Yes, you are absolutely correct. I did not realize that I did not actually
evaluate the function over a grid, makes sense that interpolation fails. I
thought that since I created the two axis vectors the function evaluation
occurs over the entire domain, meshgrid is what I was missing.
thanks,
On 9 January 2013 09:32, Shahar Shani-Kadmiel wrote:
> Hi,
>
> I'm trying to contour some data that I have and the griddata line fails. I
> tried running it on some synthetically generated data and I get the same
> IndexError. Any Ideas?
>
> Here is the example with the synthetic data:
>
> x = y =
Hi,
I'm trying to contour some data that I have and the griddata line fails. I
tried running it on some synthetically generated data and I get the same
IndexError. Any Ideas?
Here is the example with the synthetic data:
x = y = arange(-10,10,0.01)
z = x**2+y**3
xi = yi = linspace(-10.1, 10.1
Dear Benjamin,
Thanks for the reply. Apparently my Python 2.6 version was completely
removed and I only have Python 2.7. My matplotlib and numpy are the latest
versions. In matplotlib homepage, I found that meshgrid should do the same
job, but I cannot make script to run it from a file. Is there a
On Thursday, May 24, 2012, Umut Yildiz wrote:
> Dear All,
>
> I used to use griddata in order to make my contourmaps. However, after I
> updated
> my Python from 2.6 to 2.7 griddata is not working anymore.
>
> I tried some workarounds but no success.
>
> The countourmap that I produced before is h
Dear All,
I used to use griddata in order to make my contourmaps. However, after I updated
my Python from 2.6 to 2.7 griddata is not working anymore.
I tried some workarounds but no success.
The countourmap that I produced before is here.
http://dl.dropbox.com/u/17983476/matplotlib/contour_dT_wo
Jeff,
Thanks. That indeed did work (after downloading python-dev package). I just
didn't know that 'install' was the argument that I was supposed to pass to
it =)
After installing I tried griddata again. My input data was originally a
list. The new griddata didn't like this and so I simply used
a
On 12/21/11 12:31 AM, Brad Malone wrote:
Hi, I'm still working on my interpolating from an irregularly space
grid and then running pcolormesh on the resulting output. With some of
the newer data I've been plotting I've noticed that my plots are
complete garbage. I realized that this was actuall
Hi, I'm still working on my interpolating from an irregularly space grid
and then running pcolormesh on the resulting output. With some of the newer
data I've been plotting I've noticed that my plots are complete garbage. I
realized that this was actually because of the output from griddata rather
Hi all,
what are the differences between the griddata
implementations in matplotlib and scipy, i.e.
from scipy.interpolate import griddata
and
from matplotlib.mlab import griddata
Is the Shepard algorithm available in matplotlib/scipy ?
Nils
Reference:
Robert J. Renka
Algorithm 790: CSHEP2
Hello all
Can any one explain me how can I use griddata in matplotlib for 2-d
interpolation of polar to cartesian co-ordinates.
I have two 1D arrays r and theta of 128 and 64 cells respectively
I have a 2D array temperature T(r,theta) with (128, 64) cells. I would like to
have Tnew(200,200) cel
Fri, 18 Feb 2011 20:24:31 +0100, Nils Wagner wrote:
> what is the reason for the white areas in the corners of the
> interpolation domain?
> Any idea ?
Griddata does not do any extrapolation, and the corners are outside the
convex hull of the point set.
> import numpy as np
> from scipy.interpol
On Fri, Feb 18, 2011 at 1:24 PM, Nils Wagner
wrote:
> Hi all,
>
> what is the reason for the white areas in the corners of the interpolation
> domain ?
> Any idea ?
The white areas are not bounded by your data points (they are located
outside the convex hull of the data points), so the interpolat
On Mon, Feb 14, 2011 at 8:40 AM, Nils Wagner
wrote:
> Hi all,
>
> Is it possible to apply griddata to polar coordinates or
> do I need cartesian coordinates ?
>
> http://matplotlib.sourceforge.net/api/mlab_api.html#matplotlib.mlab.griddata
You can keep the data in polar coordinates, you just need
Hi all,
Is it possible to apply griddata to polar coordinates or
do I need cartesian coordinates ?
http://matplotlib.sourceforge.net/api/mlab_api.html#matplotlib.mlab.griddata
Nils
--
The ultimate all-in-one performanc
2010/4/3 ms :
> I am trying to use griddata to plot some (irregularly) spaced data as a
> contour plot, but sometimes ALL the grid it outputs is masked: so no plot.
>
> In the docs I read:
> "A masked array is returned if any grid points are outside convex hull
> defined by input data (no extrapola
Hi,
I am trying to use griddata to plot some (irregularly) spaced data as a
contour plot, but sometimes ALL the grid it outputs is masked: so no plot.
In the docs I read:
"A masked array is returned if any grid points are outside convex hull
defined by input data (no extrapolation is done)."
but
tfoutz99,
I occasionally run into this issue as well. At quick glance I suspect it
may be related to the limitation listed at:
http://www.scipy.org/scipy/scikits/ticket/61
...but I could be way off base as I'm not sure if the code is derived from
the same place.
-Erik
tfoutz99 wrote:
>
> Hi
Hi! I am basing my code off the example posted at:
http://www.scipy.org/Cookbook/Matplotlib/Gridding_irregularly_spaced_data
Gridding irregularly spaced data
When I use my own data, I am getting a KeyError (Posted below)
However, if I only use a subset of my data (for which the total
length=263
plankton wrote:
>
> Greetings all,
>
> I rotate a vector field and than I tried to interpolate it to a new grid
> using griddata.
>
> CODE:
>
> x_grid_unique = unique(x_grid)
> y_grid_unique = unique(y_grid)
> x_meshgrid, y_meshgrid = meshgrid(x_grid_unique, y_gri
plankton wrote:
> Greetings all,
>
> I rotate a vector field and than I tried to interpolate it to a new grid
> using griddata.
>
> CODE:
>
> x_grid_unique = unique(x_grid)
> y_grid_unique = unique(y_grid)
> x_meshgrid, y_meshgrid = meshgrid(x_grid_unique, y_grid_unique)
Greetings all,
I rotate a vector field and than I tried to interpolate it to a new grid
using griddata.
CODE:
x_grid_unique = unique(x_grid)
y_grid_unique = unique(y_grid)
x_meshgrid, y_meshgrid = meshgrid(x_grid_unique, y_grid_unique)
x_rot_meshgrid = reshap
I ve a problem with the use of griddata.
I have a grid of x,y with value z. the grid have 4500 points
I would like to have a rigular grid of 1500 points.
I try the function zi = griddata(x,y,z,xi,yi)
but I have an error "too many indices". I don't understant why!!!
x,y,z,xi and yi are numpy array
Jeff,
Thanks for the speedy reply. Should a warning be raised then that my
data (though in 2D form) is not 2D and therefore is invalid input?
Cheers,
Josh Lawrence
Ph.D. Student
Clemson University
On Apr 16, 2009, at 2:53 PM, Jeff Whitaker wrote:
> Josh Lawrence wrote:
>> Greetings all,
>>
>
Josh Lawrence wrote:
> Greetings all,
>
> In using the function griddata in mlab.py, I think I have found a bug.
> The following line in mlab.py errors for me.
> I supply it an xi and yi that have shape (N,1). I have surface data,
> but I only care about the variation in one direction. In mlab,
Greetings all,
In using the function griddata in mlab.py, I think I have found a bug.
The following line in mlab.py errors for me.
I supply it an xi and yi that have shape (N,1). I have surface data,
but I only care about the variation in one direction. In mlab, when it
gets to this line (29
Jeff Whitaker wrote:
> Armin Moser wrote:
>> Jeff Whitaker wrote:
>>
>>> Armin Moser wrote:
>>>
Hi,
I would like to interpolate an array of shape (801,676) to regularily
spaced datapoints using griddata. This interpolation is quick if the
(x,y) supporting points are
Eric Firing wrote:
> Jeff Whitaker wrote:
>> Armin Moser wrote:
>>> Jeff Whitaker wrote:
>>>
Armin Moser wrote:
> Hi,
>
> I would like to interpolate an array of shape (801,676) to regularily
> spaced datapoints using griddata. This interpolation is quick if the
Jeff Whitaker wrote:
> Armin Moser wrote:
>> Jeff Whitaker wrote:
>>
>>> Armin Moser wrote:
>>>
Hi,
I would like to interpolate an array of shape (801,676) to regularily
spaced datapoints using griddata. This interpolation is quick if the
(x,y) supporting points are c
Jeff Whitaker wrote:
> Armin Moser wrote:
>> Jeff Whitaker wrote:
>>
>>> Armin Moser wrote:
>>>
Hi,
I would like to interpolate an array of shape (801,676) to regularily
spaced datapoints using griddata. This interpolation is quick if the
(x,y) supporting points are
Armin Moser wrote:
> Jeff Whitaker wrote:
>
>> Armin Moser wrote:
>>
>>> Hi,
>>>
>>> I would like to interpolate an array of shape (801,676) to regularily
>>> spaced datapoints using griddata. This interpolation is quick if the
>>> (x,y) supporting points are computed as X,Y = meshgrid(x,y)
Jeff Whitaker wrote:
> Armin Moser wrote:
>> Hi,
>>
>> I would like to interpolate an array of shape (801,676) to regularily
>> spaced datapoints using griddata. This interpolation is quick if the
>> (x,y) supporting points are computed as X,Y = meshgrid(x,y). If this
>> condition is not fullfilled
On Fri, Feb 20, 2009 at 8:11 AM, Armin Moser
wrote:
> Hi,
>
> I would like to interpolate an array of shape (801,676) to regularily
> spaced datapoints using griddata. This interpolation is quick if the
> (x,y) supporting points are computed as X,Y = meshgrid(x,y). If this
> condition is not fullf
Armin Moser wrote:
> Hi,
>
> I would like to interpolate an array of shape (801,676) to regularily
> spaced datapoints using griddata. This interpolation is quick if the
> (x,y) supporting points are computed as X,Y = meshgrid(x,y). If this
> condition is not fullfilled the delaunay triangulation i
Hi,
I would like to interpolate an array of shape (801,676) to regularily
spaced datapoints using griddata. This interpolation is quick if the
(x,y) supporting points are computed as X,Y = meshgrid(x,y). If this
condition is not fullfilled the delaunay triangulation is extremely
slow, i.e. not use
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