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From: gsal <sal...@gm...> - 2012-09-14 18:51:35
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Hi, everybody:
I don't have experience with images or contours and need some help plotting
a 'z' quantity for given x,y coordinates.
What are the choices?
Here is a small sample of the data:
The first row has the i-th x-coordinate at which the field starts to have
the value [i,j].
The first column has the j-th y-coordinate at which the field starts to
have the value [i,j].
The coordinate steps are not constant, nor the same for both dimensions;
they can be anything because they come from some odd finite difference
program.
My first shot at this is getting combersome, I am hoping for a better way.
So far, because the second decimal place in the x,y coordinate is alwasy
zero, I simply turned those coordinates into integers by multiplying by ten
and truncating; then by subtracting the first value from the rest, they look
very much like matrix indeces (except for the missing ones):
Then, because I don't know any better, I broadcast the values onto another
matrix, to fill in the in-between values:
Now, I have a matrix where every i,j has its own z-value and I am supposed
to be able to plot it with ax.contourf(mymatrix)...which I can, up until
about mymatrix[:,:1900] or so, afte that, I get the following error:
Traceback (most recent call last):
File "C:\findiff\t1.py", line 73, in <module>
ax.contourf(full[:,:2000])
File "C:\Python26\lib\site-packages\matplotlib\axes.py", line 7322, in
contourf
return mcontour.QuadContourSet(self, *args, **kwargs)
File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1106, in
__init__
ContourSet.__init__(self, ax, *args, **kwargs)
File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 700, in
__init__
self._process_args(*args, **kwargs)
File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1130, in
_process_args
C = _cntr.Cntr(x, y, z.filled(), _mask)
ValueError: Arguments x, y, z, mask (if present) must be 2D arrays.
x, y, z must be castable to double.
My current matrix is about 12000x5000.
Any asistance would be greatly appreciated.
Thanks,
Germán
--
View this message in context: http://matplotlib.1069221.n5.nabble.com/Need-to-plot-z-at-given-x-y-contour-or-something-tp38926.html
Sent from the matplotlib - users mailing list archive at Nabble.com.
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From: Benjamin R. <ben...@ou...> - 2012-09-14 19:00:30
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On Fri, Sep 14, 2012 at 2:51 PM, gsal <sal...@gm...> wrote: > Hi, everybody: > > I don't have experience with images or contours and need some help plotting > a 'z' quantity for given x,y coordinates. > > What are the choices? > > Here is a small sample of the data: > > The first row has the i-th x-coordinate at which the field starts to have > the value [i,j]. > The first column has the j-th y-coordinate at which the field starts to > have the value [i,j]. > > The coordinate steps are not constant, nor the same for both dimensions; > they can be anything because they come from some odd finite difference > program. > > My first shot at this is getting combersome, I am hoping for a better way. > > So far, because the second decimal place in the x,y coordinate is alwasy > zero, I simply turned those coordinates into integers by multiplying by ten > and truncating; then by subtracting the first value from the rest, they > look > very much like matrix indeces (except for the missing ones): > > Then, because I don't know any better, I broadcast the values onto another > matrix, to fill in the in-between values: > > Now, I have a matrix where every i,j has its own z-value and I am supposed > to be able to plot it with ax.contourf(mymatrix)...which I can, up until > about mymatrix[:,:1900] or so, afte that, I get the following error: > > Traceback (most recent call last): > File "C:\findiff\t1.py", line 73, in <module> > ax.contourf(full[:,:2000]) > File "C:\Python26\lib\site-packages\matplotlib\axes.py", line 7322, in > contourf > return mcontour.QuadContourSet(self, *args, **kwargs) > File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1106, in > __init__ > ContourSet.__init__(self, ax, *args, **kwargs) > File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 700, in > __init__ > self._process_args(*args, **kwargs) > File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1130, in > _process_args > C = _cntr.Cntr(x, y, z.filled(), _mask) > ValueError: Arguments x, y, z, mask (if present) must be 2D arrays. > x, y, z must be castable to double. > > > > My current matrix is about 12000x5000. > > Any asistance would be greatly appreciated. > > Thanks, > > Germán > > > tricontourf() might be more what you are looking for. Another possibility is pcolor() (note that for irregularly spaced grids, pcolormesh() would not work). Cheers! Ben Root |
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From: Eric F. <ef...@ha...> - 2012-09-14 19:50:36
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On 2012/09/14 9:00 AM, Benjamin Root wrote: > tricontourf() might be more what you are looking for. Another > possibility is pcolor() (note that for irregularly spaced grids, > pcolormesh() would not work). Huh? I don't think there is anything pcolor can handle that pcolormesh can't handle faster. In both cases, the grids must be quadrilateral, but that's all. Eric |
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From: gsal <sal...@gm...> - 2012-09-14 20:09:12
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Wonderful...pcolor is doing the job without processing, it takes exactly what I already have...the n+1 values for x and y coordinates defining the boundaries of the cells and the nxn matix itself. pcolormesh and pcolorfast also work. Thank you very much. -- View this message in context: http://matplotlib.1069221.n5.nabble.com/Need-to-plot-z-at-given-x-y-contour-or-something-tp38926p38930.html Sent from the matplotlib - users mailing list archive at Nabble.com. |
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From: Benjamin R. <ben...@ou...> - 2012-09-14 20:16:16
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On Fri, Sep 14, 2012 at 3:50 PM, Eric Firing <ef...@ha...> wrote: > On 2012/09/14 9:00 AM, Benjamin Root wrote: > > tricontourf() might be more what you are looking for. Another > > possibility is pcolor() (note that for irregularly spaced grids, > > pcolormesh() would not work). > > Huh? I don't think there is anything pcolor can handle that pcolormesh > can't handle faster. In both cases, the grids must be quadrilateral, > but that's all. > > Eric > > Clarification: pcolormesh() must have a grid of coordinates (not necessarially equally spaced). As for pcolormesh() being able to handle anything that pcolor() can handle, I have run into situations where that was not the case. I don't remember the details, though. I think pcolorfast() operates like that (falling back to pcolor() as a last resort). Ben Root |
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From: Eric F. <ef...@ha...> - 2012-09-14 23:26:15
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On 2012/09/14 10:15 AM, Benjamin Root wrote: > > > On Fri, Sep 14, 2012 at 3:50 PM, Eric Firing <ef...@ha... > <mailto:ef...@ha...>> wrote: > > On 2012/09/14 9:00 AM, Benjamin Root wrote: > > tricontourf() might be more what you are looking for. Another > > possibility is pcolor() (note that for irregularly spaced grids, > > pcolormesh() would not work). > > Huh? I don't think there is anything pcolor can handle that pcolormesh > can't handle faster. In both cases, the grids must be quadrilateral, > but that's all. > > Eric > > > Clarification: pcolormesh() must have a grid of coordinates (not > necessarially equally spaced). > > As for pcolormesh() being able to handle anything that pcolor() can > handle, I have run into situations where that was not the case. I don't > remember the details, though. I think pcolorfast() operates like that > (falling back to pcolor() as a last resort). No, pcolorfast never falls back to pcolor. In order of fastest to slowest, it tries to use image rendering, then a variant of nonuniform image rendering, and then a quadmesh. It is a bit fussier about inputs than pcolor and pcolormesh, and does not draw lines. pcolor and pcolormesh differ in the mechanism they use (pcolor uses a PolyCollection) and in the way masked data are handled (pcolor draws nothing in masked regions). Eric > > Ben Root > |
