matplotlib-users Mailing List for matplotlib (Page 3)

On 12/18/11 1:33 PM, Alexis Praga wrote:
> Hi,
>
> I may have found an issue with some map projections.When I try to plot
> markers on a given map, and change the center ofprojection, the
> markers' coordinates are not remapped correctly.If the marker's
> coordinates are not inside the new boundaries, it issimply not
> displayed in some cases.For example, remapping does work with Robin's
> projection and does notwith Miller's.I have attached a small example.
> Thanks !
>
>
> Alexis Praga
>
Alexis: It looks like the coordinates you are plotting with scatter are 
between longitudes -180 and 180, and the map projection goes from 0 to 
360. Adjust your coordinates to lie within 0 and 360, and they will 
show up.
This behavior will be map projection dependent - it will only show up in 
the cylindrical projections where the x coordinate is longitude.
-Jeff

Exactly. But I do not understand why some map projectionsautomatically
adjust the data and others don't.
Alexis
On Sun, Dec 18, 2011 at 3:53 PM, Jeff Whitaker <js...@fa...> wrote:
> On 12/18/11 1:33 PM, Alexis Praga wrote:
>>
>> Hi,
>>
>> I may have found an issue with some map projections.When I try to plot
>> markers on a given map, and change the center ofprojection, the
>> markers' coordinates are not remapped correctly.If the marker's
>> coordinates are not inside the new boundaries, it issimply not
>> displayed in some cases.For example, remapping does work with Robin's
>> projection and does notwith Miller's.I have attached a small example.
>> Thanks !
>>
>>
>> Alexis Praga
>>
> Alexis: It looks like the coordinates you are plotting with scatter are
> between longitudes -180 and 180, and the map projection goes from 0 to 360.
> Adjust your coordinates to lie within 0 and 360, and they will show up.
>
> This behavior will be map projection dependent - it will only show up in the
> cylindrical projections where the x coordinate is longitude.
>
> -Jeff
>
>

Hi,
I may have found an issue with some map projections.When I try to plot
markers on a given map, and change the center ofprojection, the
markers' coordinates are not remapped correctly.If the marker's
coordinates are not inside the new boundaries, it issimply not
displayed in some cases.For example, remapping does work with Robin's
projection and does notwith Miller's.I have attached a small example.
Thanks !
Alexis Praga

On Thu, Dec 15, 2011 at 9:32 PM, Benjamin Root <ben...@ou...> wrote:
>
> On Thu, Dec 15, 2011 at 3:18 PM, Alex Naysmith <yeo...@gm...>wrote:
>
>>
>>
>> On Tue, Dec 13, 2011 at 7:12 PM, Alex Naysmith <yeo...@gm...>wrote:
>>
>>>
>>>
>>> On Mon, Dec 12, 2011 at 7:03 PM, Benjamin Root <ben...@ou...> wrote:
>>>
>>>> On Sat, Dec 10, 2011 at 10:36 AM, Alex Naysmith <
>>>> yeo...@gm...> wrote:
>>>>
>>>>> Hello,
>>>>>
>>>>> I'm trying to plot the stresses in colour of a strained isoparametric
>>>>> element.
>>>>>
>>>>> I have a six noded triangle with vertice coordinates
>>>>> [(xa1,ya1),(xa2,ya2),(xa3,ya3)] = pos_a
>>>>>
>>>>> This triangle deforms and the new coordinate positions are
>>>>> [(xb1,yb1),(xa2,yb2),(xb3,yb3)] = pos_b
>>>>>
>>>>> The remaining nodes are mid nodes also with rest and deformed
>>>>> coordinate positions.
>>>>>
>>>>> To plot the edges of the triangle I use a Jacobian transformation
>>>>> function so that the coordinates of the triangle are in Jacobian
>>>>> coordinates xi1 and xi2 (with xi3 = 1 - xi1 - xi2). This is required as the
>>>>> elements are quadratic with mid-nodes.
>>>>>
>>>>> Each interval is hard coded so that:
>>>>> xi1 = [1.0,0.9,0.8,0.7,0.6,0.5, etc..]
>>>>> xi2 = [0.0,0.1,0.2,0.3,0.4,0.5, etc..]
>>>>>
>>>>> I would like to plot the strains in colour so that the interior of the
>>>>> triangle is filled but I don't want to hard code the Jacobian intervals as
>>>>> this seems an awkward way of doing it.
>>>>>
>>>>> With strain as a function of xi1 and xi2, How can matplotlib provide a
>>>>> continuous interior strain plot of the triangle for all the xi1 and xi2
>>>>> values from 0 to 1?
>>>>>
>>>>> Regards
>>>>>
>>>>> Alex Naysmith
>>>>>
>>>>>
>>>> Alex,
>>>>
>>>> Perhaps if you can provide an example figure, we might be able to
>>>> better help you. Right now, I am having trouble envisioning what you
>>>> describe.
>>>>
>>>> Ben Root
>>>>
>>>> Ben,
>>>
>>> I have created a script that uses one isoparametric triangle as an
>>> example. The triangle nodes undergo a displacement, resulting in strains
>>> inside the triangle. The new script calculates the strains inside the
>>> triangle for a range of xi1 and xi2 barycentric coordinates and returns the
>>> global coordinates with the corresponding strain.
>>>
>>> I would like matplotlib to give me a nice interpolated colour plot of
>>> the strains inside the triangle, but as the output global coordinates are
>>> not aligned in neat rows and columns, I cannot do a straightforward
>>> meshgrid plot with imshow.
>>>
>>> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/files
>>>
>>> There are further comments in the script that may explain things better.
>>>
>>> I want a figure like this:
>>>
>>> http://matplotlib.sourceforge.net/examples/pylab_examples/animation_demo.html
>>> But for a 6 noded quadratic triangle instead of square. The intention is
>>> to have all the triangles in the mesh display their strains with
>>> interpolated colours.
>>>
>>> Regards
>>>
>>> Alex
>>>
>>
>> Hello,
>>
>> I tried using contour and now I have a figure!
>>
>>
>> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/figure.png
>>
>> I can now show clearly what I'm aiming for with this figure. I want the
>> contour fill to remain inside the triangle. There will be a whole mesh of
>> triangles to fill.
>>
>> The updated sample script is here:
>>
>>
>> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/triangle_fill_v2.py
>>
>> I arranged my data points into square X and Y arrays along with the
>> strains in the Z array and then simply P.contourf(X, Y, Z)
>>
>> As the figure shows, it's not there yet as a couple of the corners aren't
>> filled in and there's a big fill outside the triangle. I think this is due
>> to difficulties in translating points from the natural triangle coordinates
>> (barycentric) to the x,y coordinates. But the contour looks correct as it's
>> interpolated between the strain points.
>>
>> Regards
>>
>> Alex
>>
>>
> Alex,
>
> Just curious, have you checked to see if tricontourf() meets your needs?
>
>
> http://matplotlib.sourceforge.net/api/axes_api.html?highlight=tricontourf#matplotlib.axes.Axes.tricontourf
>
> Ben Root
>
> Ben,
I've now succeeded in producing the plot I want using just contour(X,Y,Z):
http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/figure2.png
The script used to produce figure2 is here:
http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/triangle_fill_v3.py
For the X,Y and Z arrays, the upper triangle repeats the same data point.
This doesn't seem to be a problem for the contour(X,Y,Z) function.
The tricontourf() approach may be better, but it involves creating a
triangle mesh inside each element.
The next step will be to plot the strain contours for all the elements in
the mesh.
Regards
Alex Naysmith

I have experimented with path.simplify and can't see any appreciable
improvements.
Incidently, I also experimented with other back ends. I found that all the
back ends
involving Agg behave similarly. However, using the 'GTK' backend it
renders the
whole 1 million points and does it very fast (about 5x faster than Agg
backends).
I also found that gnuplot can render the whole million points very fast
using the
'x11' terminal. I am guessing that both matplotlib's GTK backend and
gnuplot's
'x11' terminal use the hardware accelerated display driver.
David

Here is a script that solves the problem, based on Goyo's input:
******************
#!/usr/bin/env python
"""simple_matplot.py -- 
 Repeatedly plots sinusoids (shifted a bit each time).
 - control returns to the user after each plot;
 - the old plot is erased between cycles.
"""
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
x = np.arange(0.,2.*np.pi,0.02)
y = np.sin(x)
c = "go"
plt.ion()
fig = plt.figure()
ax = fig.add_subplot(1,1,1)
while c.find("go") > -1 :
 try:
 del ax.lines[0]
 except :
 pass
 line, = ax.plot(x,y,'-')
 plt.draw()
 print "stop or go?"
 c = raw_input()
 x = x + 0.5
**********
John Thorstensen wrote:
> 
> I have a sizeable number of python data-inspection scripts that work as
> follows:
> 
> - read some data, or do something with it
> - plot the data
> - query the user on the command line and get a response
> - do what the user commands.
> 
> These use the venerable PGPLOT package for the graphics, but this has been
> static for years and is getting increasingly vulnerable to obsolescence of
> the supporting packages (e.g., numarray).
> 
> I'm having a hard time converting these scripts to matplotlib because when
> you do a show(), the mainloop takes over. It's possible to work around by
> doing a show() for every plot, and then killing the plot manually, but
> after a couple of hundred manual kills this gets tiresome. It would also
> be possible to restructure the code so that the various options get
> controlled by key press events in the plot, I suppose, but the effort
> involved in converting all the scripts would be pretty large.
> 
> So my question: Is there a simple way of getting matplotlib to display a
> plot in a window and then surrender control to the main program, without
> destroying the plot? Something like a method to kill mainloop would be
> ideal. 
> 
> Thanks.
> 
> [Apologies if this is general knowledge -- I couldn't find an answer.]
> 
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How can I change linewidth of this arrow?
arrow1=pylab.annotate("", xytext=(-3500,-1e8),
xy=(-3500,8e8),fontsize=8,arrowprops=dict(arrowstyle="<->"))
Thank you in advance.
Petro

On 12/10/2011 01:12 PM, David Smith wrote:
> I have been working on a program that uses Matplotlib to plot data
> consisting of around one million points. Sometimes the plots succeed but
> often I get an exception: OverFlowError: Agg rendering complexity exceeded.
Are you sure path simplification is running? (i.e. the rcParam 
path.simplify is True)? That generally does a good job of removing 
excess points on the fly. You shouldn't need a development version for 
this to work. 0.99.x or later should be adequate. You're not going to 
"see" a million points at typical screen resolutions anyway.
>
> I can make this message go away by plotting the data in "chunks" as
> illustrated in the demo code below. However, the extra code is a chore
> which I don't think should be necessary - I hope the developers will
> be able to fix this issue sometime soon. I know that the development
> version has some modifications to addressing this issue. I wonder if it is
> expected to make the problem go away?
>
> By the way, this plot takes about 30 seconds to render on my I7 2600k.
> The main program reaches the show() statement quickly and prints
> "Done plotting?". Then I see that the program reaches 100% usage
> on one CPU core (4 real, 8 virtual on the 2600k) until the plot is
> displayed. I wonder if there is any way to persuade Matplotlib to run
> some of the chunks in parallel so as to use more CPU cores?
That would be great, but very difficult. The Python parts of the 
problem are tricky to parallelize due to the GIL. The Agg part of the 
problem will be difficult to parallelize unless there is a trivial way 
to chuck the plotted lines into parts before stroking -- each chunk 
could be rendered to its own buffer and then blended together in a final 
step. But all that is academic at this point -- there's no code to do 
such a thing now.
>
> Plotting something other than random data, the plots run faster and
> the maximum chunk size is smaller. The maximum chunk size
> also depends on the plot size - it is smaller for larger plots. I am
> wondering if I could use this to plot course and fine versions of the
> plots. The course plot is zoomed in version of the small-sized raster.
> That would be better than decimation as all the points would at least
> be there.
I think what you're seeing is the effect of the path simplification 
algorithm. The number of points that it removes depends on the density 
of the points and the resolution of the output image. It's hard to 
predict exactly how many points it will remove.
Mike
>
> Thanks in advance,
>
> David
>
> --------------------------- start code ---------------------------------
> ## Demo program shows how to "chunk" plots to avoid the exception:
> ##
> ## OverflowError: Agg rendering complexity exceeded.
> ## Consider downsampling or decimating your data.
> ##
> ## David Smith December 2011.
>
> from pylab import *
> import numpy as np
>
> nPts=600100
> x = np.random.rand(nPts)
> y = np.random.rand(nPts)
>
> ## This seems to always succeed if Npts<= 20000, but fails
> ## for Npts> 30000. For points between, it sometimes succeeds
> ## and sometimes fails.
> figure(1)
> plot (x, y)
>
> ## Chunking the plot alway succeeds.
> figure(2)
> chunk_size=20000
> iStarts=range(x.size/chunk_size)
> for iStart in iStarts:
> print "Plotting chunk starting at %d\n" % iStart
> plot(x[iStart:iStart+chunk_size], y[iStart:iStart+chunk_size], '-b')
>
> left_overs = nPts % chunk_size
> if left_overs> 0:
> print "Leftovers %d points\n" % left_overs
> plot(x[-left_overs-1:], y[-left_overs-1:], '-r')
>
> print "done plotting?"
> show()
> ---------------------------------- end code ------------------------
> Please don't reply to this post "It is rediculous to plot 1 million points on
> screen". I am routinely capturing million-point traces from oscilloscopes and
> other test equipment and to I need to be able to spot features in the
> data (glitches if you will) that may not show up plotting decimated data.
> I can then zoom the plot to inspect these features.
>
> ------------------------------------------------------------------------------
> Learn Windows Azure Live! Tuesday, Dec 13, 2011
> Microsoft is holding a special Learn Windows Azure training event for
> developers. It will provide a great way to learn Windows Azure and what it
> provides. You can attend the event by watching it streamed LIVE online.
> Learn more at http://p.sf.net/sfu/ms-windowsazure
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users

On Thu, Dec 15, 2011 at 3:18 PM, Alex Naysmith <yeo...@gm...>wrote:
>
>
> On Tue, Dec 13, 2011 at 7:12 PM, Alex Naysmith <yeo...@gm...>wrote:
>
>>
>>
>> On Mon, Dec 12, 2011 at 7:03 PM, Benjamin Root <ben...@ou...> wrote:
>>
>>> On Sat, Dec 10, 2011 at 10:36 AM, Alex Naysmith <yeo...@gm...
>>> > wrote:
>>>
>>>> Hello,
>>>>
>>>> I'm trying to plot the stresses in colour of a strained isoparametric
>>>> element.
>>>>
>>>> I have a six noded triangle with vertice coordinates
>>>> [(xa1,ya1),(xa2,ya2),(xa3,ya3)] = pos_a
>>>>
>>>> This triangle deforms and the new coordinate positions are
>>>> [(xb1,yb1),(xa2,yb2),(xb3,yb3)] = pos_b
>>>>
>>>> The remaining nodes are mid nodes also with rest and deformed
>>>> coordinate positions.
>>>>
>>>> To plot the edges of the triangle I use a Jacobian transformation
>>>> function so that the coordinates of the triangle are in Jacobian
>>>> coordinates xi1 and xi2 (with xi3 = 1 - xi1 - xi2). This is required as the
>>>> elements are quadratic with mid-nodes.
>>>>
>>>> Each interval is hard coded so that:
>>>> xi1 = [1.0,0.9,0.8,0.7,0.6,0.5, etc..]
>>>> xi2 = [0.0,0.1,0.2,0.3,0.4,0.5, etc..]
>>>>
>>>> I would like to plot the strains in colour so that the interior of the
>>>> triangle is filled but I don't want to hard code the Jacobian intervals as
>>>> this seems an awkward way of doing it.
>>>>
>>>> With strain as a function of xi1 and xi2, How can matplotlib provide a
>>>> continuous interior strain plot of the triangle for all the xi1 and xi2
>>>> values from 0 to 1?
>>>>
>>>> Regards
>>>>
>>>> Alex Naysmith
>>>>
>>>>
>>> Alex,
>>>
>>> Perhaps if you can provide an example figure, we might be able to better
>>> help you. Right now, I am having trouble envisioning what you describe.
>>>
>>> Ben Root
>>>
>>> Ben,
>>
>> I have created a script that uses one isoparametric triangle as an
>> example. The triangle nodes undergo a displacement, resulting in strains
>> inside the triangle. The new script calculates the strains inside the
>> triangle for a range of xi1 and xi2 barycentric coordinates and returns the
>> global coordinates with the corresponding strain.
>>
>> I would like matplotlib to give me a nice interpolated colour plot of the
>> strains inside the triangle, but as the output global coordinates are not
>> aligned in neat rows and columns, I cannot do a straightforward meshgrid
>> plot with imshow.
>>
>> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/files
>>
>> There are further comments in the script that may explain things better.
>>
>> I want a figure like this:
>>
>> http://matplotlib.sourceforge.net/examples/pylab_examples/animation_demo.html
>> But for a 6 noded quadratic triangle instead of square. The intention is
>> to have all the triangles in the mesh display their strains with
>> interpolated colours.
>>
>> Regards
>>
>> Alex
>>
>
> Hello,
>
> I tried using contour and now I have a figure!
>
>
> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/figure.png
>
> I can now show clearly what I'm aiming for with this figure. I want the
> contour fill to remain inside the triangle. There will be a whole mesh of
> triangles to fill.
>
> The updated sample script is here:
>
>
> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/triangle_fill_v2.py
>
> I arranged my data points into square X and Y arrays along with the
> strains in the Z array and then simply P.contourf(X, Y, Z)
>
> As the figure shows, it's not there yet as a couple of the corners aren't
> filled in and there's a big fill outside the triangle. I think this is due
> to difficulties in translating points from the natural triangle coordinates
> (barycentric) to the x,y coordinates. But the contour looks correct as it's
> interpolated between the strain points.
>
> Regards
>
> Alex
>
>
Alex,
Just curious, have you checked to see if tricontourf() meets your needs?
http://matplotlib.sourceforge.net/api/axes_api.html?highlight=tricontourf#matplotlib.axes.Axes.tricontourf
Ben Root

On Tue, Dec 13, 2011 at 7:12 PM, Alex Naysmith <yeo...@gm...>wrote:
>
>
> On Mon, Dec 12, 2011 at 7:03 PM, Benjamin Root <ben...@ou...> wrote:
>
>> On Sat, Dec 10, 2011 at 10:36 AM, Alex Naysmith <yeo...@gm...>wrote:
>>
>>> Hello,
>>>
>>> I'm trying to plot the stresses in colour of a strained isoparametric
>>> element.
>>>
>>> I have a six noded triangle with vertice coordinates
>>> [(xa1,ya1),(xa2,ya2),(xa3,ya3)] = pos_a
>>>
>>> This triangle deforms and the new coordinate positions are
>>> [(xb1,yb1),(xa2,yb2),(xb3,yb3)] = pos_b
>>>
>>> The remaining nodes are mid nodes also with rest and deformed coordinate
>>> positions.
>>>
>>> To plot the edges of the triangle I use a Jacobian transformation
>>> function so that the coordinates of the triangle are in Jacobian
>>> coordinates xi1 and xi2 (with xi3 = 1 - xi1 - xi2). This is required as the
>>> elements are quadratic with mid-nodes.
>>>
>>> Each interval is hard coded so that:
>>> xi1 = [1.0,0.9,0.8,0.7,0.6,0.5, etc..]
>>> xi2 = [0.0,0.1,0.2,0.3,0.4,0.5, etc..]
>>>
>>> I would like to plot the strains in colour so that the interior of the
>>> triangle is filled but I don't want to hard code the Jacobian intervals as
>>> this seems an awkward way of doing it.
>>>
>>> With strain as a function of xi1 and xi2, How can matplotlib provide a
>>> continuous interior strain plot of the triangle for all the xi1 and xi2
>>> values from 0 to 1?
>>>
>>> Regards
>>>
>>> Alex Naysmith
>>>
>>>
>> Alex,
>>
>> Perhaps if you can provide an example figure, we might be able to better
>> help you. Right now, I am having trouble envisioning what you describe.
>>
>> Ben Root
>>
>> Ben,
>
> I have created a script that uses one isoparametric triangle as an
> example. The triangle nodes undergo a displacement, resulting in strains
> inside the triangle. The new script calculates the strains inside the
> triangle for a range of xi1 and xi2 barycentric coordinates and returns the
> global coordinates with the corresponding strain.
>
> I would like matplotlib to give me a nice interpolated colour plot of the
> strains inside the triangle, but as the output global coordinates are not
> aligned in neat rows and columns, I cannot do a straightforward meshgrid
> plot with imshow.
>
> http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/files
>
> There are further comments in the script that may explain things better.
>
> I want a figure like this:
>
> http://matplotlib.sourceforge.net/examples/pylab_examples/animation_demo.html
> But for a 6 noded quadratic triangle instead of square. The intention is
> to have all the triangles in the mesh display their strains with
> interpolated colours.
>
> Regards
>
> Alex
>
Hello,
I tried using contour and now I have a figure!
 http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/figure.png
I can now show clearly what I'm aiming for with this figure. I want the
contour fill to remain inside the triangle. There will be a whole mesh of
triangles to fill.
The updated sample script is here:
http://bazaar.launchpad.net/~eckeroo/misc/mpl_scripts/view/head:/triangle_fill_v2.py
I arranged my data points into square X and Y arrays along with the strains
in the Z array and then simply P.contourf(X, Y, Z)
As the figure shows, it's not there yet as a couple of the corners aren't
filled in and there's a big fill outside the triangle. I think this is due
to difficulties in translating points from the natural triangle coordinates
(barycentric) to the x,y coordinates. But the contour looks correct as it's
interpolated between the strain points.
Regards
Alex

I'm not having any success in getting the matplotlib table commands to work. Here's an example of what I'd like to do:
Can anyone help with the table construction code? Thanks
import pylab as plt
plt.figure()
ax=plt.gca()
y=[1,2,3,4,5,4,3,2,1,1,1,1,1,1,1,1]
plt.plot([10,10,14,14,10],[2,4,4,2,2],'r')
col_labels=['col1','col2','col3']
row_labels=['row1','row2','row3']
table_vals=[11,12,13,21,22,23,31,32,33]
# the rectangle is where I want to place the table
plt.text(11,4.1,'Table Title',size=8)
plt.plot(y)
plt.show()

2011年12月15日 John Thorstensen <joh...@da...>:
> So my question: Is there a simple way of getting matplotlib to display a
> plot in a window and then surrender control to the main program, without
> destroying the plot? Something like a method to kill mainloop would be
> ideal.
Just make sure you use interactive mode and get rid of the show calls:
import matplotlib.pyplot as plt
plt.ion() # set interactive mode
plt.plot(foo) # plot something
bar() # do stuff while the plot is visible
plt.close() # if you want to close the plot window from the script
This works in mpl 1.1.0 al least with tkagg, gtkagg and qt4agg backends.
Goyo

Thank you Benjamin.
Yes, I understand the problem.
Thank you for the help. I am able to use Annotate for my specific problem: I
want to have a line starting from the end of the text to make the tick
labels with a left alignment.
However, a feature to get the boundary box would be useful for this, to
ensure that the plot is tight. 
Best regards 
Keld
Benjamin Root-2 wrote:
> 
> On Wednesday, December 14, 2011, Keld Lundgaard <kel...@gm...>
> wrote:
>>
>> Hi
>>
>> How do I get the information about the endpoint of a text element?
>>
>> Example:
>>
>> import matplotlib.pylab as plt
>> text = plt.text(0.5,0.5,'hi')
>> plt.show()
>>
>> In the dataTrans coordinates, where does the text string ends?
>>
>>
>> I have spent some time now trying to solve this, seemingly simple
>> problem,
>> so I appreciate any help a lot!
>>
>>
>> Thanks in advance!
>>
>> / Keld
> 
> It is actually a very difficult thing to do, and is why mpl does not
> provide a mechanism to find out until after rendering. The problem is
> that
> -- in the general case -- it is not possible to know ahead of time how
> much
> space some text will take up for an arbitrary font, style and size.
> 
> You can have the figure get rendered, and then you can query the text's
> bounding box, if I understand correctly. Maybe the code for
> tight_layout()
> might show how it does this.
> 
> Ben Root
> 
> ------------------------------------------------------------------------------
> 10 Tips for Better Server Consolidation
> Server virtualization is being driven by many needs. 
> But none more important than the need to reduce IT complexity 
> while improving strategic productivity. Learn More! 
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> 
> 
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On Wed, Dec 14, 2011 at 10:20 AM, Peter Liebetraut
<pet...@im...> wrote:
> Hi
>
> Latex rendering looks OK here.
>
>> In [9]: matplotlib.__version__
>> Out[9]: '1.0.0'
Just updated to the last version from github, and now works OK.

On Thu, Dec 15, 2011 at 12:33 PM, Chao YUE <cha...@gm...> wrote:
> Dear matplotlib users,
>
> How can I surpress the figure pop out when I make plot within the ipython
> interactive shell?
> suppose I make a figure first and I want to save it:
>
> fig=plt.figure()
> ax=fig.add_subplot(111)
> ax.plot(np.arange(10))
> fig.savefig('fig1.png')
> ###actually above is only an example and usually I use loop to make many
> figures.
>
Try adding 'plt.close(fig)' after you save the figure, or just plt.close()
if you want to close everything.

Dear matplotlib users,
How can I surpress the figure pop out when I make plot within the ipython
interactive shell?
suppose I make a figure first and I want to save it:
fig=plt.figure()
ax=fig.add_subplot(111)
ax.plot(np.arange(10))
fig.savefig('fig1.png')
###actually above is only an example and usually I use loop to make many
figures.
then I want to check another plot:
plt.plot(np.arange(10))
plt.show()
The problem is, when I use plt.show() to check the recently made plot, all
the figures I saved for the last session will pop out and I have to click
on the close icon to close them one by one,
which can be really annoying....
any idea would be greatly appreciated!
cheers,
Chao
-- 
***********************************************************************************
Chao YUE
Laboratoire des Sciences du Climat et de l'Environnement (LSCE-IPSL)
UMR 1572 CEA-CNRS-UVSQ
Batiment 712 - Pe 119
91191 GIF Sur YVETTE Cedex
Tel: (33) 01 69 08 29 02; Fax:01.69.08.77.16
************************************************************************************

Eric,
Good news I think I got it to work. So using the same code I sent you 
originally, I applied the following changes:
1. Install matplotlib from git (this did fix things that I wasn't noticing)
2. Add c.draw() before "c.copy_from_bbox"
3. Copy "f.bbox" instead of "a.bbox" (I think this makes sense since I 
want it to hold on to the title,ticks, and labels, a.bbox is only the 
content inside the axis rectangle)
4. Restore f_bbox, blit f.bbox.
...and it seems to work. The git version of mpl did fix this, although 
restoring just the a.bbox was only keeping the axis rectangle so it made 
it look like even worse of a bug. Using the new version I noticed this 
and then started using f.bbox which seems to work the way I want it.
Now, my final question is: Is this actually doing what I want 
performance-wise. When I blit just f.bbox, is it really only repainting 
the updated line or is it redrawing most of the figure? If you need a 
copy of the new version of my test code let me know. Thanks for any 
more clarity/help you can give.
-Dave
P.S. Is there a book or tutorial or website where I can learn more about 
how the rendering/painting of stuff like this works. For example, if I 
could better understand why your bug fix was needed.
On 12/15/11 10:04 AM, David Hoese wrote:
> Oops forgot to change subject.
>
> On 12/15/11 10:02 AM, David Hoese wrote:
>> Eric,
>>
>> I installed mpl from git ("git clone 
>> git://github.com/matplotlib/matplotlib.git", unless I was suppose to 
>> use one of the branches) and same problem. I looked at my code again 
>> and thought there should be a canvas.draw() before calling 
>> "c.copy_from_bbox(a.bbox)", but still the same problem. However, I 
>> did have it work the first time I added "c.draw()" and used the git 
>> mpl, and by work I mean that everything stayed visible on the figure.
>>
>> When people start coming in to my work I'll ask them to run my sample 
>> code and see what happens. It almost seems like mpl is handling the 
>> window activation event funny, is there an easy way to print out the 
>> callbacks being used by a mpl figure? For now, I will subclass 
>> QApplication, and implement "notify()" to print out events as they 
>> come in, but still...this is just weird. Thanks.
>>
>> -Dave
>>
>> On 12/14/11 10:30 PM, Eric Firing wrote:
>>> David,
>>>
>>> It works for me on linux with mpl from git. I haven't tried to figure
>>> it out, but it is conceivable that the problem you are seeing was fixed
>>> with this:
>>>
>>> commit b624546ae60dc5878e75a32f41a160d383548b8f
>>> Author: Eric Firing<ef...@ha...>
>>> Date: Tue Oct 18 08:06:21 2011 -1000
>>>
>>> backend_qt4agg: draw() immediately calls FigureCanvasAgg.draw()
>>>
>>> This is the latest in a series of modifications to the Qt4Agg
>>> drawing strategy going back several years. It simplifies the
>>> code and should solve the problem introduced in 6938b0025; that
>>> is, delaying the Agg draw operation until the paintEvent breaks
>>> code expecting that draw operation to have occurred immediately.
>>> The problem was reported here:
>>> http://sourceforge.net/mailarchive/message.php?msg_id=28245744
>>>
>>> Eric
>>
>

On Wed, Dec 14, 2011 at 11:34 PM, Benjamin Root <ben...@ou...> wrote:
>
>
> On Wednesday, December 14, 2011, Jason Grout <jas...@cr...>
> wrote:
> > On 12/14/11 6:33 PM, Justin wrote:
> >
> >> Any suggestions or places to find a gorgeous pie chart, let me know...
> >
> > I'd probably use Excel or OpenOffice if I were you.
> >
> > Jason
> >
>
> There are some neat effects that can be done with the AGG filter. There
> should be a page about AGG filters on the website.
>
> As for text placement, if it is just a one-off figure, you could
> explicitly code the placement yourself. It is always hard to judge when to
> get the general solution, and when to hack out the one-offs.
>
> Ben Root
>
The page I was referring to:
http://matplotlib.sourceforge.net/examples/pylab_examples/demo_agg_filter.html
I could have sworn there was another page that talks about agg filters in
further details, but I can't seem to find it.
Ben Root

Oops forgot to change subject.
On 12/15/11 10:02 AM, David Hoese wrote:
> Eric,
>
> I installed mpl from git ("git clone 
> git://github.com/matplotlib/matplotlib.git", unless I was suppose to 
> use one of the branches) and same problem. I looked at my code again 
> and thought there should be a canvas.draw() before calling 
> "c.copy_from_bbox(a.bbox)", but still the same problem. However, I 
> did have it work the first time I added "c.draw()" and used the git 
> mpl, and by work I mean that everything stayed visible on the figure.
>
> When people start coming in to my work I'll ask them to run my sample 
> code and see what happens. It almost seems like mpl is handling the 
> window activation event funny, is there an easy way to print out the 
> callbacks being used by a mpl figure? For now, I will subclass 
> QApplication, and implement "notify()" to print out events as they 
> come in, but still...this is just weird. Thanks.
>
> -Dave
>
> On 12/14/11 10:30 PM, Eric Firing wrote:
>> David,
>>
>> It works for me on linux with mpl from git. I haven't tried to figure
>> it out, but it is conceivable that the problem you are seeing was fixed
>> with this:
>>
>> commit b624546ae60dc5878e75a32f41a160d383548b8f
>> Author: Eric Firing<ef...@ha...>
>> Date: Tue Oct 18 08:06:21 2011 -1000
>>
>> backend_qt4agg: draw() immediately calls FigureCanvasAgg.draw()
>>
>> This is the latest in a series of modifications to the Qt4Agg
>> drawing strategy going back several years. It simplifies the
>> code and should solve the problem introduced in 6938b0025; that
>> is, delaying the Agg draw operation until the paintEvent breaks
>> code expecting that draw operation to have occurred immediately.
>> The problem was reported here:
>> http://sourceforge.net/mailarchive/message.php?msg_id=28245744
>>
>> Eric
>

Eric,
I installed mpl from git ("git clone 
git://github.com/matplotlib/matplotlib.git", unless I was suppose to use 
one of the branches) and same problem. I looked at my code again and 
thought there should be a canvas.draw() before calling 
"c.copy_from_bbox(a.bbox)", but still the same problem. However, I did 
have it work the first time I added "c.draw()" and used the git mpl, and 
by work I mean that everything stayed visible on the figure.
When people start coming in to my work I'll ask them to run my sample 
code and see what happens. It almost seems like mpl is handling the 
window activation event funny, is there an easy way to print out the 
callbacks being used by a mpl figure? For now, I will subclass 
QApplication, and implement "notify()" to print out events as they come 
in, but still...this is just weird. Thanks.
-Dave
On 12/14/11 10:30 PM, Eric Firing wrote:
> David,
>
> It works for me on linux with mpl from git. I haven't tried to figure
> it out, but it is conceivable that the problem you are seeing was fixed
> with this:
>
> commit b624546ae60dc5878e75a32f41a160d383548b8f
> Author: Eric Firing<ef...@ha...>
> Date: Tue Oct 18 08:06:21 2011 -1000
>
> backend_qt4agg: draw() immediately calls FigureCanvasAgg.draw()
>
> This is the latest in a series of modifications to the Qt4Agg
> drawing strategy going back several years. It simplifies the
> code and should solve the problem introduced in 6938b0025; that
> is, delaying the Agg draw operation until the paintEvent breaks
> code expecting that draw operation to have occurred immediately.
> The problem was reported here:
> http://sourceforge.net/mailarchive/message.php?msg_id=28245744
>
> Eric

On Tue, Dec 13, 2011 at 1:12 PM, Martella, C. <c.m...@vu...> wrote:
> Hello,
>
> I have 2D array where each (x, y) cell represents the height of that point
> on the Z axis (you can think of it as the map of some mountain chain).
> I'd like to get the plot_surface() of this data. What I don't understand
> in particular is the content of the X, Y ,Z array parameter in my
> particular example.
>
> Can anybody elaborate on this please?
>
>
Claudio,
X and Y represent the coordinates of the height data (which is Z). X and Y
must be the same shape as the Z data. For example:
# Create 1-D coordinate arrays
x = np.arange(-10, 10, 0.5) # length 40
y = np.arange(-25, 25, 2.5) # length 20
# Create 2-D arrays from the coordinate arrays
# The shape of X and Y will be (20, 40)
# i.e., 20 rows and 40 columns.
X, Y = np.meshgrid(x, y)
# This will also have shape of (20, 40)
Z = np.sqrt(X**2 + Y**2)
ax.plot_surface(X, Y, Z)
Now, a more complicated example shows that the X and Y coordinates do not
have to be monotonic, but the coordinates must be neighboring in both
euclidean space and in the data array, or else the surface will look
"torn". This example on the website shows how to plot a sphere by using a
parametric representation of the surface:
http://matplotlib.sourceforge.net/examples/mplot3d/surface3d_demo2.html
I hope this helps!
Ben Root

This software can be intersting for you:
http://www.sofastatistics.com/home.php
2011年12月15日 Benjamin Root <ben...@ou...>:
>
>
> On Wednesday, December 14, 2011, Jason Grout <jas...@cr...>
> wrote:
>> On 12/14/11 6:33 PM, Justin wrote:
>>
>>> Any suggestions or places to find a gorgeous pie chart, let me know...
>>
>> I'd probably use Excel or OpenOffice if I were you.
>>
>> Jason
>>
>
> There are some neat effects that can be done with the AGG filter. There
> should be a page about AGG filters on the website.
>
> As for text placement, if it is just a one-off figure, you could explicitly
> code the placement yourself. It is always hard to judge when to get the
> general solution, and when to hack out the one-offs.
>
> Ben Root
>
> ------------------------------------------------------------------------------
> 10 Tips for Better Server Consolidation
> Server virtualization is being driven by many needs.
> But none more important than the need to reduce IT complexity
> while improving strategic productivity. Learn More!
> http://www.accelacomm.com/jaw/sdnl/114/51507609/
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
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>

I have a sizeable number of python data-inspection scripts that work as
follows:
- read some data, or do something with it
- plot the data
- query the user on the command line and get a response
- do what the user commands.
These use the venerable PGPLOT package for the graphics, but this has been
static for years and is getting increasingly vulnerable to obsolescence of
the supporting packages (e.g., numarray).
I'm having a hard time converting these scripts to matplotlib because when
you do a show(), the mainloop takes over. It's possible to work around by
doing a show() for every plot, and then killing the plot manually, but after
a couple of hundred manual kills this gets tiresome. It would also be
possible to restructure the code so that the various options get controlled
by key press events in the plot, I suppose, but the effort involved in
converting all the scripts would be pretty large.
So my question: Is there a simple way of getting matplotlib to display a
plot in a window and then surrender control to the main program, without
destroying the plot? Something like a method to kill mainloop would be
ideal. 
Thanks.
[Apologies if this is general knowledge -- I couldn't find an answer.]
-- 
View this message in context: http://old.nabble.com/Non-interactive-plotting-to-window--tp32981792p32981792.html
Sent from the matplotlib - users mailing list archive at Nabble.com.

On Wednesday, December 14, 2011, Jason Grout <jas...@cr...>
wrote:
> On 12/14/11 6:33 PM, Justin wrote:
>
>> Any suggestions or places to find a gorgeous pie chart, let me know...
>
> I'd probably use Excel or OpenOffice if I were you.
>
> Jason
>
There are some neat effects that can be done with the AGG filter. There
should be a page about AGG filters on the website.
As for text placement, if it is just a one-off figure, you could explicitly
code the placement yourself. It is always hard to judge when to get the
general solution, and when to hack out the one-offs.
Ben Root

On 12/14/11 6:33 PM, Justin wrote:
> Any suggestions or places to find a gorgeous pie chart, let me know...
I'd probably use Excel or OpenOffice if I were you.
Jason