Message Boards Message Boards

0
|
2536 Views
|
8 Replies
|
7 Total Likes
View groups...
Share
Share this post:

How to change the default graphic shape in ContourPlot output?

Tony Coliderse
Posted 1 year ago

The density figure like this can be plotted in Python with Contour, and I can't do this with Mathematica in a elliptical shape. The same command Contour in Mathematica give a square picture, can Mathematica plot the same shape like Python?
And the picture is a simple example, the data can be chosen freely.

Attachment

Attachments:
截屏2022-09-13 22.00.33.png
POSTED BY: Tony Coliderse
8 Replies
Sort By:

Tony,

Use RegionFunction to specify the region. For example, this specifies a ring:

ListDensityPlot[test, AspectRatio -> Automatic, 
 RegionFunction -> Function[{x, y, z}, 2 < x^2 + y^2 < 6]]

Regards

Neil
POSTED BY: Neil Singer

Thank you sir! This way can help me solve this problem greatly!

Regards

Tony
POSTED BY: Tony Coliderse

By the way, I with I knew why

ContourPlot[x, Element[{x, y}, Disk[]]]
Plot3D[x, Element[{x, y}, Disk[]]]

draw the plot on a disk, but the following does not:

ParametricPlot3D[{x, y, x}, Element[{x, y}, Disk[]]]

Actually, it did up to version 12.0. But then that capability was inexplicably removed.
POSTED BY: Gianluca Gorni

Gianluca,

This works:

ParametricPlot3D[{x, y, x}, Element[{x, y}, Disk[]], 
 RegionFunction -> Function[{x, y, z}, x^2 + y^2 < 1]]

or better yet,

ParametricPlot3D[{x, y, x}, Element[{x, y}, Disk[]], 
 RegionFunction -> Function[{x, y, z}, Element[{x, y}, Disk[]]]]

but you should report that the element syntax no longer automatically limits the region.

Regards

Neil
POSTED BY: Neil Singer

Yes, we can do it with RegionFunction, but it is not so convenient for Polygons, for example. The new behaviour of ParametricPlot3D is declared explicitly in the documentation, and I cannot report it as a bug, maybe as a complaint?
POSTED BY: Gianluca Gorni

If you want to do the mesh lines and other details, you can use more options. These make no sense to the plot because I just made things up and copied the MeshFunction from an example in the documentation to show you.

DensityPlot[Sin[x + Cos[y]], {x, y} \[Element] Disk[{0, 0}, {6, 3}], 
 AspectRatio -> Automatic, BoundaryStyle -> Directive[Red, Thick], 
 Frame -> None, Mesh -> 10, 
 MeshFunctions -> {Re[Sqrt[#1 + I #2]] &, Im[Sqrt[#1 + I #2]] &}]

gives

enter image description here
POSTED BY: Neil Singer

Thanks for your help! But if I want to do this with the data in a list shape, rather than a continous function. How to achieve it in this case? I read the documentation in detail, but can't solve it.

test = Table[{x, y, x + y}, {x, 0, 10}, {y, 0, 10}];
test = Flatten[test, 1];
ListDensityPlot[test,AspectRatio -> Automatic]
POSTED BY: Tony Coliderse

Tony,

You need to plot over a region. For example,

DensityPlot[Sin[x + Cos[y]], {x, y} \[Element] Disk[{0, 0}, {6, 3}], 
 AspectRatio -> Automatic]

Gives

enter image description here

Regards,

Neil
POSTED BY: Neil Singer
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract