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From: Tom Vaughan <tom@so...>  20100907 23:55:11

Hello, Today we have made Grima available as free software under the MIT license. Grima is a pygtk+ widget that embeds matplotlib. Basically, this means that Grima allows matplotlib to play nicely with the GTK+ main loop. Grima is hosted on GitHub at http://github.com/cdsi/grima. Please note that this is a very early alpha release. There is very little documentation on how to use Grima, or on its future plans. Our needs are related to being able to visualize arbitrary sets of time series data (like device measurements), as well as store and retrieve this data in a modular way. We plan to provide a mechanism to work with structured JSON data in couchdb or redis. For now, applications simply pass x and y values per the current matplotlib API. We have decided to make Grima available at this point so that others have the opportunity to evolve it beyond our own limited scope. Contributions (ideas, critiques, patches) are welcomed. To start, please take a look at: http://github.com/cdsi/grima/blob/master/bin/grimasubplot.py. I am more than happy to answer any questions. You may contact me directly at tom@..., or at cdsilotr@... The latter is a Google Group that covers Grima as well as some other bits of free software also released today. A list of these are up at http://github.com/cdsi. Thank you, Tom PS  A special thanks to the matplotlib community, and to http://unpythonic.blogspot.com/2007/08/usingthreadsinpygtk.html for all of the excellent work upon which Grima is based.  Visit our website: http://software6.net/ Follow us on Twitter: http://twitter.com/software6 
From: Brian Larsen <balarsen@la...>  20100907 23:50:38

Hey all, I think I know the answer here as "no" or something, but say I have a curve I want to plot and I want the color to change along the curve to denote the 3rd variable is there anyway to do this is matplotlib? What I mean is take the simple plot from pylab import * plot(range(30), range(30, 60), lw=10) and say that the 3rd variable is inten = [val ** 2 for val in range(30)] then can the line change color along its length according to a specified color table? In IDL this is done by just giving a color array with the same length as the data then the line changes with the current colortable. Thanks much, Brian  Brian A. Larsen Space Science and Applications Group ISR1 Los Alamos National Laboratory PO Box 1663, MSD466 Los Alamos, NM 87545 USA (For overnight add: SM30, Bikini Atoll Road) Phone: 5056657691 Fax: 5056657395 email: balarsen@... Correspondence / Technical data or Software Publicly Available 
From: Joshua Holbrook <josh.holbrook@gm...>  20100907 21:34:45

Hey y'all, I recently read about Chernoff faces (http://en.wikipedia.org/wiki/Chernoff_face) in one of Edward Tufte's books (great read btw) and would like to mess around with them in matplotlib. My current approach is to generate the faces as images, and then use them as markers on an xy plot (like the example I found in the Tufte book). I just realized, though, that I have no idea how to incorporate images as position markers in matplotlib, or if it's even possible. My search of the mpl docs didn't turn up much. Any ideas? Joshua Holbrook 
From: Amenity Applewhite <amenity@en...>  20100907 18:45:53

Enthought Python Distribution Webinar September 10 This Friday,Warren Weckesser will host the first of three webinars in a series on solving differential equations in Python. We will take a close look at the two tools available for solving ordinary differential equations in SciPy: the "odeint" function and the "ode" class. Two examples will be discussed: (1) the famous Lorenz equations that exhibit chaos, and (2) the GrayScott reactiondiffusion equations in 1D, from which we will obtain a system of ordinary differential equations by using the "Method of Lines". The demonstrations also include 3D plots and animation using Matplotlib. Enthought Python Distribution Webinar How do I...solve differential equations with Python? Part I: SciPy Tools Friday, September 10: 1pm CST/6pm UTC Wait list (for non EPD subscribers): email:amenity@... Early in 2011, Warren will host Part II: boundary value problems, and in the spring he'll follow up with a third installment to the series. Have a fantastic September, The Enthought Team 
From: Aman Thakral <aman.thakral@gm...>  20100907 17:37:28

Hi, I seem to be encountering a strange problem. I'm using a SpanSelector and a RectangularSelector in my application and they seem to be working in Linux but not in Windows. I'm using wxpython as the gui layer. Has anyone else encountered similar issues? Thanks, Aman 
From: Francesco Montesano <franz.bergesund@go...>  20100907 16:14:49

Dear Joe, finally I had time to come back to my python scritp for the contour plots. You're code works very nicelly and does exactly what I need. Thank you for the help Francesco 2010/7/26 Joe Kington <jkington@...>: > It sounds like you're wanting a gaussian kernel density estimate (KDE) (not > the desktop!). The other options you mentioned are for interpolation, and > are not at all what you're wanting to do. > > You can use scipy.stats.kde.gaussian_kde(). However, it currently doesn't > take a weights array, so you'll need to modify it for your use case. > > If you prefer, I have faster version of a gaussian KDE that can take a > weights array. It's actually slower than the scipy's gaussian kde for a low > number of points, but for hundreds, thousands, or millions of points, it's > several orders of magnitude faster. (Though the speedup depends on the > covariance of the points... higher covariance = slower, generally speaking) > > Here's a quick pastebin of the code. http://pastebin.com/LNdYCZgw > > To use it, you do something like the below... (assuming the code in the > pastebin is saved in a file called fast_kde.py) > > import numpy as np > import matplotlib.pyplot as plt > from fast_kde import fast_kde > > # From your description of your data... > weights, x, y = np.loadtxt('chain.txt', usecols=(0,4,6)).T > > kde_grid = fast_kde(x, y, gridsize=(200,200), weights=weights) > > # Plot the grid > plt.figure() > plt.imshow(kde_grid, extent=(x.min(), x.max(), y.max(), y.min()) > > # Reverse the yaxis > plt.gca().invert_yaxis() > > plt.show() > > Hope that helps a bit, > Joe > > > > > On Sat, Jul 24, 2010 at 3:56 AM, montefra <franz.bergesund@...> > wrote: >> >> Hi, >> >> I am writing a program that reads three columns (one column containing the >> weights, the other two containing the values I want to plot) from a file >> containing the results from a MonteCarlo Markov Chain. The file contains >> thousends of lines. Then create the 2D histogram and make contourplots. >> Here >> is a sample of the code (I don't know if is correct, it's just to show >> what >> I do) >> >> >>> import numpy as np >> >>> import matplotlib.pyplot as mplp >> >>> chain = np.loadtxt("chain.txt", usecols=[0,4,6]) #read columns 0 >> >>> (the >> >>> weights), 4 and 6 (the data), from the file "chain.txt" >> >>> h2D, xe, ye = np.histogram2D(chain[:,1],chain[:,2], >> >>> weights=chain[:,0]) >> >>> #create the 2D histogram >> >>> x = (xe[:1] + xe[1:])/2. #x and y values for the plot (I use the mean >> >>> of each bin) >> >>> y = (ye[:1] + ye[1:])/2. >> >>> mplp.figure() #open the figure >> >>> mplp.contourf(x, y, h2D.T, origin='lower') #contour plot >> >> As it is the contours are not smooth and they look not that nice. After >> days >> of searches I've found three methods and tried, unsuccesfully, to apply >> them >> 1) 2d interpolation: I got "segmentation fault" (on a quadcore machine >> with >> 8Gb of RAM) >> 2) Rbf (radial basis functions): I got wrong contours >> 3) ndimage: it creates spurious features (like secondary peaks parallel to >> the direction of the main one) >> >> Before beginning with Python, I used to use IDL to plot, and there is a >> function 'smooth' that smooth for you 2D histograms. I haven't found >> anything similar for Python. >> Does anyone have an idea or suggestion on how to do it? >> >> Thank in advance >> Francesco >> >>  >> View this message in context: >> http://old.nabble.com/Smoothcontourplotstp29253884p29253884.html >> Sent from the matplotlib  users mailing list archive at Nabble.com. >> >> >> >>  >> The Palm PDK Hot Apps Program offers developers who use the >> PlugIn Development Kit to bring their C/C++ apps to Palm for a share >> of $1 Million in cash or HP Products. Visit us here for more details: >> http://ad.doubleclick.net/clk;226879339;13503038;l? >> http://clk.atdmt.com/CRS/go/247765532/direct/01/ >> _______________________________________________ >> Matplotlibusers mailing list >> Matplotlibusers@... >> https://lists.sourceforge.net/lists/listinfo/matplotlibusers > >  personals: montyfra@..., monte_fra@... (messenger), franz.bergesund@... work: montefra@... http://picasaweb.google.it/franz.bergesund 
From: Benjamin Root <ben.root@ou...>  20100907 16:06:14

2010/9/7 Guillaume Chérel <guillaume.c.cherel@...> > Hello, > > I'm trying to draw circles with the scatter function. They are supposed > to represent trees in the savannah. It is thus important that they are > displayed with a proper size, that is, one which represents their actual > size on the field. After quite some confusion, I've found out (I think) > that the size argument one can specify with the scatter function is > given as a disk's surface in pixels square (I think that's what means > the "points^2" in the documentation and from my own tests) > > What I would like is to give a surface in unit^2, where "unit" is the > unit of my data, and which you can read on the plot's axes ticks. For > example, each tree has coordinates like x=3500, y=2210. (The unit here > is centimeters but we don't really need to know this). Say I want to > draw a tree which canopy is 200 cm wide. That makes a disk which radius > is 100, or surface 100^2*PI. How can I draw this? > > Many thanks, > Guillaume > > > Guillaume, Using scatter is probably not the way to go about what you want. The circles for scatter are a fixed size and if you zoom in, they will not scale accordingly. You probably want to create patches of Circles: http://matplotlib.sourceforge.net/api/artist_api.html#matplotlib.patches.Circle http://www.mailarchive.com/matplotlibusers@.../msg06786.html Or even utilize a collection of Circles: http://matplotlib.sourceforge.net/api/collections_api.html#matplotlib.collections.CircleCollection Note that for a CircleCollection, you would use 'offset' to indicate the center of each circle. After creating the collection, you would then use ax.add_collection() function to add that collection to the axes. I hope that is helpful. Ben Root 
From: Guillaume Chérel <guillaume.c.cherel@gm...>  20100907 14:08:41

Hello, I'm trying to draw circles with the scatter function. They are supposed to represent trees in the savannah. It is thus important that they are displayed with a proper size, that is, one which represents their actual size on the field. After quite some confusion, I've found out (I think) that the size argument one can specify with the scatter function is given as a disk's surface in pixels square (I think that's what means the "points^2" in the documentation and from my own tests) What I would like is to give a surface in unit^2, where "unit" is the unit of my data, and which you can read on the plot's axes ticks. For example, each tree has coordinates like x=3500, y=2210. (The unit here is centimeters but we don't really need to know this). Say I want to draw a tree which canopy is 200 cm wide. That makes a disk which radius is 100, or surface 100^2*PI. How can I draw this? Many thanks, Guillaume 
From: Torsten Hahn <Torsten.Hahn@ph...>  20100907 12:26:10

Hi all, is there an easy way to draw an arrow in a 3D Plot (just a single arrow). I couldn't find any useful example in the 1.0.0 docu. Best regards, Torsten. 
From: JaeJoon Lee <lee.j.joon@gm...>  20100907 03:00:54

On Tue, Sep 7, 2010 at 11:04 AM, Paul Ivanov <pivanov314@...> wrote: > Is this a reasonable way of achieving the desired result? > Yes. You may take a look at the legend guide. http://matplotlib.sourceforge.net/users/legend_guide.html For your original question, it is not possible to do that with the current legend implementation. However, you may put the legend inside the AnnotationBbox, which enables this. I'm posting the example for any future reference. Regards, JJ # small example ax = plt.subplot(1,1,1) ax.plot([0,1], label='ax1') leg = ax.legend() ax.legend_ = None # remove the legend from the axes. ax2 = ax.twinx() ax2.plot([1,0], 'r',label='ax2') leg2 = ax2.legend() # create a legend # add leg as AnnotationBbox from matplotlib.offsetbox import AnnotationBbox leg3 = AnnotationBbox(leg._legend_box, (0, 1), xybox=(5, 0), xycoords=leg2.legendPatch, boxcoords="offset points", box_alignment=(1., 1.), pad=0, ) # adjust zorder so that leg3 is drawn after leg2 leg3.zorder = leg2.zorder+0.1 ax2.add_artist(leg3) 
From: Paul Ivanov <pivanov314@gm...>  20100907 02:04:43

Paul Ivanov, on 20100906 18:01, wrote: > I want to have two legends (from different axes) positioned right up > against on another. > > Here's a static example, except I want the second legend to be defined > relative to the first (if leg is moved, I want leg2 to move as well). I > can't seem to figure out the proper bbox_to_anchor and bbox_transform > parameters to pass to the second legend() to make this work. > > # small example > ax = plt.subplot(1,1,1) > ax2 = ax.twinx() > ax.plot([0,1], label='ax1') > ax2.plot([1,0], 'r',label='ax2') > leg = ax.legend(loc='lower left', borderaxespad=0, > bbox_to_anchor=(.85,.85)) > leg2 = ax2.legend(loc='upper left', borderaxespad=0, > bbox_to_anchor=(.85,.85)) > I guess I really just want one legend, so I figured out an alternative solution: # alternative to having two legends ax = plt.subplot(1,1,1) ax2 = ax.twinx() lines= ax.plot([0,1], label='ax1') lines2= ax2.plot([4,3], 'r',label='ax2') lines.extend(lines2) labels = [l.get_label() for l in lines] leg = ax.legend(lines, labels) Is this a reasonable way of achieving the desired result? thanks, Paul 
From: Paul Ivanov <pivanov314@gm...>  20100907 01:01:31

I want to have two legends (from different axes) positioned right up against on another. Here's a static example, except I want the second legend to be defined relative to the first (if leg is moved, I want leg2 to move as well). I can't seem to figure out the proper bbox_to_anchor and bbox_transform parameters to pass to the second legend() to make this work. # small example ax = plt.subplot(1,1,1) ax2 = ax.twinx() ax.plot([0,1], label='ax1') ax2.plot([1,0], 'r',label='ax2') leg = ax.legend(loc='lower left', borderaxespad=0, bbox_to_anchor=(.85,.85)) leg2 = ax2.legend(loc='upper left', borderaxespad=0, bbox_to_anchor=(.85,.85)) thanks in advance, Paul Ivanov 