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In the code below, If I chenage (4/([Zeta]*[Omega])) to, say 20, nothing is plotted. Why?
If I remove the two sliders at the beginning, nothing is plotted
ClearAll[\[Zeta], \[Omega]]
{Slider[ Dynamic[\[Zeta]], {0.1, 1.4, 0.1}], Dynamic[\[Zeta]]}
{Slider[ Dynamic[\[Omega]], {1, 5, 0.1}], Dynamic[\[Omega]]}
tf[\[Omega]_, \[Zeta]_] :=
TransferFunctionModel[\[Omega]^2/(s^2 +
2 \[Zeta] \[Omega] s + \[Omega]^2), s]
f[t_] = OutputResponse[tf[\[Omega], \[Zeta]], UnitStep[t], t];
Manipulate[
Plot[f[t], {t, 0, (4/(\[Zeta]*\[Omega]))},
PlotRange -> {{0, (4/(\[Zeta]*\[Omega]))}, {0, 2}}], {{\[Zeta],
0.2}, 0.1, 1.4}, {{\[Omega], 1}, 0.5, 4}
]
tf[o_, z_] := TransferFunctionModel[o^2/(s^2 + 2 z o s + o^2), s]
f[t_, o_, z_] = OutputResponse[tf[o, z], UnitStep[t], t];
Manipulate[Plot[f[t, o, z], {t, 0, 20}, PlotRange -> {0, 2}],
{{z, 0.2}, 0.1, 1.4}, {{o, 1}, 0.5, 4}]
I want to generate a plot like the following
I am not sure how to generate a shading even though I can get the frame done. I'd like to know the general approach to shade certain areas in a plot in Mathematica. Please help. Thank you.
Perhaps you are looking for RegionPlot?
RegionPlot[(-1 + x)^2 + (-1 + y)^2 < 1 &&
x^2 + (-1 + y)^2 < 1 && (-1 + x)^2 + y^2 < 1 && x^2 + y^2 < 1,
{x, 0, 1}, {y, 0, 1}]
Note the use of op_ in the following (only one set of equations for the curves and the intersection!):
t[op_] :=Reduce[op[(x - #[[1]])^2 + (y - #[[2]])^2, 1], y] & /# Tuples[{0, 1}, 2]
tx = Texture[Binarize#RandomImage[NormalDistribution[1, .005], 1000 {1, 1}]];
Show[{
Plot[y /. ToRules /# #, {x, 0, 1}, PlotRange -> {{0, 1}, {0, 1}}] &# t[Equal],
RegionPlot[And ## #, {x, 0, 1}, {y, 0, 1}, PlotStyle -> tx] &# t[Less]},
Frame->True,AspectRatio->1,FrameStyle->Directive[Blue, Thick],FrameTicks->None]
If, for any particular reason, you want the dotted effect in your picture, you can achieve this like so:
pts = RandomReal[{0, 1}, {10000, 2}];
pts = Select[pts,
And ## Table[Norm[# - p] < 1, {p,
{{0, 0}, {1, 0}, {1, 1}, {0, 1}}}] &];
Graphics[{Thick,
Line[{{0, 0}, {1, 0}, {1, 1}, {0, 1}, {0, 0}}],
Circle[{0, 0}, 1, {0, Pi/2}],
Circle[{1, 0}, 1, {Pi/2, Pi}],
Circle[{1, 1}, 1, {Pi, 3 Pi/2}],
Circle[{0, 1}, 1, {3 Pi/2, 2 Pi}],
PointSize[Small], Point[pts]
}]
I'm trying to combine 3 functions graphed on a Plot[] and 1 function graphed on a ParametricPlot[]. My equations are as follows:
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1}, PlotLegend -> {"-2 x", "-2 \!\(\*SqrtBox[\(x\)]\)", "-2 \!\(\*SuperscriptBox[\(x\), \(3/5\)]\)"}]
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u,0, 1.40138}, PlotLegend -> {"Problem 3"}]
Show[plota, plotb]
This is the image it gives:
As yoda said, PlotLegends is terrible. However, if you don't mind setting the plot styles manually and repeating them lateron, ShowLegend can help.
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1},
PlotStyle -> {{Red}, {Blue}, {Orange}}];
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u, 0, 1.40138},
PlotStyle -> {{Black}}];
And now
ShowLegend[Show[plota, plotb],
{{{Graphics[{Red, Line[{{0, 0}, {1, 0}}]}], Label1},
{Graphics[{Blue, Line[{{0, 0}, {1, 0}}]}], Label2},
{Graphics[{Orange, Line[{{0, 0}, {1, 0}}]}], Label3},
{Graphics[{Black, Line[{{0, 0}, {1, 0}}]}], Label4}},
LegendSize -> {0.5, 0.5}, LegendPosition -> {0.5, -0.2}}]
which will give you this:
You can also write some simple functions to make this a little less cumbersome, if you deal with this problem often.
Well, the root cause of the error is the PlotLegends package, which is a terrible, buggy package. Removing that, Show combines them correctly:
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1}]
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u,
0, 1.40138}]
Show[plota, plotb]
You can see Simon's solution here for ideas to label your different curves without using PlotLegends. This answer by James also demonstrates why PlotLegends has the reputation it has...
You can still salvage something with the PlotLegends package. Here's an example using ShowLegends that you can modify to your tastes
colors = {Red, Green, Blue, Pink};
legends = {-2 x, -2 Sqrt[x], -2 x^(3/5), "Problem 3"};
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1},
PlotStyle -> colors[[1 ;; 3]]];
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u,
0, 1.40138}, PlotStyle -> colors[[4]]];
ShowLegend[
Show[plota,
plotb], {Table[{Graphics[{colors[[i]], Thick,
Line[{{0, 0}, {1, 0}}]}], legends[[i]]}, {i, 4}],
LegendPosition -> {0.4, -0.15}, LegendSpacing -> 0,
LegendShadow -> None, LegendSize -> 0.6}]
As the other answers pointed out, the culprit is PlotLegend. So, sometimes is useful to be able to roll your own plot legends:
plotStyle = {Red, Green, Blue};
labls = {"a", "b", "Let's go"};
f[i_, s_] := {Graphics[{plotStyle[[i]], Line[{{0, 0}, {1, 0}}]},
ImageSize -> {15, 10}], Style[labls[[i]], s]};
Plot[{Sin[x], Sin[2 x], Sin[3 x]}, {x, 0, 2 Pi},
PlotStyle -> plotStyle,
Epilog ->
Inset[Framed[Style#Column[{Grid[Table[f[i, 15], {i, 1, 3}]]}]],
Offset[{-2, -2}, Scaled[{1, 1}]], {Right, Top}],
PlotRangePadding -> 1
]
Given a set of points in the plane T={a1,a2,...,an} then Graphics[Polygon[T]] will plot the polygon generated by the points. How can I add labels to the polygon's vertices? Have merely the index as a label would be better then nothing. Any ideas?
pts = {{1, 0}, {0, Sqrt[3]}, {-1, 0}};
Graphics[
{{LightGray, Polygon[pts]},
{pts /. {x_, y_} :> Text[Style[{x, y}, Red], {x, y}]}}
]
To add point also
pts = {{1, 0}, {0, Sqrt[3]}, {-1, 0}};
Graphics[
{{LightGray, Polygon[pts]},
{pts /. {x_, y_} :> Text[Style[{x, y}, Red], {x, y}, {0, -1}]},
{pts /. {x_, y_} :> {Blue, PointSize[0.02], Point[{x, y}]}}
}
]
update:
Use the index:
pts = {{1, 0}, {0, Sqrt[3]}, {-1, 0}};
Graphics[
{{LightGray, Polygon[pts]},
{pts /. {x_, y_} :>
Text[Style[Position[pts, {x, y}], Red], {x, y}, {0, -1}]}
}
]
Nasser's version (update) uses pattern matching. This one uses functional programming. MapIndexed gives you both the coordinates and their index without the need for Position to find it.
pts = {{1, 0}, {0, Sqrt[3]}, {-1, 0}};
Graphics[
{
{LightGray, Polygon[pts]},
MapIndexed[Text[Style[#2[[1]], Red], #1, {0, -1}] &, pts]
}
]
or, if you don't like MapIndexed, here's a version with Apply (at level 1, infix notation ###).
pts = {{1, 0}, {0, Sqrt[3]}, {-1, 0}};
idx = Range[Length[pts]];
Graphics[
{
{LightGray, Polygon[pts]},
Text[Style[#2, Red], #1, {0, -1}] & ### ({pts, idx}\[Transpose])
}
]
This can be expanded to arbitrary labels as follows:
pts = {{1, 0}, {0, Sqrt[3]}, {-1, 0}};
idx = {"One", "Two", "Three"};
Graphics[
{
{LightGray, Polygon[pts]},
Text[Style[#2, Red], #1, {0, -1}] & ### ({pts, idx}\[Transpose])
}
]
You can leverage the options of GraphPlot for this. Example:
c = RandomReal[1, {3, 2}]
g = GraphPlot[c, VertexLabeling -> True, VertexCoordinateRules -> c];
Graphics[{Polygon#c, g[[1]]}]
This way you can also make use of VertexLabeling -> Tooltip, or VertexRenderingFunction if you want to. If you do not want the edges overlaid, you may add EdgeRenderingFunction -> None to the GraphPlot function. Example:
c = RandomReal[1, {3, 2}]
g = GraphPlot[c, VertexLabeling -> All, VertexCoordinateRules -> c,
EdgeRenderingFunction -> None,
VertexRenderingFunction -> ({White, EdgeForm[Black], Disk[#, .02],
Black, Text[#2, #1]} &)];
Graphics[{Brown, Polygon#c, g[[1]]}]
Say I have three lists: a={1,5,10,15} b={2,4,6,8} and c={1,1,0,1,0}. I want a plot which has a as the x axis, b as the y axis and a red/black dot to mark 1/0. For. e.g. The coordinate (5,4) will have a red dot.
In other words the coordinate (a[i],b[i]) will have a red/black dot depending on whether c[i] is 1 or 0.
I have been trying my hand with ListPlot but can't figure out the options.
I suggest this.
a = {1, 5, 10, 15};
b = {2, 4, 6, 8};
c = {1, 1, 0, 1};
Graphics[
{#, Point#{##2}} & ###
Thread#{c /. {1 -> Red, 0 -> Black}, a, b},
Axes -> True, AxesOrigin -> 0
]
Or shorter but more obfuscated
Graphics[
{Hue[1, 1, #], Point#{##2}} & ### Thread#{c, a, b},
Axes -> True, AxesOrigin -> 0
]
Leonid's idea, perhaps more naive.
f[a_, b_, c_] :=
ListPlot[Pick[Transpose[{a, b}], c, #] & /# {0, 1},
PlotStyle -> {PointSize[Large], {Blue, Red}}]
f[a, b, c]
Edit: Just for fun
f[h_, a_, b_, c_, opt___] :=
h[Pick[Transpose[{a, b}], c, #] & /# {0, 1},
PlotStyle -> {PointSize[Large], {Blue, Red}}, opt]
f[ ListPlot,
Sort#RandomReal[1, 100],
Sin[(2 \[Pi] #)/100] + RandomReal[#/100] & /# Range[100],
RandomInteger[1, 100],
Joined -> True,
InterpolationOrder -> 2,
Filling -> Axis]
Here are your points:
a = {1, 5, 10, 15};
b = {2, 4, 6, 8};
c = {1, 1, 0, 1};
(I deleted the last element from c to make it the same length as a and b). What I'd suggest is to separately make images for points with zeros and ones and then combine them - this seems easiest in this situation:
showPoints[a_, b_, c_] :=
With[{coords = Transpose[{a, b}]},
With[{plotF = ListPlot[Pick[coords, c, #], PlotMarkers -> Automatic, PlotStyle -> #2] &},
Show[MapThread[plotF, {{0, 1}, {Black, Red}}]]]]
Here is the usage:
showPoints[a, b, c]
One possibility:
ListPlot[List /# Transpose[{a, b}],
PlotMarkers -> {1, 1, 0, 1} /. {1 -> { Style[\[FilledCircle], Red], 10},
0 -> { { Style[\[FilledCircle], Black], 10}}},
AxesOrigin -> {0, 0}]
Giving as output:
You could obtain similar results (to those of Leonid) using Graphics:
Graphics[{PointSize[.02], Transpose[{(c/. {1 -> Red, 0 -> Black}),
Point /# Transpose[{a, b}]}]},
Axes -> True, AxesOrigin -> {0, 0}]