Re: [Matplotlib-users] Polar 3D plot?
Hi,
you can create your supporting points on a regular r, phi grid and
transform them then to cartesian coordinates:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
step = 0.04
maxval = 1.0
fig = plt.figure()
ax = Axes3D(fig)
# create supporting points in polar coordinates
r = np.linspace(0,1.25,50)
p = np.linspace(0,2*np.pi,50)
R,P = np.meshgrid(r,p)
# transform them to cartesian system
X,Y = R*np.cos(P),R*np.sin(P)
Z = ((R**2 - 1)**2)
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.jet)
ax.set_zlim3d(0, 1)
ax.set_xlabel(r'$\phi_\mathrm{real}$')
ax.set_ylabel(r'$\phi_\mathrm{im}$')
ax.set_zlabel(r'$V(\phi)$')
ax.set_xticks([])
plt.show()
hth
Armin
klukas schrieb:
I'm guessing this is currently impossible with the current mplot3d
functionality, but I was wondering if there was any way I could generate a
3d graph with r, phi, z coordinates rather than x, y, z?
The point is that I want to make a figure that looks like the following:
http://upload.wikimedia.org/wikipedia/commons/7/7b/Mexican_hat_potential_polar.svg
Using the x, y, z system, I end up with something that has long tails like
this:
http://upload.wikimedia.org/wikipedia/commons/4/44/Mecanismo_de_Higgs_PH.png
If I try to artificially cut off the data beyond some radius, I end up with
jagged edges that are not at all visually appealing.
I would appreciate any crazy ideas you can come up with.
Thanks,
Jeff
P.S. Code to produce the ugly jaggedness is included below:
---
from mpl_toolkits.mplot3d import Axes3D
import matplotlib
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
step = 0.04
maxval = 1.0
fig = plt.figure()
ax = Axes3D(fig)
X = np.arange(-maxval, maxval, step)
Y = np.arange(-maxval, maxval, step)
X, Y = np.meshgrid(X, Y)
R = np.sqrt(X**2 + Y**2)
Z = ((R**2 - 1)**2) * (R 1.25)
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.jet)
ax.set_zlim3d(0, 1)
#plt.setp(ax.get_xticklabels(), visible=False)
ax.set_xlabel(r'$\phi_\mathrm{real}$')
ax.set_ylabel(r'$\phi_\mathrm{im}$')
ax.set_zlabel(r'$V(\phi)$')
ax.set_xticks([])
plt.show()
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Institute of Solid State Physics
Graz University of Technology
Petersgasse 16
8010 Graz
Austria
Tel.: 0043 316 873 8477
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Re: [Matplotlib-users] Can matplotlib generate charts like this?
Erik Wickstrom schrieb: Hi all, Can matplotlib (or any other Python charting library) generate charts like this: (also attached if you prefer) Here is a demo-script. The second way is inspired by matlab. Can this be done more easily in python? Can X and Y be built more elegantly with numpy? Armin ---8 from pylab import * # data generation x = linspace(0,10,100) y = exp(-x) ye = y + rand(y.size)-0.5 # plot the vertical lines # with loop for xl,yl,yel in zip(x,y,ye): plot([xl,xl],[yl,yel],'r') plot(x,y,x,ye,'d') # plot by separating with NaN figure() X = zeros((x.size,3)) Y = zeros((x.size,3)) X[:,0],X[:,1],X[:,2] = x,x,NaN Y[:,0],Y[:,1],Y[:,2] = y,ye,NaN plot(X.flatten(),Y.flatten(),x,y,x,ye,'d') show() ---8 -- Let Crystal Reports handle the reporting - Free Crystal Reports 2008 30-Day trial. Simplify your report design, integration and deployment - and focus on what you do best, core application coding. Discover what's new with Crystal Reports now. http://p.sf.net/sfu/bobj-july ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] zero division: warning and exception
darkside schrieb: Hi list, I have to make a division that sometimes yields and inf, and I want to replace it by 0. I have try this: --- import pylab as p p.seterr(divide='raise') l = array vector defined along the program try: a = (dr*R*dl)/(1.-((R0/R)*p.sin(l))**2)**(1./2) except FloatingPointError: a=0 - It works, but it doesn't return an array as expect, if some of the values are zero, then a = 0. So I tried: -- a = p.zeros(len(l)) for i in range(len(l)): try: a[i] = (dr*R*dl)/(1.-((R0/R)*p.sin(l[i]))**2)**(1./2) except FloatingPointError: a[i]=0 But doing it this way I'm not able to get an exception: array([ Inf]) And I don't know what I have to change to get an exception doing things this way. You can do it that way: a = rand(5,5) b = a.round() c = a/b c[isinf(c)] = 0 # use indexing with boolean array [1] to set zero In your case: a = (dr*R*dl)/(1.-((R0/R)*p.sin(l))**2)**(1./2) a[isinf(a)] = 0 HTH Armin [1]http://www.scipy.org/Tentative_NumPy_Tutorial#head-0dffc419afa7d77d51062d40d2d84143db8216c2 -- Crystal Reports - New Free Runtime and 30 Day Trial Check out the new simplified licensing option that enables unlimited royalty-free distribution of the report engine for externally facing server and web deployment. http://p.sf.net/sfu/businessobjects ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] contour overlapping
Sebastian Busch schrieb: Armin Moser wrote: Sebastian Busch wrote: ... array([list(a[i,:i])+list(b[i,i:]) for i in range(a.shape[0])]) It seems that I did not understand what you tried to reach. ... Sorry. I wanted to do the same as Matthias -- taking his example: I meant I did not understand in the first what Bala tried to reach. I have answered to the wrong mail and quoted badly. Sorry Armin -- The NEW KODAK i700 Series Scanners deliver under ANY circumstances! Your production scanning environment may not be a perfect world - but thanks to Kodak, there's a perfect scanner to get the job done! With the NEW KODAK i700 Series Scanner you'll get full speed at 300 dpi even with all image processing features enabled. http://p.sf.net/sfu/kodak-com ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] contour overlapping
Bala subramanian schrieb: Dear Matthias, Thank you for the information. Could you please provide me a small example of such overlapping. Look at http://matplotlib.sourceforge.net/examples/pylab_examples/contour_image.html or any other contour example from this page: http://matplotlib.sourceforge.net/examples/index.html Armin -- The NEW KODAK i700 Series Scanners deliver under ANY circumstances! Your production scanning environment may not be a perfect world - but thanks to Kodak, there's a perfect scanner to get the job done! With the NEW KODAK i700 Series Scanner you'll get full speed at 300 dpi even with all image processing features enabled. http://p.sf.net/sfu/kodak-com ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] contour overlapping
Bala subramanian schrieb: hai Armin, I looked through the examples. I could not find any example of overlapping two differnet countours on the same plot. I think the first example filled contours does exactly that. You want to show two contours over each other in the same plot. You just have to substitute the Z in cset_1 with matrix_1 and in cset_2 with matrix_2. Of course it will be helpful to use different colormaps. E.g. a grey one for the underlying contour and a colored for the top one. x = arange(5) y = arange(5) x,y = meshgrid(x,y) Z = x**2+y**2 #contourf(Z,cmap=cm.binary) # filled contours gray contour(Z) # not filled contours colored error = rand(x.shape[0],x.shape[1]) # to generate a new Z Z = (x+error)**2+(y+error)**2 contour(Z) # colored not filled contours Armin -- The NEW KODAK i700 Series Scanners deliver under ANY circumstances! Your production scanning environment may not be a perfect world - but thanks to Kodak, there's a perfect scanner to get the job done! With the NEW KODAK i700 Series Scanner you'll get full speed at 300 dpi even with all image processing features enabled. http://p.sf.net/sfu/kodak-com ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] contour overlapping
Sebastian Busch wrote:
Matthias Michler wrote:
...
for i in xrange(len(matrix3[:, 0])): # all rows
for j in xrange(len(matrix3[0, :])):# all columns
...
if your matrices a and b are rectangular (and i think the diagonal
makes only sense in this case), you can also say:
array([list(a[i,:i])+list(b[i,i:]) for i in range(a.shape[0])])
It seems that I did not understand what you tried to reach. Sorry for
pointing into the wrong direction.
Another possibility would be to use masked arrays:
8-
from pylab import *
x = arange(100)
y = arange(100)
x,y = meshgrid(x,y)
Z = x**2+y**2
mask = ones(Z.shape)
# contour, imshow, pcolor do not show values at positions
# where the mask is True
lower_left_masked = triu(mask)==0 # lower left part in matrix masked
Z = ma.masked_array(Z,mask=lower_left_masked)
contourf(Z,origin='lower')
error = rand(x.shape[0],x.shape[1])
upper_right_masked = tril(mask)==0
Z = (x+error)**2+(y+error)**2
Z = ma.masked_array(Z,mask=upper_right_masked)
contourf(Z,cmap=cm.binary)
axis('tight')
show()
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Re: [Matplotlib-users] Tight axis after deletion of image
Armin Moser schrieb:
Hi,
i have an application showing some pseudo color plots as images using
imshow. After deleting an image and calling ax.axis('tight') the limits
of the axes are not updated correctly. Is this a bug, incorrect use or
intended? The behavior is demonstrated in the appended script.
It looks like calling ax.relim() is the preferred way of setting limits
after deleting artists. Unfortunately this method does only consider
patches and lines. I added some lines in axes.py to support images to
(the patch against the current svn-version is appended).
If this is not the correct place for submitting patches please tell me
where I should put it instead.
Thanks and best regards
Armin
--- C:\DOKUME~1\ArminM\LOKALE~1\Temp\axes.py-revBASE.svn001.tmp.py Mi Apr
22 11:40:58 2009
+++ C:\Dokumente und Einstellungen\ArminM\Eigene
Dateien\matplotlib_trunk\matplotlib\lib\matplotlib\axes.py Mi Apr 22
11:40:49 2009
@@ -1430,6 +1430,14 @@
for p in self.patches:
self._update_patch_limits(p)
+for im in self.images:
+self._update_image_limits(im)
+
+def _update_image_limits(self,im):
+xmin,xmax,ymin,ymax = im.get_extent()
+corners = (xmin,ymin),(xmax,ymax)
+self.update_datalim(corners)
+
def update_datalim(self, xys, updatex=True, updatey=True):
'Update the data lim bbox with seq of xy tups or equiv. 2-D array'
# if no data is set currently, the bbox will ignore its
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[Matplotlib-users] Tight axis after deletion of image
Hi,
i have an application showing some pseudo color plots as images using
imshow. After deleting an image and calling ax.axis('tight') the limits
of the axes are not updated correctly. Is this a bug, incorrect use or
intended? The behavior is demonstrated in the appended script.
Best regards,
Armin
-8-
#!/usr/bin/env python
from pylab import *
a = rand(5,5)
ax = axes()
ax.imshow(a,extent=(0,5,0,5))
ax.imshow(a,extent=(5,10,5,10))
ax.axis('tight')
waitforbuttonpress()
ax.images.pop(0)
ax.axis('tight')
show()
-8-
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[Matplotlib-users] griddata performance
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 useable. Is this a known property of the used triangulation? The triangulation can be performed with matlab without any problems. Armin -- Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] griddata performance
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 the delaunay triangulation is extremely slow, i.e. not useable. Is this a known property of the used triangulation? The triangulation can be performed with matlab without any problems. Armin Armin: You could try installing the natgrid toolkit and see if that speeds up griddata at all. If not, please post a test script with data and maybe we can figure out what is going on. I have already tried natgrid and it didn't improve the situation. As suggested I append a script demonstrating the problem. Thanks Armin --8- from numpy import * from pylab import * import time deg2rad = pi/180.0 ai = 0.12*deg2rad x = linspace(13,40,676) y = linspace(10,22,801) x = x*deg2rad y = y*deg2rad [x,y] = meshgrid(x,y) z = (x**2+y**2) xi = linspace(x.min(),x.max(),x.shape[1]) yi = linspace(y.min(),y.max(),y.shape[0]) tic= time.time() zi = griddata(x.flatten(),y.flatten(),z.flatten(),xi,yi) toc = time.time() print toc-tic fac = 2*pi/1.2681 nx = fac * (cos(y)*cos(x) - cos(ai)) ny = fac * (cos(y)*sin(x)) nz = fac * (sin(y) + sin(ai)) np = sqrt(nx**2 + ny**2) z = (np**2+nz**2)*exp(-0.001*nz) xi = linspace(np.min(),np.max(),x.shape[1]) yi = linspace(nz.min(),nz.max(),y.shape[0]) tic = time.time() zi = griddata(np.flatten(),nz.flatten(),z.flatten(),xi,yi) toc = time.time() print toc-tic -- Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] griddata performance
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 computed as X,Y = meshgrid(x,y). If this condition is not fullfilled the delaunay triangulation is extremely slow, i.e. not useable. Is this a known property of the used triangulation? The triangulation can be performed with matlab without any problems. Armin Armin: You could try installing the natgrid toolkit and see if that speeds up griddata at all. If not, please post a test script with data and maybe we can figure out what is going on. I have already tried natgrid and it didn't improve the situation. As suggested I append a script demonstrating the problem. Thanks Armin Armin: On my mac, your two benchmarks take 15 and 14 seconds. Do you consider that too slow? No definitely not but up to now I always killed the process (after more than 10 minutes). I tried the script again and it worked... I have to check the original on monday at work. Perhaps this is just a toy example to test griddata, but I assume you realize that you wouldn't normally use griddata to interpolate data on one regular grid to another regular grid. griddata is strictly for interpolating scatter data (not on a regular mesh) to a regular mesh. Yes I do, but the supporting points of the second example are not on a regular grid. Thanks a lot Armin -- Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] griddata performance
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 (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 useable. Is this a known property of the used triangulation? The triangulation can be performed with matlab without any problems. Armin Armin: You could try installing the natgrid toolkit and see if that speeds up griddata at all. If not, please post a test script with data and maybe we can figure out what is going on. I have already tried natgrid and it didn't improve the situation. As suggested I append a script demonstrating the problem. Thanks Armin Armin: On my mac, your two benchmarks take 15 and 14 seconds. Do you consider that too slow? I got 10 s and 8 s on my Thinkpad--but that was without natgrid. When I installed natgrid, the benchmark was taking so long I killed the window. Thats the behaviour I observed (on Windows and Linux) with and without natgrid. I'm very puzzled at the moment. Thanks a lot Armin -- Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
