matplotlib-users Mailing List for matplotlib (Page 9)

Hi,
I would like to set the zorder on a collection e.g. PatchCollection. By looking at the matplotlib source code in collections.py, I figured that it would be possible to set the zorder attribute of a collection manually, e.g.
p = PatchCollection(...)
p.zorder = ...
but I was wondering whether it would make sense to allow zorder to be passed as a keyword argument when initializing a collection, or when running ax.add_collection, or is there a reason that this was not done?
Cheers,
Tom

Ok, it works ok with pcolormesh, sorry for bothering you.
--
Oleeksandr
2011年4月1日 sanGuziy <guz...@gm...>
>
> Dear list,
>
> I have a question concerning pcolor.
> I am trying to plot a huge 2d array 15000x10000.
> And in top I see that the program already uses 31Gb of ram and it grows.
> Is this expected behaviour?
> Why it needs so much memory?
> the sctipt is simple:
>
> import matplotlib
> import numpy as np
> import matplotlib.pyplot as plt
>
> from numpy.random import rand
>
> import mpl_toolkits.basemap as bm
>
>
> if __name__ == "__main__":
>
> print matplotlib.__version__
> print matplotlib.__revision__
> print matplotlib.__date__
>
>
> print 'basemap'
> print bm.__version__
>
> print 'numpy'
> print np.__version__
>
>
> x = rand(15000,10000)
> #print x
> plt.pcolor(x)
> plt.savefig('test_pcolor.png', bbox_inches = 'tight')
> print "Hello World"
>
> matplotlib version is:
> 1.0.0
> $Revision: 8503 $
> $Date: 2010年07月06日 08:56:31 -0500 (2010年7月06日) $
>
> should I update?
>
> thank you for any help
> --
> Oleksandr Huziy
> --
> View this message in context:
> http://old.nabble.com/pcolor-tp31300362p31300362.html
> Sent from the matplotlib - users mailing list archive at Nabble.com.
>
>
>
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Dear list,
I have a question concerning pcolor.
I am trying to plot a huge 2d array 15000x10000.
And in top I see that the program already uses 31Gb of ram and it grows.
Is this expected behaviour?
Why it needs so much memory?
the sctipt is simple:
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
from numpy.random import rand
import mpl_toolkits.basemap as bm 
if __name__ == "__main__":
 print matplotlib.__version__
 print matplotlib.__revision__
 print matplotlib.__date__
 
 
 print 'basemap'
 print bm.__version__
 
 print 'numpy'
 print np.__version__
 
 x = rand(15000,10000)
 #print x
 plt.pcolor(x)
 plt.savefig('test_pcolor.png', bbox_inches = 'tight')
 print "Hello World"
matplotlib version is:
1.0.0
$Revision: 8503 $
$Date: 2010年07月06日 08:56:31 -0500 (2010年7月06日) $
should I update?
thank you for any help
--
Oleksandr Huziy
-- 
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Sent from the matplotlib - users mailing list archive at Nabble.com.

Thanks so much JJ! It works great.
carlo
Jae-Joon Lee wrote:
> 
> If you want full control of label coordinates, you need to use
> "Axis.set_label_coords" method. For example,
> 
> ax = gca()
> ax.xaxis.set_label_coords(0.5, -0.1)
> 
> And alternative way is to adjust the padding between the axis and the
> label.
> 
> ax.xaxis.labelpad = 0
> 
> Regards,
> 
> -JJ
> 
> 
> On Mon, Mar 21, 2011 at 3:27 AM, andes <czu...@ya...> wrote:
>> x = linspace(0,1,10)
>> y = x**2
>> plot(x, y)
>> xlabel('xname', position=(0.5,0.1)) #<------
>> ylabel('yname', position=(0.1,0.5)) #<------
> 
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Using some real world measurement data that has an underlying comb spectrum with specgram, I have noticed an issue with the amplitude of narrowband signals being affected by the number of FFT points used (NTTF). For areas where there is no signal (in this case near 0 Hz) I see the noise floor (level) drop by about 3 dB as I double the value of NTTF, which is expected. A narrower "resolution" bandwidth will provide lower broadband noise amplitude. However, the signal has a comb spectrum present, and as I increase the value of NTTF the amplitude values of each component of the comb spectrum increases by about 3 dB when I double NFFT. I would expect that the narrowband levels would remain the same and not increase. I expect that the underlying FFT function in specgram scales by 1/NFFT, or maybe it does not (the FFT function in MATLAB does not included the 1/NFFT scaling). To get a "true" amplitude for each frequency bin does the output of specgram need to be scaled by 1/NFFT (or 1/NFFT*1/NFFT since it's a power spectrum)?
Lee

From: Nat Echols [mailto:nat...@gm...] 
Sent: Thursday, March 31, 2011 16:47
 
I'd like to divide the line segments up to get a smoother color gradient, but
the values are dictated by the experiment, not a mathematical function.
<snip>
so I guess what I'm really asking for is a way to add intermediate X,Y values
between every pair of values in 'points'. I can do this myself in Python, but
I assume that's going to be pretty sluggish.
Perhaps (for convenience if not speed) you could use a routine from
scipy.interpolate [1,2], such as interp1d for piecewise linear interpolation.
I imagine that, within each linear piece, you would want the density of x
values to be roughly proportional to the slope.
[1] http://docs.scipy.org/doc/scipy-0.8.x/reference/tutorial/interpolate.html
[2] http://docs.scipy.org/doc/scipy-0.8.x/reference/interpolate.html

Andrew,
After you are done modifying your figure (either in your code or in the
GUI), run the following line:
savefig('example.png', bbox_inches='tight')
Does it save the figure the way you want?
Best,
carlo
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Hi all,
I was surprised today to notice that "subplot" was the slowest part of some plotting code of mine.
On my machine, the last line of the following code puts ten subplots on a figure and records the amount of time it took to make them:
>>> import matplotlib>>> matplotlib.use('AGG')>>> import time>>> from matplotlib.pyplot import *>>> def f():... t = time.time()... clf()... for i in xrange(10):... subplot(5,2,i+1)... return time.time()-t... >>> >>> times = [f() for x in xrange(10)]
This code gives me a bunch of times that are on average about half a second. I expected it to be much faster as I wasn't actually plotting anything.
Is this expected? Can I choose a faster backend or something? I've experimented a little but without success. I realize that 5 hundredths of a second per subplot isn't terrifically slow, ... but I guess I make a lot of plots.
Thanks a lot,
Eddie Schlafly