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On Thursday, June 14, 2012, Daπid wrote:

> First, this is another topic, so please, change the subject of the
> message so it doesn't get messed up with others (and possible help
> lost in the process).
>
> Now, you are indeed plotting one dot at the time and generating a
> label for it. If you don't want that, you have to plot the whole list
> at the time:
>
>
> x=[rand() for i in xrange(10)]
> y=[rand() for i in xrange(10)]
>
> scatter(x,y, label='points')
> legend()
> show()
>
> where the definition of x includes a list comprehension (equivalent at
> "for i in xrange(10): x.append(rand())" ).
>
> On another topic, people are not usually fan of using  from MODULEX
> import *, as it can turn into poor code and name collisions. It is
> nicer if you write "import pylab as plt", and refer to the functions
> as plt.scatter and so on.
>
>
> Regards.


Point of style: in general, yes, but pylab was intended for that to help
transition matlab users. Pylab really shouldn't be loaded as plt, because
that is what pyplot is usually imported as.  Of course, this is all just a
matter of style and preference.
Cheers
!
Ben Root

First, this is another topic, so please, change the subject of the
message so it doesn't get messed up with others (and possible help
lost in the process).

Now, you are indeed plotting one dot at the time and generating a
label for it. If you don't want that, you have to plot the whole list
at the time:


x=[rand() for i in xrange(10)]
y=[rand() for i in xrange(10)]

scatter(x,y, label='points')
legend()
show()

where the definition of x includes a list comprehension (equivalent at
"for i in xrange(10): x.append(rand())" ).

On another topic, people are not usually fan of using  from MODULEX
import *, as it can turn into poor code and name collisions. It is
nicer if you write "import pylab as plt", and refer to the functions
as plt.scatter and so on.


Regards.

On Wed, Jun 13, 2012 at 9:21 PM, Steven Boada <bo...@ph...> wrote:
> List,
>
> I'm making a scatter plot using a for loop. Here's a simple example..
>
> for i in range(10):
>     x=rand()
>     y=rand()
>     scatter(x,y,label='point')
>
> legend()
> show()
>
>
> When you do this, you get a legend entry for every single point. In this
> case, I get 9 entries in my legend.
>
> Is there a way to only get a single entry? I have looked into creating
> the legends by hand, but I'm not having much luck. Googling, only turned
> up a single example of someone else with the same problem.
>
> Help me list, you're my only hope.
>
> Steven
>
> ------------------------------------------------------------------------------
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On 6/14/12 7:52 AM, Yasin Selçuk Berber wrote:
> I want to set a projection area larger than data's geographical coverage.
> But data must still be georeferenced correctly inside map region
> and coastlines in projection area should still be visible outside data 
> area.
> right now, the data is always stretched inside whole projection area.
> trying to mimic with characters:
> --right now--
> ulcorner
> _______________________
> |   p r o j.  a r e a  |
> |                      |
> |     w h o l e        |
> |      d a t a         |
> |        i s           |
> |   s t r e t c h e d  |
> |                      |
> -----------------------| lrcorner
> --what i try to do--
> ulcorner
> _______________________
> |                      |
> |__________    this    |
> |          |   is      |
> |   this   |projection |
> |   is     |   area    |
> |   data   |           |
> |          |           |
> -----------------------| lrcorner
> Code below certainly does not work and stretch data to full projection 
> region:
> -------------------------
> *code skipped*
> m = Basemap(projection='merc',llcrnrlat=37,urcrnrlat=42,\
>             llcrnrlon=24,urcrnrlon=34,resolution='i')
> m.imshow(data, cmap=plt.cm.jet, interpolation='nearest')
> plt.show()
> *code skipped*
> ---------------------------
> i fiddled with imshow's extent and clip_box keywords but since could 
> not get it work. and googling for a serious amount of time didnt help 
> either.
> Some posts on web mentioned bbox or set_autoscale_on related things 
> but i cant seem to get it. Any ideas ?
> thanks.
>
>
> -- 
> Yasin Selçuk Berber
> "Bismillah, her hayr?n bas,?d?r."

Yasin:  If you use pcolor or contourf, you can specify the x,y (map 
projection) coordinates of the data grid, and the data will be plotted 
on a subset of the map projection region.  imshow (which doesn't take 
x,y coordinates) just fills the whole plotting region with an image.  If 
you want to use imshow, perhaps you could define an inset map (with no 
coastlines drawn) inside your larger map, and plot the image on the 
inset map.

-Jeff

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On Tue, Jun 12, 2012 at 11:59 AM, Paul Hobson <pmh...@gm...> wrote:

> On Mon, Jun 11, 2012 at 11:03 PM, Justin R <jus...@gm...> wrote:
> > operating system Windows 7
> > matplotlib version : 1.1.0
> > obtained from sourceforge
> >
> > the class seems to generate the same Wt matrix for every input. The
> > every element of the weight matrix is either +sqrt(1/2) or -sqrt(1/2).
> >
> > dat1 = 4*np.random.randn(200,1) + 2
> > dat2 = dat1*.25 + 1*np.random.randn(200,1)
> > pcaObj1 = PCA(np.hstack((dat1,dat2)))
> > print pcaObj1.Wt
> >
> > dat3 = 2*np.random.randn(200,1) + 2
> > dat4 = dat3*2 + 3*np.random.randn(200,1)
> > pcaObj2 = PCA(np.hstack((dat1,dat2)))
> > print pcaObj2.Wt
> >
> > The output Y seems to be correct, and the projection function works.
> > only the Wt matrix seems to be messed up. Am I using this class
> > incorrectly, or could this be a bug?
> > thanks,
> > Justin
>
> Justin, could you post a self-contained script that demonstrates the
> issue? Where does this PCA function come from?
>
> In [1]: from pylab import *
>
> In [2]: PCA
> ---------------------------------------------------------------------------
> NameError                                 Traceback (most recent call last)
> C:\Users\phobson\<ipython-input-2-dcf6991f51c0> in <module>()
> ----> 1 PCA
>
> NameError: name 'PCA' is not defined
>
>

Paul,

In case you never got an answer to this: PCA is in the mlab submodule, so
if you do "from pylab import *", you would use mlab.PCA.  (At least that's
the case in matplotlib 1.1.0).

Warren



> -paul
>
>
> ------------------------------------------------------------------------------
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> threat landscape has changed and how IT managers can respond. Discussions
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> threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/
> _______________________________________________
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>

http://matplotlib.sourceforge.net/examples/axes_grid/scatter_hist.html

This is something I'd like to be able to produce, but for millions of
records, where the center will look like one undifferentiated mass of dark
blue.

Is there a way to let something like that scale better to large sizes? For
instance, is it possible to have a point be one pixel and an RGBA color of
00000001 (black color one notch away from complete transparency)? Or for
another approach, is it possible to say "I want the highest value to be the
darkest color, e.g. black and everything else to be greyscale against its
percentage of the maximum value?

Other suggestions?

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I want to set a projection area larger than data's geographical coverage.

But data must still be georeferenced correctly inside map region
and coastlines in projection area should still be visible outside data area.

right now, the data is always stretched inside whole projection area.

trying to mimic with characters:


--right now--

ulcorner
_______________________
|   p r o j.  a r e a  |
|                      |
|     w h o l e        |
|      d a t a         |
|        i s           |
|   s t r e t c h e d  |
|                      |
-----------------------| lrcorner


--what i try to do--

ulcorner
_______________________
|                      |
|__________    this    |
|          |   is      |
|   this   |projection |
|   is     |   area    |
|   data   |           |
|          |           |
-----------------------| lrcorner


Code below certainly does not work and stretch data to full projection
region:

-------------------------
*code skipped*

m = Basemap(projection='merc',llcrnrlat=37,urcrnrlat=42,\
            llcrnrlon=24,urcrnrlon=34,resolution='i')

m.imshow(data, cmap=plt.cm.jet, interpolation='nearest')

plt.show()

*code skipped*
---------------------------

i fiddled with imshow's extent and clip_box keywords but since could not
get it work. and googling for a serious amount of time didnt help either.
Some posts on web mentioned bbox or set_autoscale_on related things but i
cant seem to get it. Any ideas ?

thanks.


-- 
Yasin Selçuk Berber
"Bismillah, her hayrın başıdır."

using axes.set_anchor will do the trick.

chao


wiswit wrote:
> 
> Dear all,
> 
> I think this is quite easy but I searched the internet and mailing list
> and
> not able to find an answer.
> ax2 is an inset axes within the "ax" axes in figure "fig", which I make
> following here
> http://matplotlib.sourceforge.net/examples/pylab_examples/axes_demo.html
> 
> but now my problem is that I cannot fix the ax2 the exact position I want,
> it seems that draw() command change this:
> 
> 
> In [352]:
> 
> ax2.set_position([0.125,0.63,0.25,0.25])
> 
> ax2.set_position([0.125,0.63,0.25,0.25])
> 
> In [353]:
> 
> ax2.get_position()
> 
> ax2.get_position()
> 
> Out[353]:
> 
> Bbox(array([[ 0.125,  0.63 ],
>        [ 0.375,  0.88 ]]))
> 
> In [354]:
> 
> draw()
> 
> draw()
> 
> In [355]:
> 
> ax2.get_position()
> 
> ax2.get_position()
> 
> Out[355]:
> 
> Bbox(array([[ 0.15625,  0.63   ],
>        [ 0.34375,  0.88   ]]))
> 
> 
> could anyone give any hints? thanks!
> 
> 
> 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
> ************************************************************************************
> 
> ------------------------------------------------------------------------------
> Live Security Virtual Conference
> Exclusive live event will cover all the ways today's security and 
> threat landscape has changed and how IT managers can respond. Discussions 
> will include endpoint security, mobile security and the latest in malware 
> threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
> 
> 

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so, I didn't notice the axes.set_anchor method. This will do the trick. 

Chao


wiswit wrote:
> 
> Dear all,
> 
> asking question in a good way is art and I am trying to do that :-). I
> spent whole day trying to put an inset axes within another hosting axes
> the
> exact position I want.
> and from here
> http://old.nabble.com/Adding-custom-axes-within-a-subplot-td22159536.html
>  Jae-Joon
> Lee <http://old.nabble.com/user/UserProfile.jtp?user=1141641>  gave a good
> answer using only four lines:
> 
>     Bbox = matplotlib.transforms.Bbox.from_bounds(.4, .1, .5, .3)
> #numbers in fraction of hosting axes
>     trans = ax.transAxes + fig.transFigure.inverted()
>     l, b, w, h = matplotlib.transforms.TransformedBbox(Bbox, trans).bounds
>     axins = fig.add_axes([l, b, w, h])
> 
> It works fine. Now my question is I want inset axes to have 'equal' aspect
> because I want 1:1 ratio plot. and I found that using
> axins.set_aspect('equal')
> will change the position of the inset axes. Then I tried to adjust the
> width and height of inset axes with the hosting axes aspect ratio before I
> draw it so that
> I would expect they look already "aspect-equal" before I feed data to it.
> 
> So my first question is,  How can I get the axes aspect ratio,
> axes.get_aspect() and  axes._aspect both give only 'auto' but not
> numerical
> value.
> (I assume it's height/width ratio in terms of figure fraction or it's
> inverse?, I tried this but it doesn't work.)
> 
> another side-question, I have a feeling that understanding transform is of
> great value working with matplotlib. But I don't understand the
> four lines above, and I can not find further information either in the
> matplotlib document or online. Is there any source except having
> dig into source code? thanks!!!!!!!!
> 
> I make an example script below to show the problem (long but easy). I hope
> someone could offer some help. :-)
> 
> ###script showing the problem
> fig=plt.figure()
> #plot two subplot to have ax aspect far from 'equal'
> ax=fig.add_subplot(211)
> a=np.arange(0,2*np.pi,0.1)
> ax.plot(a,np.sin(a))
> 
> def create_inset_axes(x0,y0,width,height):  #the four numbers are
> x0,y0,width,height
>     Bbox = matplotlib.transforms.Bbox.from_bounds(x0,y0,width,height)
>     trans = ax.transAxes + fig.transFigure.inverted()
>     l, b, w, h = matplotlib.transforms.TransformedBbox(Bbox, trans).bounds
>     return fig.add_axes([l, b, w, h])
> 
> def get_axes_aspect_ratio(ax):
>     box=ax.get_position()
>     ratio=(box.x1-box.x0)/(box.y1-box.y0)
>     return ratio
> 
> axins=create_inset_axes(0.1,0.05,0.2,0.2)
> axins.plot(np.arange(10),'ro')
> ax.text(0.35,0.15,'no any adjustment',transform=ax.transAxes)
> 
> axins=create_inset_axes(0.1, 0.3, 0.2, 0.2)
> axins.plot(np.arange(10),'ro')
> axins.set_aspect('equal')
> ax.text(0.35,0.4,'explicitly set aspect as equal',transform=ax.transAxes)
> 
> axins=create_inset_axes(0.1, 0.55, 0.2, 0.2*ratio) #adjust the height by
> ax
> axes width/height ratio
> axins.plot(np.arange(10),'ro')
> ax.text(0.35,0.7,'adjust with hosting axes width/length
> ratio',transform=ax.transAxes)
> 
> 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
> ************************************************************************************
> 
> ------------------------------------------------------------------------------
> Live Security Virtual Conference
> Exclusive live event will cover all the ways today's security and 
> threat landscape has changed and how IT managers can respond. Discussions 
> will include endpoint security, mobile security and the latest in malware 
> threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
> 
> 

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