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Hi Jae-Joon On Mon, Jun 22, 2009 at 21:14, Jae-Joon Lee<lee...@gm...> wrote: > Without actual code, it is difficult to figure out what the real problem is. sorry I didn't attach a complete example but I would have to strip down a lot :) > Anyhow, did you check the below animation example? > > http://matplotlib.sourceforge.net/examples/animation/animation_blit_gtk.html yes, but I didn't have the illumination; until now :) > In the example, the grid is static (i.e., not animated). If what you that's perfect, I do not want it to be anymated. > want is to have the grid animated, then an explicit draw_artist call > is required (note that grid is drawn by axis). yeah, I want it to be only on Y (only horizonatal lines). The solution is: self.ax = self.fig.add_subplot(111) ... self.ax.grid(True) self.canvas.draw() # <-- this was missing self.bg = self.canvas.copy_from_bbox(self.ax.bbox) adding an explicit draw() call let the grid() be drawn and then the result is what's expected. Thanks a lot! Cheers, -- Sandro Tosi (aka morph, morpheus, matrixhasu) My website: http://matrixhasu.altervista.org/ Me at Debian: http://wiki.debian.org/SandroTosi
Without actual code, it is difficult to figure out what the real problem is. Anyhow, did you check the below animation example? http://matplotlib.sourceforge.net/examples/animation/animation_blit_gtk.html In the example, the grid is static (i.e., not animated). If what you want is to have the grid animated, then an explicit draw_artist call is required (note that grid is drawn by axis). Regards, -JJ On Sun, Jun 21, 2009 at 11:54 AM, Sandro Tosi<mat...@gm...> wrote: > Hello, > I'm using blit to animate my plot, and I'd like to add a grid on it > but I failed at it. > > What I do is: > > 1. prepare Figure, Axes, etc without plotting anything > 2. save teh background with > self.bg = self.canvas.copy_from_bbox(self.ax.bbox) > > (on graph update) > > 3. restore background with > self.canvas.restore_region(self.bg) > 4. <line>.set_ydata(...) > 5. self.ax.draw_artist(self.<line>) > 6. self.canvas.blit(self.ax.bbox) > > Now, I tried to place a "self.ax.grid(True)" in almost every possible > position, but never a grid came out :( > > What is the right approach to solve it? Thanks in advance > > Regards, > -- > Sandro Tosi (aka morph, morpheus, matrixhasu) > My website: http://matrixhasu.altervista.org/ > Me at Debian: http://wiki.debian.org/SandroTosi > > ------------------------------------------------------------------------------ > Are you an open source citizen? Join us for the Open Source Bridge conference! > Portland, OR, June 17-19. Two days of sessions, one day of unconference: 250ドル. > Need another reason to go? 24-hour hacker lounge. Register today! > http://ad.doubleclick.net/clk;215844324;13503038;v?http://opensourcebridge.org > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users >
Tony, My understanding is that (which might be wrong) drawing collections involves (at least) 2 transforms. The first transform is (mostly) for scaling, and the second transform is for offset. And this seems to be true for PolygonCollection (which scatter creates) as far as I can see. set_transform() method sets the transform for scaling, which means that (0,0) should transform to (0,0). Otherwise the polygon is not drawn where you intended. And, obviously transData does NOT, which I think is the root of the problem. Unfortunately, the current PolygonCollection class does not seem to have any support for what you want. On the other hand, it seems that EllipseCollection lets you specify the ellipse size in data coordinate. The easiest solution I can think of is doing some monkey patching. import matplotlib.pyplot as plt import matplotlib.transforms as transforms import numpy as np fig = plt.figure() ax = fig.add_subplot(111) x = [0.25, 0.75, 0.25, 0.75] y = [0.25, 0.25, 0.75, 0.75] r = 0.1 * np.ones(4) col = plt.scatter(x, y, np.pi*r**2) from matplotlib.collections import RegularPolyCollection class RegularPolyCollection2(RegularPolyCollection): def get_transform(self): ax = self.axes sc_x = ax.bbox.width / ax.viewLim.width sc_y = ax.bbox.height / ax.viewLim.height return transforms.Affine2D().scale(sc_x, sc_y) col.__class__ = RegularPolyCollection2 plt.axis('equal') plt.show() Alternatively, you may modify your code to use EllipseCollection (if what you want are just cricles). I hope this solution fits your need. Regards, -JJ On Thu, Jun 18, 2009 at 7:24 PM, Tony S Yu<to...@mi...> wrote: > I'd like to plot a collection and scale the size of the collection > elements in relation to the data. My guess is that I need to use the > data transformation (ax.transData) since I would like the size of the > collection elements altered when zooming in/out. > > Unfortunately, my attempt has led to weird results: the collection > offsets are shifted from the desired coordinates when using > ax.transData. Weirder still: the collection elements move *relative to > the data coordinates* when panning the figure. > > I suspect that setting the collection transform to ax.transData is > somehow applying some part of the transform twice. Does anyone know > what I'm doing wrong here and how I can fix this? > > Thanks! > -Tony > > Attached is a toy example of what I'm trying to do. The radii of the > circles are plotted correctly, but notice the x, y coordinates don't > match the circle centers in the plot. Also, try panning the plot and > watch as the circles move relative to the tick marks. > > > >>> import matplotlib.pyplot as plt > >>> import matplotlib.transforms as transforms > >>> import numpy as np > >>> fig = plt.figure() > >>> ax = fig.add_subplot(111) > >>> x = [0.25, 0.75, 0.25, 0.75] > >>> y = [0.25, 0.25, 0.75, 0.75] > >>> r = 0.1 * np.ones(4) > >>> col = plt.scatter(x, y, np.pi*r**2) > >>> pts2pixels = transforms.Affine2D().scale(72.0/fig.dpi) > >>> col.set_transform(pts2pixels + ax.transData) > >>> plt.axis('equal') > >>> plt.show() > > ------------------------------------------------------------------------------ > 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 > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users >
Anyone else have ideas on how to display large images? Thanks, Adam -- View this message in context: http://www.nabble.com/Segmentation-fault-using-imshow-on-large-image-tp23207792p24152022.html Sent from the matplotlib - users mailing list archive at Nabble.com.
Hi everybody, I have some probleme with Matplot/numpy ! I'm using matplot v0.91.2 and I'm trying to get the hist function working !! I'm using the sample code of matplot: ---------------------------------------------------------- #!/usr/bin/env python import numpy as np import matplotlib.mlab as mlab import matplotlib.pyplot as plt mu, sigma = 100, 15 x = mu + sigma*np.random.randn(10000) # the histogram of the data n, bins, patches = plt.hist(x, 50, normed=1, facecolor='green', alpha=0.75) print len(n) print len(bins) print n print bins # add a 'best fit' line y = mlab.normpdf( bins, mu, sigma) l = plt.plot(bins, y, 'r--', linewidth=1) plt.xlabel('Smarts') plt.ylabel('Probability') plt.title(r'$\mathrm{Histogram\ of\ IQ:}\ \mu=100,\ \sigma=15$') plt.axis([40, 160, 0, 0.03]) plt.grid(True) plt.show() ------------------------------------------------------------ With the same matplot lib but a different version of numpy I get 2 different number of bins !!! How is that possible ??? using the 1.0.4 version of numpy, I get len(n) = 50 and len(bins) = 50 ... what sounds pretty right to me !! but using the version 1.2.1 (the last one I guess) I got len(n) = 50 and len(bins) = 51 What looks weird, doesn't it ??? do you guys have an explanation ?? Anybody faced the same problem ? thx, Jerome -- View this message in context: http://www.nabble.com/Matplot-Numpy-possible-bug-%21-tp24145696p24145696.html Sent from the matplotlib - users mailing list archive at Nabble.com.
On Sun, Jun 21, 2009 at 4:42 PM, Thomas Robitaille<tho...@gm...> wrote: > Hi Jae-Joon, > > Thanks for the quick fix! Just looked in the svn browser, and noticed you > changed line 5290 of axes.py to > > 'o' : (0,0,3), > > Should this not be > > 'o' : (0,3,0), > > ? No, somehow the meaning of each item is different there. The last item is the symbol style, unlike the input parameter for scatter where the second one is the symbol style. For example, 's' : (4,math.pi/4.0,0), # square Regards, -JJ > > Thanks, > > Tom > > On Jun 21, 2009, at 2:59 PM, Jae-Joon Lee wrote: > >> Thanks for the report. >> And, this turned out to be a bug. The symbol style code was simply >> ignored when its value is 3. >> >> While the bug should now be fixed (both in the trunk and the maint. >> branch), you may use marker style like (20,0,0) (or increase the first >> number when symbol is large) for a workaround. >> >> Regards, >> >> -JJ >> >> >> On Sun, Jun 21, 2009 at 1:09 PM, Thomas >> Robitaille<tho...@gm...> wrote: >>> >>> Hi, >>> >>> I'm trying to use the scatter method, making use of the option to specify >>> the marker as a tuple. From the documentation, it would seem that >>> specifying >>> marker=(0,3,0) should draw a circle. However, this is not the case. If >>> you >>> consider the following code: >>> >>> import matplotlib >>> matplotlib.use('Agg') >>> import matplotlib.pyplot as mpl >>> >>> fig = mpl.figure() >>> ax = fig.add_subplot(111) >>> #ax.scatter([x],[y],c='red',marker=(0,3,0.)) >>> ax.scatter([10.],[10.],c='red',marker=(3,3,0.)) >>> ax.scatter([11.],[10.],c='red',marker=(6,3,0.)) >>> ax.set_xlim(5.,15.) >>> fig.savefig('scatter.png') >>> >>> The first ax.scatter causes an error, the second plots a triangle, and >>> the >>> third a hexagon. However, the documentation states that (a) setting the >>> second element to '3' should plot a circle, and (b) the other arguments >>> should be ignored, so the first ax.scatter should not cause an error. >>> >>> Is this a bug, or am I misunderstanding the documentation? >>> >>> Thanks, >>> >>> Thomas >>> -- >>> View this message in context: >>> http://www.nabble.com/possible-bug-with-scatter-tp24136534p24136534.html >>> Sent from the matplotlib - users mailing list archive at Nabble.com. >>> >>> >>> >>> ------------------------------------------------------------------------------ >>> Are you an open source citizen? Join us for the Open Source Bridge >>> conference! >>> Portland, OR, June 17-19. Two days of sessions, one day of unconference: >>> 250ドル. >>> Need another reason to go? 24-hour hacker lounge. Register today! >>> >>> http://ad.doubleclick.net/clk;215844324;13503038;v?http://opensourcebridge.org >>> _______________________________________________ >>> Matplotlib-users mailing list >>> Mat...@li... >>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users >>> > >