matplotlib-users Mailing List for matplotlib

Hi Users,

What is the best way to add a subscript to a tick label using the 
default font?

-- 
Chris

You didn't mention the version of matplotlib you are using.  There are 
some known bugs using NaNs on 0.90.x and 0.91.x that have been fixed on 
the trunk.  Unfortunately, it will be difficult to backport these fixes.

Have you tried using masked arrays instead of arrays with NaNs?

Also, can you provide a standalone example that reproduces this error 
(including specific data?) so that we can further investigate?

Cheers,
Mike

Fabrice Silva wrote:
> Hello,
>
> I want to plot data containing nan values, but it seems that matplotlib
> forget some non-nan values near nan values.
> In the attached image, I plotted in blue the array for increasing of
> x-data, and in red for decreasing x-data : 
>
> import numpy as NX
>
>         In [1]: from pylab import figure, plot, show
>         In [2]: figure(2)
>         Out[2]: <matplotlib.figure.Figure instance at 0x8445c8c>
>         
>         In [3]: type(Freq)
>         Out[3]: <type 'numpy.ndarray'>
>         
>         In [4]: NX.any(NX.isnan(Freq))
>         Out[4]: True
>         
>         In [5]: type(Grillefr)
>         Out[5]: <type 'numpy.ndarray'>
>         
>         In [6]: NX.any(NX.isnan(Grillefr))
>         Out[6]: False
>         
>         In [7]: plot(Grillefr, Freq, 'b', lw=2.)
>         Out[7]: [<matplotlib.lines.Line2D instance at 0x9245e8c>]
>         
>         In [8]: plot(Grillefr[::-1], Freq[::-1], 'r', lw=0.5)
>         Out[8]: [<matplotlib.lines.Line2D instance at 0x9245fec>]
>         
>         In [9]: show()
>         
> with Grillefr containing values from 100 to 2000.
> In the matplotlibrc, I have : 
> maskedarray  : True
>
> Can anybody explain me why some points are not drawn by matplotlib ?
>   
>
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On Wed, May 21, 2008 at 10:57 AM, PaterMaximus
<Pat...@go...> wrote:
> I want to make one scatter plot and use the same scales on another. I think
> I seen getting the Axes from the first scatter plot using v=axis() and then
> setting them on second with axis(v) but I can not get to work. Any help
> appreciated

Try using shared scales:

ax1 = subplot(211)
ax2 = subplot(212, sharex=ax1, sharey=ax1)

JDH

On Thu, May 22, 2008 at 9:45 AM, Jouni K. Seppänen <jk...@ik...> wrote:

> I think this behavior of legend is suboptimal, but there is a logic to
> it: by default you label objects in the order you draw them, and the
> histogram plot happens to consist of several similarly-colored objects.

The best solution would be to rewrite the hist to use a collection.
This would make legend behave properly, would be a lot faster, but
would break a fair amount of legend code.  Probably a good thing to do
for 0.98

JDH

Correction :
> In the matplotlibrc, I have : 
> maskedarray  : False

-- 
Fabrice Silva <si...@lm...>
LMA UPR CNRS 7051 - équipe S2M

On Thu, 22 May 2008, Friedrich Hagedorn apparently wrote:
> I hope I can change my habit to type 'g' instead of 'r'. 

Or use an email client that allows you to register
your email lists to address this problem.
(Such as Mahogany mail.)

Cheers,
Alan Isaac

Hello,

I want to plot data containing nan values, but it seems that matplotlib
forget some non-nan values near nan values.
In the attached image, I plotted in blue the array for increasing of
x-data, and in red for decreasing x-data : 

import numpy as NX

        In [1]: from pylab import figure, plot, show
        In [2]: figure(2)
        Out[2]: <matplotlib.figure.Figure instance at 0x8445c8c>
        
        In [3]: type(Freq)
        Out[3]: <type 'numpy.ndarray'>
        
        In [4]: NX.any(NX.isnan(Freq))
        Out[4]: True
        
        In [5]: type(Grillefr)
        Out[5]: <type 'numpy.ndarray'>
        
        In [6]: NX.any(NX.isnan(Grillefr))
        Out[6]: False
        
        In [7]: plot(Grillefr, Freq, 'b', lw=2.)
        Out[7]: [<matplotlib.lines.Line2D instance at 0x9245e8c>]
        
        In [8]: plot(Grillefr[::-1], Freq[::-1], 'r', lw=0.5)
        Out[8]: [<matplotlib.lines.Line2D instance at 0x9245fec>]
        
        In [9]: show()
        
with Grillefr containing values from 100 to 2000.
In the matplotlibrc, I have : 
maskedarray  : True

Can anybody explain me why some points are not drawn by matplotlib ?
-- 
Fabrice Silva <si...@lm...>
LMA UPR CNRS 7051 - équipe S2M

> p.setp(ax1.get_yticklabels(), color='r')
This worked!

Thanks a lot!
Chiara

> To: mat...@li...
> From: jk...@ik...
> Date: Thu, 22 May 2008 17:52:09 +0300
> Subject: Re: [Matplotlib-users] changing ticklabels color
> 
> Chiara Caronna <chi...@ho...> writes:
> 
> > ax1=p.subplot(212)
> > ax1.set_xlabel('t [sec]')
> > ax1.set_ylabel('g^2(q,t)')
> > ax1.set_yticklabels(color='r')
> 
> You could do
> 
> for label in ax1.get_yticklabels(): label.set_color('r')
> 
> or use the setp shortcut:
> 
> p.setp(ax1.get_yticklabels(), color='r')
> 
> -- 
> Jouni K. Seppänen
> http://www.iki.fi/jks
> 
> 
> -------------------------------------------------------------------------
> This SF.net email is sponsored by: Microsoft
> Defy all challenges. Microsoft(R) Visual Studio 2008.
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> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users

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On Thu, May 22, 2008 at 05:59:23PM +0300, Jouni K. Seppänen wrote:
> Friedrich Hagedorn <fri...@gm...> writes:
> 
> > could you set the right reply-adress in the mails from the 
> > matplotlib mailinglist?
> 
> This is something of a controversial issue. One side is presented at
> 
> http://www.unicom.com/pw/reply-to-harmful.html
> 
> and I'm sure the other one has a web page somewhere, too. On lists that
> do not munge the reply-to address, such as the matplotlib lists, you
> have two options: you can reply to the sender only ('r' in mutt) or
> reply to everyone ('g'). On lists that do munge the address, both
> commands reply to the complete list.

Ok, I test it with 'g' by this reply, and it works. I hope I can change
my habit to type 'g' instead of 'r'.


On Thu, May 22, 2008 at 04:59:39PM +0200, Johann Rohwer wrote:
> This has been discussed before. See:
>
> http://sourceforge.net/mailarchive/message.php?msg_id=4756A726.1080104%40gmx.net

Ah, I see the discussion. I am for 'r' :-)


By,  Friedrich

Chiara Caronna <chi...@ho...> writes:

> ax1=p.subplot(212)
> ax1.set_xlabel('t [sec]')
> ax1.set_ylabel('g^2(q,t)')
> ax1.set_yticklabels(color='r')

You could do

for label in ax1.get_yticklabels(): label.set_color('r')

or use the setp shortcut:

p.setp(ax1.get_yticklabels(), color='r')

-- 
Jouni K. Seppänen
http://www.iki.fi/jks

This has been discussed before. See:

http://sourceforge.net/mailarchive/message.php?msg_id=4756A726.1080104%40gmx.net

Johann

Friedrich Hagedorn wrote:
> Hello List-Admin,
> 
> could you set the right reply-adress in the mails from the 
> matplotlib mailinglist?
> 
> Everytime when I what to reply to a message I have to set the right
> email-adress to
> 
>   mat...@li...
> 
> otherwise the email would sent only as a privat mail.
> 
> I use mutt and on all other mailinglist I have no problem. So I think
> this is a matplotlib-ml problem.

On Thu, May 22, 2008 at 02:30:24PM +0000, Chiara Caronna wrote:
>    Hello,
>    I would like to change the color of the yticklabels. I tried to use this
>    command:
> 
>    ax1=p.subplot(212)
[...]
>    ax1.set_yticklabels(color='r')

* Solution 1:

In [1]: ax1=subplot(111)

In [2]: setp(ax1.get_yaxis().get_major_ticklabels(), color='r')

* Solution 2:

In [1]: ax1=subplot(111)

In [2]: for label in ax1.get_yaxis().get_majorticklabels():
   ...:     label.set_color('r')
   ...:     
   ...:     

In [3]: draw()

May be there is a shorter solution but I dont know of it :-)


By,  Friedrich

Friedrich Hagedorn <fri...@gm...> writes:

> could you set the right reply-adress in the mails from the 
> matplotlib mailinglist?

This is something of a controversial issue. One side is presented at

http://www.unicom.com/pw/reply-to-harmful.html

and I'm sure the other one has a web page somewhere, too. On lists that
do not munge the reply-to address, such as the matplotlib lists, you
have two options: you can reply to the sender only ('r' in mutt) or
reply to everyone ('g'). On lists that do munge the address, both
commands reply to the complete list.

-- 
Jouni K. Seppänen
http://www.iki.fi/jks

David Simpson <dav...@ch...> writes:

> This is probably my lack of knowledge of python, but how do I set up
> legend labels for some bar-plots that have been produced inside a 
> function. For example, the following will nicely plot my bar-plots, but 
> then legend doesn't know about the colours used, so here just uses black 
> for both labels.

Legends for multiple histograms don't behave the way people expect,
because the colors are taken bar by bar, first from one histogram, then
from the second histogram, etc. You need to take just one bar from each
histogram and pass them as the first argument to legend:

------------------------------------------------------------------------
from pylab import *

x=arange(0,5)
y=array([ 1.2, 3.4, 5.4, 2.3, 1.0])
z=array([ 2.2, 0.7, 0.4, 1.3, 1.2])

def plotb(x,y,col):
         p=bar(x,y,color=col)
         return p[0]

p1 = plotb(x,y,'k')
p2 = plotb(x+0.4,z,'y')

legend((p1, p2), ('YYY','ZZZ'))
show()
------------------------------------------------------------------------

I think this behavior of legend is suboptimal, but there is a logic to
it: by default you label objects in the order you draw them, and the
histogram plot happens to consist of several similarly-colored objects.

-- 
Jouni K. Seppänen
http://www.iki.fi/jks

Hello List-Admin,

could you set the right reply-adress in the mails from the 
matplotlib mailinglist?

Everytime when I what to reply to a message I have to set the right
email-adress to

  mat...@li...

otherwise the email would sent only as a privat mail.

I use mutt and on all other mailinglist I have no problem. So I think
this is a matplotlib-ml problem.


Thanks,  Friedrich

On Thu, May 22, 2008 at 02:52:13PM +0200, David Simpson wrote:
> This is probably my lack of knowledge of python, but how do I set up
> legend labels for some bar-plots that have been produced inside a 
> function. For example, the following will nicely plot my bar-plots, but 
> then legend doesn't know about the colours used, so here just uses black 
> for both labels. I'd like the labels to have the same colour as the bars
> generated inside plotb. (I am using a function here as my real code has 
> extra stuff to calculate error-bars and suchlike for each data set.)
> 
> x=arange(0,5)
> y=array([ 1.2, 3.4, 5.4, 2.3, 1.0])
> z=array([ 2.2, 0.7, 0.4, 1.3, 1.2])
> 
> def plotb(x,y,col):
>          p=bar(x,y,color=col)
> 
> plotb(x,y,'k')
> plotb(x+0.4,z,'y')
> 
> legend(('YYY,'ZZZ'))
>
> I tried passing the object "p" through the plotb argument list, but
> python didn't like that. (I am just learning python, and so far haven't
> seen how to pass such objects around.

You could return the plotted lines from the function plotb. 
Here is my attempt:

In [1]: x=arange(0,5)

In [2]: y=array([ 1.2, 3.4, 5.4, 2.3, 1.0])

In [3]: z=array([ 2.2, 0.7, 0.4, 1.3, 1.2])

In [4]: def plotb(x,y,col):
   ...:     lines = bar(x,y,color=col)
   ...:     return lines
   ...: 

In [5]: l1 = plotb(x,y,'k')

In [6]: l2 = plotb(x+0.4,z,'y')

In [7]: legend((l1[0], l2[0]), ('YYY','ZZZ'))
Out[7]: <matplotlib.legend.Legend object at 0x908dc6c>


The legend() function could label any line object. So every single bar-line
could be listed in the legend. But I think you would only have one from 
each color, so I have choosen the first: l1[0] and l2[0].

Is this what you what?


By,  Friedrich

Hello,
I would like to change the color of the yticklabels. I tried to use this command:

ax1=p.subplot(212)
ax1.set_xlabel('t [sec]')
ax1.set_ylabel('g^2(q,t)')
ax1.set_yticklabels(color='r')

but it gives an error:

TypeError: set_yticklabels() takes at least 2 non-keyword arguments (1 given)
if I write instead 
ax1.set_yticklabels(['1','2'],color='r')
it works, but it puts clearly labels 1 and 2... I don't want to change the labels, only the color! is there anyway of doing it?
Hope you can help me
Chiara

> Date: Thu, 22 May 2008 14:52:13 +0200
> From: dav...@ch...
> To: mat...@li...
> Subject: [Matplotlib-users] Legend labels - interaction with functions
> 
> This is probably my lack of knowledge of python, but how do I set up
> legend labels for some bar-plots that have been produced inside a 
> function. For example, the following will nicely plot my bar-plots, but 
> then legend doesn't know about the colours used, so here just uses black 
> for both labels. I'd like the labels to have the same colour as the bars
> generated inside plotb. (I am using a function here as my real code has 
> extra stuff to calculate error-bars and suchlike for each data set.)
> 
> x=arange(0,5)
> y=array([ 1.2, 3.4, 5.4, 2.3, 1.0])
> z=array([ 2.2, 0.7, 0.4, 1.3, 1.2])
> 
> def plotb(x,y,col):
>          p=bar(x,y,color=col)
> 
> plotb(x,y,'k')
> plotb(x+0.4,z,'y')
> 
> legend(('YYY,'ZZZ'))
> 
> 
> I tried passing the object "p" through the plotb argument list, but
> python didn't like that. (I am just learning python, and so far haven't
> seen how to pass such objects around.
> 
> Thanks for any tips,
> 
> Dave
> 
> -------------------------------------------------------------------------
> This SF.net email is sponsored by: Microsoft
> Defy all challenges. Microsoft(R) Visual Studio 2008.
> http://clk.atdmt.com/MRT/go/vse0120000070mrt/direct/01/
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users

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This is probably my lack of knowledge of python, but how do I set up
legend labels for some bar-plots that have been produced inside a 
function. For example, the following will nicely plot my bar-plots, but 
then legend doesn't know about the colours used, so here just uses black 
for both labels. I'd like the labels to have the same colour as the bars
generated inside plotb. (I am using a function here as my real code has 
extra stuff to calculate error-bars and suchlike for each data set.)

x=arange(0,5)
y=array([ 1.2, 3.4, 5.4, 2.3, 1.0])
z=array([ 2.2, 0.7, 0.4, 1.3, 1.2])

def plotb(x,y,col):
         p=bar(x,y,color=col)

plotb(x,y,'k')
plotb(x+0.4,z,'y')

legend(('YYY,'ZZZ'))


I tried passing the object "p" through the plotb argument list, but
python didn't like that. (I am just learning python, and so far haven't
seen how to pass such objects around.

Thanks for any tips,

Dave

Hello,

On Wednesday 21 May 2008 17:57:28 PaterMaximus wrote:
> I want to make one scatter plot and use the same scales on another. I think
> I seen getting the Axes from the first scatter plot using v=axis() and then
> setting them on second with axis(v) but I can not get to work. Any help
> appreciated

Maybe I don't understand correctly, but what you can do with v=axis() and 
axis(v) is to get the current axes limits and apply them to another axes. 
Doing so you need to build the second axes by yourself (with axes oder 
subplot - command) and afterwards apply the limits.

regards Matthias