4
Try this
vec[t_, x_, y_, z_] := {x t, y t, z t}
x0[n_] := RandomReal[{-10, 10}]
y0[n_] := RandomReal[{-10, 10}]
z0[n_] := RandomReal[{-10, 10}]
vectors[t_] := Table[vec[t, x0[n], y0[n], z0[n]], {n, 1, 10}];
styles = Table[{Hue[0.1*n], Thickness[0.005]}, {n, 1, 10}];
randVectors =
ParametricPlot3D[Evaluate@vectors[t], {t, -30, 30},
PlotStyle ->...
4
Ok, I've found the solution: Just use BaseStyle -> Thick, instead of PlotStyle -> Thick.
3
You can achieve this using the option ColorFunctionof ListPlot3D.
We first create test some data, then define the color function and finally make the plot:
dat = Flatten[
Table[{x, y, Sin[x y], Sin[x ]^2 Sin[y ]^2}, {x, 0, Pi, Pi/20}, {y,
0, Pi, Pi/20}], 1];
col = Interpolation[dat[[All, {1, 2, 4}]]];
colfunction := Function[{x, y, z}, RGBColor[col[...
2
New Edit
Maybe use With
Manipulate[With[{data = RandomReal[{0, 1}, {2, 2}]},
Dynamic@Show[ArrayPlot[data^p], Graphics[Circle[]]]], {p, 0.1, 10},
SynchronousUpdating -> False]
Edit
Manipulate[data = RandomReal[{0, 1}, {2, 2}];
Dynamic@Show[ArrayPlot[data^p], Graphics[Circle[]]], {p, 0.1, 10},
SynchronousUpdating -> False]
Original
Manipulate[...
1
To simplify the generation of vectors
$Version
(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)
Clear["Global`*"]
vec[t_, x_, y_, z_] := {x t, y t, z t}
SeedRandom[1234];
xyz = RandomReal[{-10, 10}, {10, 3}];
randVectors = ParametricPlot3D[
Evaluate[vec[t, ##] & @@@ xyz], {t, -30, 30},
PlotStyle -> Thick];
...
1
kglr's solution is what I would have done in old Mathematica. Nowadays, I would use Riffle[]:
With[{n = 5},
Graphics[Riffle[Flatten[{Circle[{0, 0}, 2 #1 - 1], Circle[{0, 0}, 2 #1]} & /@
Range[n]], {Blue, Red}, {1, -2, 2}],
Axes -> True]]
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