T2"},
{{p5, {-1, 1}}, Locator, Appearance -> "T3"}
]
lineInTri will be used to see what portion of the line will not be drawn. This line will most likely be covered by many triangles. For this reason, we need to keep a list of all the portions of each line that will not be drawn. These lists will not have an order. All we know is that this lists are one dimensional segments. Each one consisting of numbers in the [0,1] interval. I'm not aware of a union function for one dimensional segments so here is my implementation.
union[obj_] := Module[{p, tmp, dummy, newp, EPS = 10^-3},
p = Sort[obj];
tmp = p[[1]];
If[tmp[[1]] < EPS, tmp[[1]] = 0];
{dummy, newp} = Reap[
Do[
If[(p[[i, 1]] - tmp[[2]]) > EPS && (tmp[[2]] - tmp[[1]]) > EPS,
Sow[tmp]; tmp = p[[i]],
tmp[[2]] = Max[p[[i, 2]], tmp[[2]]]
];
, {i, 2, Length#p}
];
If[1 - tmp[[2]] < EPS, tmp[[2]] = 1];
If[(tmp[[2]] - tmp[[1]]) > EPS, Sow[tmp]];
];
If[Length#newp == 0, {}, newp[[1]]]
]
This function would be shorter but here I have included some if statements to check if a number is close to zero or one. If one number is EPS apart from zero then we make this number zero, the same applies for one. Another aspect that I'm covering here is that if there is a relatively small portion of the segment to be displayed then it is most likely that it needs to be deleted. For instance if we have {{0,.5}, {.500000000001}} this means that we need to draw {{.5, .500000000001}}. But this segment is very small to be even noticed specially in a large line segment, for all we know those two numbers are the same. All of this things need to be taken into account when implementing union.
Now we are ready to see what needs to be deleted from a line segment. The next requires the list of objects generated from G3ToG2Info, an object from this list and an index.
getSections[L_, obj_, start_ ] := Module[{dummy, p, seg},
{dummy, p} = Reap[
Do[
If[Length#obj[[i]] == 3,
seg = lineInTri[L, obj[[i]]];
If[Length#seg != 0, Sow[seg]];
]
, {i, start, Length#obj}
]
];
If[Length#p == 0, Return[{}], Return[union[First#p]]];
]
getSections returns a list containing the portions that need to be deleted from L. We know that obj is the list of triangles, lines and colors, we know that objects in the list with a higher index will be drawn on top of ones with lower index. For this reason we need the index start. This is the index we will start looking for triangles in obj. Once we find a triangle we will obtain the portion of the segment that lies in the triangle using the function lineInTri. At the end we will end up with a list of sections which we can combine by using union.
Finally, we get to getHiddenLines. All this requires is to look at each object in the list returned by G3ToG2Info and apply the function getSections. getHiddenLines will return a list of lists. Each element is a list of sections that need to be deleted.
getHiddenLines[obj_] := Module[{pts},
pts = Table[{}, {Length#obj}];
Do[
If[Length#obj[[j]] == 2,
pts[[j]] = getSections[obj[[j]], obj, j + 1]
];
, {j, Length#obj}
];
Return[pts];
]
getFrame
If you have manage to understand the concepts up to here I'm sure you know what will be done next. If we have the list of triangles, lines and colors and the sections of the lines that need to be deleted we need to draw only the colors and the sections of the lines that are visible. First we make a complement function, this will tell us exactly what to draw.
complement[obj_] := Module[{dummy, p},
{dummy, p} = Reap[
If[obj[[1, 1]] != 0, Sow[{0, obj[[1, 1]]}]];
Do[
Sow[{obj[[i - 1, 2]], obj[[i, 1]]}]
, {i, 2, Length#obj}
];
If[obj[[-1, 2]] != 1, Sow[{obj[[-1, 2]], 1}]];
];
If[Length#p == 0, {}, Flatten# First#p]
]
Now the getFrame function
getFrame[obj_, pts_] := Module[{dummy, lines, L, u, d},
{dummy, lines} = Reap[
Do[
L = obj[[i]];
If[Length#L == 2,
If[Length#pts[[i]] == 0, Sow[L]; Continue[]];
u = complement[pts[[i]]];
If[Length#u > 0,
Do[
d = L[[2]] - L[[1]];
Sow[{L[[1]] + u[[j - 1]] d, L[[1]] + u[[j]] d}]
, {j, 2, Length#u, 2 }]
];
];
If[Length#L == 1, Sow[L]];
, {i, Length#obj}]
];
First#lines
]
Final words
I'm somewhat happy with the results of the algorithm. What I do not like is the execution speed. I have written this as I would in C/C++/java using loops. I tried my best to use Reap and Sow to create growing lists instead of using the function Append. Regardless of all of this I still had to use loops. It should be noted that the wire frame picture posted here took 63 seconds to generate. I tried doing a wire frame for the picture in the question but this 3D object contains about 32000 objects. It was taking about 13 seconds to compute the portions that need to be displayed for a line. If we assume that we have 32000 lines and it takes 13 seconds to do all the computations that will be about 116 hours of computational time.
I'm sure this time can be reduced if we use the function Compile on all of the routines and maybe finding a way not to use the Do loops. Can I get some help here Stack Overflow?
For your convinience I have uploaded the code to the web. You can find it here. If you can apply a modified version of this code to the plot in the question and show the wire frame I will mark your solution as the answer to this post.
Best,
J Manuel Lopez
This isn't right, but somewhat interesting:
Plot3D[Sin[x + y^2], {x, -3, 3}, {y, -2, 2}, Boxed -> False,
PlotStyle -> {EdgeForm[None], FaceForm[Red, None]}, Mesh -> False]
With a FaceForm of None, the polygon isn't rendered. I'm not sure there's a way to do this with the Mesh lines.
I am looking to plot something like the whispering gallery modes -- a 2D cylindrically symmetric plot in polar coordinates. Something like this:
I found the following code snippet in Trott's symbolics guidebook. Tried running it on a very small data set; it ate 4 GB of memory and hosed my kernel:
(* add points to get smooth curves *)
addPoints[lp_][points_, \[Delta]\[CurlyEpsilon]_] :=
Module[{n, l}, Join ## (Function[pair,
If[(* additional points needed? *)
(l = Sqrt[#. #]&[Subtract ## pair]) < \[Delta]\[CurlyEpsilon], pair,
n = Floor[l/\[Delta]\[CurlyEpsilon]] + 1;
Table[# + i/n (#2 - #1), {i, 0, n - 1}]& ## pair]] /#
Partition[If[lp === Polygon,
Append[#, First[#]], #]&[points], 2, 1])]
(* Make the plot circular *)
With[{\[Delta]\[CurlyEpsilon] = 0.1, R = 10},
Show[{gr /. (lp : (Polygon | Line))[l_] :>
lp[{#2 Cos[#1], #2 Sin[#1]} & ###(* add points *)
addPoints[lp][l, \[Delta]\[CurlyEpsilon]]],
Graphics[{Thickness[0.01], GrayLevel[0], Circle[{0, 0}, R]}]},
DisplayFunction -> $DisplayFunction, Frame -> False]]
Here, gr is a rectangular 2D ListContourPlot, generated using something like this (for example):
data = With[{eth = 2, er = 2, wc = 1, m = 4},
Table[Re[
BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[
I m phi]], {r, 0, 10, .2}, {phi, 0, 2 Pi, 0.1}]];
gr = ListContourPlot[data, Contours -> 50, ContourLines -> False,
DataRange -> {{0, 2 Pi}, {0, 10}}, DisplayFunction -> Identity,
ContourStyle -> {Thickness[0.002]}, PlotRange -> All,
ColorFunctionScaling -> False]
Is there a straightforward way to do cylindrical plots like this?.. I find it hard to believe that I would have to turn to Matlab for my curvilinear coordinate needs :)
Previous snippets deleted, since this is clearly the best answer I came up with:
With[{eth = 2, er = 2, wc = 1, m = 4},
ContourPlot[
Re[BesselJ[(Sqrt[eth] m)/Sqrt[er], Sqrt[eth] r wc] Exp[I phi m]]/.
{r ->Norm[{x, y}], phi ->ArcTan[x, y]},
{x, -10, 10}, {y, -10, 10},
Contours -> 50, ContourLines -> False,
RegionFunction -> (#1^2 + #2^2 < 100 &),
ColorFunction -> "SunsetColors"
]
]
Edit
Replacing ContourPlot by Plot3D and removing the unsupported options you get:
This is a relatively straightforward problem. The key is that if you can parametrize it, you can plot it. According to the documentation both ListContourPlot and ListDensityPlot accept data in two forms: an array of height values or a list of coordinates plus function value ({{x, y, f} ..}). The second form is easier to deal with, such that even if your data is in the first form, we'll transform it into the second form.
Simply, to transform data of the form {{r, t, f} ..} into data of the form {{x, y, f} ..} you doN[{#[[1]] Cos[ #[[2]] ], #[[1]] Sin[ #[[2]] ], #[[3]]}]& /# data, when applied to data taken from BesselJ[1, r/2] Cos[3 t] you get
What about when you just have an array of data, like this guy? In that case, you have a 2D array where each point in the array has known location, and in order to plot it, you have to turn it into the second form. I'm partial to MapIndexed, but there are other ways of doing it. Let's say your data is stored in an array where the rows correspond to the radial coordinate and the columns are the angular coordinate. Then to transform it, I'd use
R = 0.01; (*radial increment*)
T = 0.05 Pi; (*angular increment*)
xformed = MapIndexed[
With[{r = #2[[1]]*R, t = #2[[1]]*t, f = #1},
{r Cos[t], r Sin[t], f}]&, data, {2}]//Flatten[#,1]&
which gives the same result.
If you have an analytic solution, then you need to transform it to Cartesian coordinates, like above, but you use replacement rules, instead. For instance,
ContourPlot[ Evaluate[
BesselJ[1, r/2]*Cos[3 t ] /. {r -> Sqrt[x^2 + y^2], t -> ArcTan[x, y]}],
{x, -5, 5}, {y, -5, 5}, PlotPoints -> 50,
ColorFunction -> ColorData["DarkRainbow"], Contours -> 25]
gives
Two things to note: 1) Evaluate is needed to ensure that the replacement is performed correctly, and 2) ArcTan[x, y] takes into account the quadrant that the point {x,y} is found in.