matplotlib-users Mailing List for matplotlib (Page 2)

So what is to be done here? It seems to me that at least the factor of two should be
fixed for one-sided PSDs, and the 1/fs normalization difference with Matlab
documented. Ideally, I'd think this normalization would be on by default, with the
option to turn it off.

Are you planning to submit a patch, or shall I look into it? If I submit a patch, does matplotlib
require a copyright assignment?

Thanks,
Matt

________________________________________
From: Ryan May [rm...@gm...]
Sent: Monday, December 01, 2008 3:31 PM
To: Fago, Matt - AES
Cc: Matplotlib Users
Subject: Re: [Matplotlib-users] Matplotlib PSD bug?

Fago, Matt - AES wrote:
> I suppose the issue is: what is correct? Or is it a matter of definition?
>
> I don't have Stoica and Moses, but Bendat and Piersol eqn 11.102:
>
>     One_Sided_PSD = 2/(n_d * N * dt) * Sum(FFT^2)
>
> where there are n_d is the number of averages and N is the number of points in the FFT.
> That seems to be scaling by the length? I'm fairly certain that the factor of two (as shown
> above) is required for a one-sided PSD, as that comes from 'removing' the negative
> frequency range.
>
> Note that Matlab shows such scaling (by 2/L) even when computing the power spectra directly
> from a raw (unaveraged) FFT:
>
>    http://www.mathworks.com/support/tech-notes/1700/1702.html
>
> To me, as Matplotlib is striving to be Matlab-compatible, the default behaviour should be to
> give results as close to the Matlab implementation as possible. One could always have an
> option to turn off the scaling.

Yeah, scaling by a factor of two for one-sided is definitely correct now
that I think about it.  Note, however, that the scaling by the length is
not the problem. In fact, the current psd implementation does this
correctly when it corrects for the power in the window.  The problem
stems from matlab scaling by the sampling frequency.  Stoica and Moses
don't do this, nor does Welch 1967.  I'm all for doing things in a
Matlab-compatible way--when they're correct.  One of the reasons I
stopped using Matlab was that it tried to be too smart for me.

> Note that the Matplotlib results also seem to have significantly less frequency resolution than
> the Matlab results. Is this the case, or am I not using noffset, nfft, and pad_to correctly?

It's not that you're using those incorrectly, but rather that you failed
to notice that the default window is the Hanning window, not
rectangular.  Try adding window=mlab.window_none to the call. (Sorry, I
meant to mention that before in my reply.

Ryan

--
Ryan May
Graduate Research Assistant
School of Meteorology
University of Oklahoma

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Cool!  Thanks.  pyplot.subplots_adjust seems to be some kind of "super"
function that controls margins for the entire figure as well as spacing
between the subplots.

On Mon, Dec 8, 2008 at 1:32 AM, Pierre GM <pgm...@gm...> wrote:

>
> On Dec 8, 2008, at 1:09 AM, Roy H. Han wrote:
>
> > I figured it out.
> > Specifically, I need to use pylab.Axes(figure, [left, bottom, width,
> > height]) where each value in the frame is between 0 and 1
> > [.2,.1,.7,.8]
> > means
> > 20 percent margin on left
> > 10 percent margin on bottom
> > 10 percent margin on right (1 - .2 - .7)
> > 10 percent margin on top (1 - .1 - .8)
>
>
> Yep, you got it.
> Now, you can also use the simpler `pyplot.subplots_adjust` as a
> function, or as a method for your figure. That way, you're not tied to
> defining the positions at the creation of your subplot.
>
>
>
>
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Dear Jeff,

2008/12/8 Jeff Whitaker <js...@fa...>:
> Mauro:  I realized that this is actually possible with the Cassini
> projection (the transverse aspect of the cylindrical equidistant).
>
> Don't know why you would actually do it though, if you want to show polar
> regions you should probably use the polar stereographic maps (npstere,
> spstere).

I had not yet had time to try Cassini's before receiving your posting.
As Scott already pointed out, no doubt the polar projections are more
adequate to representing the poles (pretty obvious!). I just intended
to try making the most parsimonious use of the Basemap parameters :-).
The polar stereographic are nice, but what about the polar azimuthal,
that also look good? Just curious.

Best regards,

-- 
Dr. Mauro J. Cavalcanti
Ecoinformatics Studio
P.O. Box 46521, CEP 20551-970
Rio de Janeiro, RJ, BRASIL
E-mail: mau...@gm...
Web: http://studio.infobio.net
Linux Registered User #473524 * Ubuntu User #22717
"Life is complex. It consists of real and imaginary parts."

Dear Jeff,

2008/12/8 Jeff Whitaker <js...@fa...>:
> Mauro:  I just updated SVN basemap so you can specify just lon_0  for all
> cylindrical projections (cyl,gall,merc,mill) to get a global map centered on
> lon_0 (implying
> llcrnrlat=-90,urcrnrlat=90,llcrnrlon=lon_0-180,urcrnrlon=lon_0+180).

Excellent! Will however have to change my Basemap installation to the
SVN repository.

> No.  Cylindrical projections can't cross the poles.

Just figured this out, by trial-and-error. BTW, these explorations led
me to comment that biogeographers (almost all of them biologists like
myself) have seldom (or never!) paid attention to the question of map
projections! This by itself should merit a scientific paper!

So, Basemap may also be lending to scientific insights... Exciting, isn't it?

More soon.... ;-)

Best wishes,

-- 
Dr. Mauro J. Cavalcanti
Ecoinformatics Studio
P.O. Box 46521, CEP 20551-970
Rio de Janeiro, RJ, BRASIL
E-mail: mau...@gm...
Web: http://studio.infobio.net
Linux Registered User #473524 * Ubuntu User #22717
"Life is complex. It consists of real and imaginary parts."

Mauro Cavalcanti wrote:
> Dear Jeff,
>
> 2008/12/7 Jeff Whitaker <js...@fa...>:
>   
>> Mauro:  Just set the llrncrlat,urcrnrlon appropriately.  For instance,
>> llcrnrlon=0, urcrnrlon=360 will produce a map centered on the dateline while
>> llcrnrlon=-180, urcrnrlon=180 will produce a map centered on Greenwich.
>>     
>
> Thanks! I had not figured out that the decimal coordinates used by
> Basemap range from 0 to 360 (and not just from -180 to 180). Using
> this tip, I could also create a map centered on the Indian Ocean. BTW,
> by varying the llcrnrlat and urcrnrlat, can one also create maps
> centered on the North and South poles using a Equirectangular
> projection?
>   

Mauro:  I realized that this is actually possible with the Cassini 
projection (the transverse aspect of the cylindrical equidistant).

from mpl_toolkits.basemap import Basemap
import matplotlib.pyplot as plt
width=20000000
m = Basemap(lon_0=0,lat_0=0,projection='cass',\
            width=0.75*width,height=2*width)
m.drawcoastlines()
m.fillcontinents(color='coral',lake_color='aqua')
m.drawparallels(range(-80,81,21))
m.drawmeridians(range(-180,181,60))
m.drawmapboundary(fill_color='aqua')
plt.show()

Don't know why you would actually do it though, if you want to show 
polar regions you should probably use the polar stereographic maps 
(npstere, spstere).


-Jeff
> With best regards,
>
>   


-- 
Jeffrey S. Whitaker         Phone : (303)497-6313
NOAA/OAR/CDC  R/PSD1        FAX   : (303)497-6449
325 Broadway                Boulder, CO, USA 80305-3328

Mauro Cavalcanti wrote:
> Dear Jeff,
>
> 2008/12/7 Jeff Whitaker <js...@fa...>:
>   
>> Mauro:  Just set the llrncrlat,urcrnrlon appropriately.  For instance,
>> llcrnrlon=0, urcrnrlon=360 will produce a map centered on the dateline while
>> llcrnrlon=-180, urcrnrlon=180 will produce a map centered on Greenwich.
>>     
>
> Thanks! I had not figured out that the decimal coordinates used by
> Basemap range from 0 to 360 (and not just from -180 to 180). 

Mauro:  I just updated SVN basemap so you can specify just lon_0  for 
all cylindrical projections (cyl,gall,merc,mill) to get a global map 
centered on lon_0 (implying 
llcrnrlat=-90,urcrnrlat=90,llcrnrlon=lon_0-180,urcrnrlon=lon_0+180).
> Using
> this tip, I could also create a map centered on the Indian Ocean. BTW,
> by varying the llcrnrlat and urcrnrlat, can one also create maps
> centered on the North and South poles using a Equirectangular
> projection?
>   
No.  Cylindrical projections can't cross the poles.

-Jeff
> With best regards,
>
>   


-- 
Jeffrey S. Whitaker         Phone : (303)497-6313
NOAA/OAR/CDC  R/PSD1        FAX   : (303)497-6449
325 Broadway                Boulder, CO, USA 80305-3328

Dear Jeff,

2008/12/7 Jeff Whitaker <js...@fa...>:
> Mauro:  Just set the llrncrlat,urcrnrlon appropriately.  For instance,
> llcrnrlon=0, urcrnrlon=360 will produce a map centered on the dateline while
> llcrnrlon=-180, urcrnrlon=180 will produce a map centered on Greenwich.

Thanks! I had not figured out that the decimal coordinates used by
Basemap range from 0 to 360 (and not just from -180 to 180). Using
this tip, I could also create a map centered on the Indian Ocean. BTW,
by varying the llcrnrlat and urcrnrlat, can one also create maps
centered on the North and South poles using a Equirectangular
projection?

With best regards,

-- 
Dr. Mauro J. Cavalcanti
Ecoinformatics Studio
P.O. Box 46521, CEP 20551-970
Rio de Janeiro, RJ, BRASIL
E-mail: mau...@gm...
Web: http://studio.infobio.net
Linux Registered User #473524 * Ubuntu User #22717
"Life is complex. It consists of real and imaginary parts."

On Dec 8, 2008, at 1:09 AM, Roy H. Han wrote:

> I figured it out.
> Specifically, I need to use pylab.Axes(figure, [left, bottom, width,  
> height]) where each value in the frame is between 0 and 1
> [.2,.1,.7,.8]
> means
> 20 percent margin on left
> 10 percent margin on bottom
> 10 percent margin on right (1 - .2 - .7)
> 10 percent margin on top (1 - .1 - .8)


Yep, you got it.
Now, you can also use the simpler `pyplot.subplots_adjust` as a  
function, or as a method for your figure. That way, you're not tied to  
defining the positions at the creation of your subplot.

I figured it out.
Specifically, I need to use pylab.Axes(figure, [left, bottom, width,
height]) where each value in the frame is between 0 and 1
[.2,.1,.7,.8]
means
20 percent margin on left
10 percent margin on bottom
10 percent margin on right (1 - .2 - .7)
10 percent margin on top (1 - .1 - .8)

    figure = pylab.figure()
    axes = pylab.Axes(figure, [.2,.1,.7,.8]) # [left, bottom, width, height]
where each value is between 0 and 1
    figure.add_axes(axes)

    axes.barh(selectedFeatureNumbers, alphas, align='center')
    axes.set_xlabel('Weights set by ViolaJones')
    axes.set_title(title)
    axes.set_yticks(selectedFeatureNumbers)
    axes.set_yticklabels([dictionary[x][0] for x in selectedFeatureNumbers])
    for tick in axes.yaxis.get_major_ticks(): tick.label1.set_fontsize(6)
    pylab.savefig('%s.png' % title)


On Mon, Dec 8, 2008 at 12:56 AM, Roy H. Han <
sta...@gm...> wrote:

> Hi,
>
> Is there a way to increase the left-margin of pylab plots?  Specifically
> I'm trying to make enough space so that the text labels on the left are
> visible.  I tried decreasing the font size of the labels but it isn't
> enough.
>
> Thanks,
> RHH
>

Hi,

Is there a way to increase the left-margin of pylab plots?  Specifically I'm
trying to make enough space so that the text labels on the left are
visible.  I tried decreasing the font size of the labels but it isn't
enough.

Thanks,
RHH

Thomas Pfaff-3 wrote:
> 
> Hi,
> 
> does that mean that it works for you with PyScripter?
> Right now the only way to do more than one plot with MPL with PyScripter
> is 
> to reinitialize the remote python interpreter over and over again.
> If you found a way around that I would be glad to hear about it.
> 
> ...
> 

I can get two plots from Pyscripter as follows:
Find out where your matplotlibrc file is

>>> import matplotlib
>>>matplotlib.get_configdir()
'H:\\.matplotlib'

edit the matplotlib file in this directory (or copy the example from the
matplotlib site)to set interactive to True and the backend to one of the Tk
or wx options:

backend      : WXAgg
...
interactive  : True      # see
http://matplotlib.sourceforge.net/interactive.html

make sure there is no # at the front of the interactive line

Choose the matching external Python Engine in PyScripter ie. Run>Python
Engine > Remote (Wx)

Then this example should give two interactive plots:

>>> import pylab
>>> pylab.figure(1)

>>> pylab.plot([1,2,3])

>>> pylab.xlabel('X label for 1st plot')

>>> pylab.figure(2)

>>> pylab.plot([10,20,30])

>>> pylab.xlabel('X label for 2nd plot')

go back and add a y label to the first plot

>>> pylab.figure(1)

>>> pylab.ylabel('Y label for 1st plot')




-- 
View this message in context: http://www.nabble.com/Can-matplotlib-be-run-from-PythonWin-IDE-in-interactive-mode--tp20822638p20888060.html
Sent from the matplotlib - users mailing list archive at Nabble.com.

Mauro Cavalcanti wrote:
> Dear Jeffrey,
>
> 2008/12/6 Jeff Whitaker <js...@fa...>:
>
>   
>> Mauro:  There never actually was a Basemap 0.90 - could you check that
>> version number again with
>>     
>
> Sorry, my reference to a "0.90" version was a typo! Indeed, I have version 0.99
>
> Thanks for the information on the drawing of rivers; I will see how I
> can provide this feature to users, warning them that the first plot
> may take some time to complete.
>
> BTW, even if there are reasons for the slowness of river plotting and
> Blue Marble superimposition, have you any information on the most
> appropriate hardware requirements for getting the fastest plots in
> these cases?
>   

Mauro:  All I can say is, the faster the better.
> As of volunteers for helping with the Basemap documentation, count me
> in, just after I have my current project finished (hopefully by the
> next month). One idea I have is creating a "Basemap Cookbook" website,
> essentially presenting all the examples included in the current
> distribution, plus a few others (as the additional example for
> plotting shapefiles that you posted on the list a few days ago).
>   

Great. 

-Jeff
> With warmest regards,
>
>   


-- 
Jeffrey S. Whitaker         Phone : (303)497-6313
NOAA/OAR/CDC  R/PSD1        FAX   : (303)497-6449
325 Broadway                Boulder, CO, USA 80305-3328

Mauro Cavalcanti wrote:
> Dear Jeffrey & ALL,
>
> Are there any way to plot a Basemap using a standard Equidistant
> Cylindrical Projection, but centered on the Pacific Basin instead of
> the default plot? This can be done easily with all projections that
> require just a center point at lat_0, long_0 (Robinson, Mollweide,
> Orthographic, Geostationary) but could not figure out how to do this
> in projections that require a bounding rectangle to be defined (as is
> the case with the Equidistant Cylindrical).
>
> Thanks in advance!
>
> With best regards,
>
>   
Mauro:  Just set the llrncrlat,urcrnrlon appropriately.  For instance, 
llcrnrlon=0, urcrnrlon=360 will produce a map centered on the dateline 
while llcrnrlon=-180, urcrnrlon=180 will produce a map centered on 
Greenwich.

-Jeff

-- 
Jeffrey S. Whitaker         Phone : (303)497-6313
NOAA/OAR/CDC  R/PSD1        FAX   : (303)497-6449
325 Broadway                Boulder, CO, USA 80305-3328