That is very cool, I hadn't thought of it!
 
So what you're saying is that any transformation (a complex distortion) of a 
regular rectangular grid is fine. The fact that the grid's 'pixels' are four 
sided quadrilaterals satisfies this condition and the contour algorithm works...
 
Cheers,
Scott

>>> "Mark Bakker" <[EMAIL PROTECTED]> 7/11/2007 11:36 >>>
Viraj and Jeff -

Maybe one extension of Jeff's answer.
The process works as long as x, y, and z are 2D arrays of the same size and 
shape.
Hence, x and y don't have to form a rectangular grid. 
I have used this feature regularly for conformal mapping. 
And it makes a lot of sense. 
The contour routine simply looks for intersections between x and y values.
Then when it plots it uses the x and y values in the arrays. 
So when those are not a rectangular grid, it doesn't care.
It's a cool feature.
I can give an example if you want,

Mark

 



Viraj Vajratkar wrote:
> hey guys... i got it... u can use contour(x,y,z)... as in
> x=load('urfile1.dat'), y=load('urfile2.dat), z=load('urfile3.dat
> ').... and then type out the above.... for details about the 
> parameters x,y,z see... .
> http://www.scilab.org/product/man-eng/graphics/contour.htm .... so
> matplotlib CAN plot a contour from discrete points!!!.... ive tried it 
> and it works...
Viraj:  That only works because x and y describe a rectangular grid.  If
x and y described irregularly spaced points, you would need to grid the
data first using one of the methods described on that Cookbook page. 

-Jeff



Please find our Email Disclaimer here: http://www.ukzn.ac.za/disclaimer/
-------------------------------------------------------------------------
This SF.net email is sponsored by DB2 Express
Download DB2 Express C - the FREE version of DB2 express and take
control of your XML. No limits. Just data. Click to get it now.
http://sourceforge.net/powerbar/db2/
_______________________________________________
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/matplotlib-users

Reply via email to