Numerical contour plot of compiled function
$begingroup$
I have come across a problem in Mathematica when I have tried to plot the contours of a compiled function.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this works perfectly but suppose I want to only plot a single contour:
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this then returns the error
but still plots the contour correctly. Evaluating f[x, y] == 0.1 gives x^2 + y^2 == 0.1 along with the same error message as the output so it's clear that the problem is that in order to plot the contour specified Mathematica requires the equation for the contour to be in symbolic form suggesting that ContourPlot is using a symbolic method to solve the problem.
However this is no good for anything difficult. I have a function longFunc[x,y] which is an extremely complicated expression and hence must be compiled and evaluated numerically. When I do a simple ContourPlot it works perfectly but if I specify a contour longFunc[x,y]==0.5 it falls back to a symbolic method and doesn't return an answer for a very long time.
So the question is: Is there an option within ContourPlot to force Mathematica to search for a solution to the equality using a numerical method?
Thank you!
Related:
ContourPlot is slow and unwieldy and generates a large-data graphic
Plot compiled function with LogLinearPlot
plotting compile
asked 3 hours ago
TakodaTakoda
1055
$endgroup$
add a comment |
$begingroup$
I have come across a problem in Mathematica when I have tried to plot the contours of a compiled function.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this works perfectly but suppose I want to only plot a single contour:
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this then returns the error
but still plots the contour correctly. Evaluating f[x, y] == 0.1 gives x^2 + y^2 == 0.1 along with the same error message as the output so it's clear that the problem is that in order to plot the contour specified Mathematica requires the equation for the contour to be in symbolic form suggesting that ContourPlot is using a symbolic method to solve the problem.
However this is no good for anything difficult. I have a function longFunc[x,y] which is an extremely complicated expression and hence must be compiled and evaluated numerically. When I do a simple ContourPlot it works perfectly but if I specify a contour longFunc[x,y]==0.5 it falls back to a symbolic method and doesn't return an answer for a very long time.
So the question is: Is there an option within ContourPlot to force Mathematica to search for a solution to the equality using a numerical method?
Thank you!
Related:
ContourPlot is slow and unwieldy and generates a large-data graphic
Plot compiled function with LogLinearPlot
plotting compile
asked 3 hours ago
TakodaTakoda
1055
$endgroup$
add a comment |
$begingroup$
I have come across a problem in Mathematica when I have tried to plot the contours of a compiled function.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this works perfectly but suppose I want to only plot a single contour:
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this then returns the error
but still plots the contour correctly. Evaluating f[x, y] == 0.1 gives x^2 + y^2 == 0.1 along with the same error message as the output so it's clear that the problem is that in order to plot the contour specified Mathematica requires the equation for the contour to be in symbolic form suggesting that ContourPlot is using a symbolic method to solve the problem.
However this is no good for anything difficult. I have a function longFunc[x,y] which is an extremely complicated expression and hence must be compiled and evaluated numerically. When I do a simple ContourPlot it works perfectly but if I specify a contour longFunc[x,y]==0.5 it falls back to a symbolic method and doesn't return an answer for a very long time.
So the question is: Is there an option within ContourPlot to force Mathematica to search for a solution to the equality using a numerical method?
Thank you!
Related:
ContourPlot is slow and unwieldy and generates a large-data graphic
Plot compiled function with LogLinearPlot
plotting compile
asked 3 hours ago
TakodaTakoda
1055
$endgroup$
I have come across a problem in Mathematica when I have tried to plot the contours of a compiled function.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this works perfectly but suppose I want to only plot a single contour:
ContourPlot[f[x, y] == 0.1, {x, y} [Element] Disk[{0, 0}, 1]]
this then returns the error
but still plots the contour correctly. Evaluating f[x, y] == 0.1 gives x^2 + y^2 == 0.1 along with the same error message as the output so it's clear that the problem is that in order to plot the contour specified Mathematica requires the equation for the contour to be in symbolic form suggesting that ContourPlot is using a symbolic method to solve the problem.
However this is no good for anything difficult. I have a function longFunc[x,y] which is an extremely complicated expression and hence must be compiled and evaluated numerically. When I do a simple ContourPlot it works perfectly but if I specify a contour longFunc[x,y]==0.5 it falls back to a symbolic method and doesn't return an answer for a very long time.
So the question is: Is there an option within ContourPlot to force Mathematica to search for a solution to the equality using a numerical method?
Thank you!
Related:
ContourPlot is slow and unwieldy and generates a large-data graphic
Plot compiled function with LogLinearPlot
plotting compile
plotting compile
asked 3 hours ago
TakodaTakoda
1055
asked 3 hours ago
TakodaTakoda
1055
asked 3 hours ago
TakodaTakoda
1055
asked 3 hours ago
TakodaTakoda
1055
asked 3 hours ago
TakodaTakoda
1055
1055
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Ok I figured it out. You specify the contours you want through the Contours option of the function, surprisingly.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y], {x, y} [Element] Disk[{0, 0}, 1],
Contours -> {0.1}, ContourShading -> False]
Works beautifully. If you specify an equality in the argument ContourPlot defaults to symbolic solving which is usually way too slow.
answered 2 hours ago
TakodaTakoda
1055
$endgroup$
add a comment |
$begingroup$
Try
f = Compile[{{x, _Real}, {y, _Real}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element]Disk[{0, 0}, 1]]
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
$endgroup$
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
add a comment |
$begingroup$
Here are a couple more ideas for displaying the contour at $z=1/10$
You could also define a helper function g that forces Plot3D to work with numeric values. Note: this will give a symbolic rather than a numeric tooltip for the contour.
g[x_?NumericQ, y_?NumericQ] := f[x, y];
ContourPlot[g[x, y] == 1/10, {x, -1, 1}, {y, -1, 1}, ContourShading -> False]
You could also show the contour on the 3D surface defined by f:
Plot3D[f[x, y], {x, y} ∈ Disk,
MeshFunctions -> {#3 &},
Mesh -> {{{1/10, Directive[Black, AbsoluteThickness[3.5]]}}},
BoxRatios -> {1, 1, 1/3},
ColorFunction -> (White &),
Lighting -> "Neutral"]
answered 19 mins ago
m_goldbergm_goldberg
85k872196
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Ok I figured it out. You specify the contours you want through the Contours option of the function, surprisingly.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y], {x, y} [Element] Disk[{0, 0}, 1],
Contours -> {0.1}, ContourShading -> False]
Works beautifully. If you specify an equality in the argument ContourPlot defaults to symbolic solving which is usually way too slow.
answered 2 hours ago
TakodaTakoda
1055
$endgroup$
add a comment |
$begingroup$
Ok I figured it out. You specify the contours you want through the Contours option of the function, surprisingly.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y], {x, y} [Element] Disk[{0, 0}, 1],
Contours -> {0.1}, ContourShading -> False]
Works beautifully. If you specify an equality in the argument ContourPlot defaults to symbolic solving which is usually way too slow.
answered 2 hours ago
TakodaTakoda
1055
$endgroup$
add a comment |
$begingroup$
Ok I figured it out. You specify the contours you want through the Contours option of the function, surprisingly.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y], {x, y} [Element] Disk[{0, 0}, 1],
Contours -> {0.1}, ContourShading -> False]
Works beautifully. If you specify an equality in the argument ContourPlot defaults to symbolic solving which is usually way too slow.
answered 2 hours ago
TakodaTakoda
1055
$endgroup$
Ok I figured it out. You specify the contours you want through the Contours option of the function, surprisingly.
f = Compile[{{x}, {y}}, x^2 + y^2];
ContourPlot[f[x, y], {x, y} [Element] Disk[{0, 0}, 1],
Contours -> {0.1}, ContourShading -> False]
Works beautifully. If you specify an equality in the argument ContourPlot defaults to symbolic solving which is usually way too slow.
answered 2 hours ago
TakodaTakoda
1055
answered 2 hours ago
TakodaTakoda
1055
answered 2 hours ago
TakodaTakoda
1055
answered 2 hours ago
TakodaTakoda
1055
1055
add a comment |
add a comment |
$begingroup$
Try
f = Compile[{{x, _Real}, {y, _Real}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element]Disk[{0, 0}, 1]]
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
$endgroup$
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
add a comment |
$begingroup$
Try
f = Compile[{{x, _Real}, {y, _Real}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element]Disk[{0, 0}, 1]]
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
$endgroup$
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
add a comment |
$begingroup$
Try
f = Compile[{{x, _Real}, {y, _Real}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element]Disk[{0, 0}, 1]]
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
$endgroup$
Try
f = Compile[{{x, _Real}, {y, _Real}}, x^2 + y^2];
ContourPlot[f[x, y] == 0.1, {x, y} [Element]Disk[{0, 0}, 1]]
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
edited 2 hours ago
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
answered 3 hours ago
Ulrich NeumannUlrich Neumann
8,141516
8,141516
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
add a comment |
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Doesn't change anything. The problem is that when you specify a contour to plot, Mathematica reverts back to trying to symbolically solve the equality.
$endgroup$
– Takoda
3 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Contour is evaluated without error message, so obviously somthing changed.
$endgroup$
– Ulrich Neumann
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
$begingroup$
Not for me. I am using Mathematica 11.2 so that might be why.
$endgroup$
– Takoda
2 hours ago
add a comment |
$begingroup$
Here are a couple more ideas for displaying the contour at $z=1/10$
You could also define a helper function g that forces Plot3D to work with numeric values. Note: this will give a symbolic rather than a numeric tooltip for the contour.
g[x_?NumericQ, y_?NumericQ] := f[x, y];
ContourPlot[g[x, y] == 1/10, {x, -1, 1}, {y, -1, 1}, ContourShading -> False]
You could also show the contour on the 3D surface defined by f:
Plot3D[f[x, y], {x, y} ∈ Disk,
MeshFunctions -> {#3 &},
Mesh -> {{{1/10, Directive[Black, AbsoluteThickness[3.5]]}}},
BoxRatios -> {1, 1, 1/3},
ColorFunction -> (White &),
Lighting -> "Neutral"]
answered 19 mins ago
m_goldbergm_goldberg
85k872196
$endgroup$
add a comment |
$begingroup$
Here are a couple more ideas for displaying the contour at $z=1/10$
You could also define a helper function g that forces Plot3D to work with numeric values. Note: this will give a symbolic rather than a numeric tooltip for the contour.
g[x_?NumericQ, y_?NumericQ] := f[x, y];
ContourPlot[g[x, y] == 1/10, {x, -1, 1}, {y, -1, 1}, ContourShading -> False]
You could also show the contour on the 3D surface defined by f:
Plot3D[f[x, y], {x, y} ∈ Disk,
MeshFunctions -> {#3 &},
Mesh -> {{{1/10, Directive[Black, AbsoluteThickness[3.5]]}}},
BoxRatios -> {1, 1, 1/3},
ColorFunction -> (White &),
Lighting -> "Neutral"]
answered 19 mins ago
m_goldbergm_goldberg
85k872196
$endgroup$
add a comment |
$begingroup$
Here are a couple more ideas for displaying the contour at $z=1/10$
You could also define a helper function g that forces Plot3D to work with numeric values. Note: this will give a symbolic rather than a numeric tooltip for the contour.
g[x_?NumericQ, y_?NumericQ] := f[x, y];
ContourPlot[g[x, y] == 1/10, {x, -1, 1}, {y, -1, 1}, ContourShading -> False]
You could also show the contour on the 3D surface defined by f:
Plot3D[f[x, y], {x, y} ∈ Disk,
MeshFunctions -> {#3 &},
Mesh -> {{{1/10, Directive[Black, AbsoluteThickness[3.5]]}}},
BoxRatios -> {1, 1, 1/3},
ColorFunction -> (White &),
Lighting -> "Neutral"]
answered 19 mins ago
m_goldbergm_goldberg
85k872196
$endgroup$
Here are a couple more ideas for displaying the contour at $z=1/10$
You could also define a helper function g that forces Plot3D to work with numeric values. Note: this will give a symbolic rather than a numeric tooltip for the contour.
g[x_?NumericQ, y_?NumericQ] := f[x, y];
ContourPlot[g[x, y] == 1/10, {x, -1, 1}, {y, -1, 1}, ContourShading -> False]
You could also show the contour on the 3D surface defined by f:
Plot3D[f[x, y], {x, y} ∈ Disk,
MeshFunctions -> {#3 &},
Mesh -> {{{1/10, Directive[Black, AbsoluteThickness[3.5]]}}},
BoxRatios -> {1, 1, 1/3},
ColorFunction -> (White &),
Lighting -> "Neutral"]
answered 19 mins ago
m_goldbergm_goldberg
85k872196
edited 14 mins ago
answered 19 mins ago
m_goldbergm_goldberg
85k872196
answered 19 mins ago
m_goldbergm_goldberg
85k872196
answered 19 mins ago
m_goldbergm_goldberg
85k872196
85k872196
add a comment |
add a comment |
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