Re: [Matplotlib-users] Memory leaks in a web application

2009-02-08 Thread Jouni K . Seppänen
A B python6...@gmail.com writes:

f = pyplot.figure()

When you do this, matplotlib retains a reference to the figure until you
close it so that you can go back to it with e.g. figure(4). So add
pyplot.close(f) to your script. Or, even better, use the object-oriented
API. To get started with that, see

http://matplotlib.sourceforge.net/leftwich_tut.txt

-- 
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http://www.iki.fi/jks


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[Matplotlib-users] controlling tick label size

2009-02-08 Thread Alan Jackson
Trying to find a simple way to shrink the tick labels for this plot -
since I can have many tiny histograms, the labels need to be smaller,
but it isn't obvious to me how to simply shrink them.

code snippet...

num = len(datasets)
fig = plt.figure()
rows = np.int(np.sqrt(num))
cols = np.int(float(num)/float(rows)+.9)
for i in range(num):
ax = fig.add_subplot(rows, cols, i+1)
ax.hist(datasets[i][np.isfinite(datasets[i])], bins=60 )
ax.set_xlabel(labels[i], size = 9)
ax.set_ylabel('Counts', size = 9)
plt.show()


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---
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| a...@ajackson.org  | And a Heaven in a Wild Flower, |
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Re: [Matplotlib-users] controlling tick label size

2009-02-08 Thread Alan Jackson
Never mind - I just saw the very timely e-mail from Jouni.

Thanks!

On Sun, 8 Feb 2009 14:16:59 -0600
Alan Jackson a...@ajackson.org wrote:

 Trying to find a simple way to shrink the tick labels for this plot -
 since I can have many tiny histograms, the labels need to be smaller,
 but it isn't obvious to me how to simply shrink them.
 
 code snippet...
 
 num = len(datasets)
 fig = plt.figure()
 rows = np.int(np.sqrt(num))
 cols = np.int(float(num)/float(rows)+.9)
 for i in range(num):
 ax = fig.add_subplot(rows, cols, i+1)
 ax.hist(datasets[i][np.isfinite(datasets[i])], bins=60 )
 ax.set_xlabel(labels[i], size = 9)
 ax.set_ylabel('Counts', size = 9)
 plt.show()
 
 
 -- 
 ---
 | Alan K. Jackson| To see a World in a Grain of Sand  |
 | a...@ajackson.org  | And a Heaven in a Wild Flower, |
 | www.ajackson.org   | Hold Infinity in the palm of your hand |
 | Houston, Texas | And Eternity in an hour. - Blake   |
 ---
 
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-- 
---
| Alan K. Jackson| To see a World in a Grain of Sand  |
| a...@ajackson.org  | And a Heaven in a Wild Flower, |
| www.ajackson.org   | Hold Infinity in the palm of your hand |
| Houston, Texas | And Eternity in an hour. - Blake   |
---

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[Matplotlib-users] release candidate

2009-02-08 Thread John Hunter
We have accumulated a number of bug fixes in our stable release
branch, so I would like to release the 3rd bugfix release.  Please
test and report any problems

  
http://matplotlib.sourceforge.net/release-candidates/matplotlib-0.98.5.3.tar.gz

Charlie, will you have any time early next week for the windows
builds?  I can handle the OSX and src release.

Thanks,
JDH

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[Matplotlib-users] different Locator for opposite axes

2009-02-08 Thread Thomas Robitaille
Hi everyone,

I am plotting a figure where I need two independent x axes and two  
independent y axes. I've tried to use both twinx and twiny at the  
same time, and this works to some extent, but it looks like it is  
plotting the labels for the bottom x axis and the right-hand y axis  
twice, which makes me think that I must be doing something wrong (the  
numbers appear more 'bold'). The code is below. Is there a better way  
to do this?

In reality, I don't need a different scale for the opposite axes, but  
I want to specify different Locator functions, but I assume that  
creating a new axes instance as done below is the only way to do this?

Thanks for any advice,

Thomas

###
fig = figure()
ax  = fig.add_subplot(111)
ax2 = ax.twinx().twiny()
for tick in ax.yaxis.get_major_ticks():
tick.label1On = True
tick.label2On = False
tick.tick1On = True
tick.tick2On = False
for tick in ax.xaxis.get_major_ticks():
tick.label1On = True
tick.label2On = False
tick.tick1On = True
tick.tick2On = False
for tick in ax2.yaxis.get_major_ticks():
tick.label1On = False
tick.label2On = True
tick.tick1On = False
tick.tick2On = True
for tick in ax2.xaxis.get_major_ticks():
tick.label1On = False
tick.label2On = True
tick.tick1On = False
tick.tick2On = True
ax.scatter([0.4],[0.6])
ax2.scatter([10.],[10.])
draw()
###


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Re: [Matplotlib-users] release candidate

2009-02-08 Thread Charlie Moad
Sure.  Just send me a note when the src is ready.

- Charlie

On Sun, Feb 8, 2009 at 3:46 PM, John Hunter jdh2...@gmail.com wrote:
 We have accumulated a number of bug fixes in our stable release
 branch, so I would like to release the 3rd bugfix release.  Please
 test and report any problems

  
 http://matplotlib.sourceforge.net/release-candidates/matplotlib-0.98.5.3.tar.gz

 Charlie, will you have any time early next week for the windows
 builds?  I can handle the OSX and src release.

 Thanks,
 JDH


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Re: [Matplotlib-users] rc params strangeness/error

2009-02-08 Thread Leo Trottier
Hi,

Did anyone have any insight on this?  Alternatively, anyone know why there
has been little uptake on this question ...(apologies that it was
accidentally sent twice)?

Leo

On Tue, Feb 3, 2009 at 11:42 PM, Leo Trottier l...@cogsci.ucsd.edu wrote:

 Matplotlib 0.98.5.2
 Location: C:\leo\.matplotlib
 Running Windows XP SP2
 Obtained from pythonxy v. 2.1.10
 No rc customizations ...

 I've been having a devil of a time getting my rcParams to update
 mid-script.  Perhaps this isn't possible, but if that's so, it's not
 entirely clear.  Even if it is so, this is seemingly a defect, because in an
 interactive session one expects quite different behavior (and no amount of
 iPython %reset-ing seems to be able to help).

 Note that manually setting the font (or what have you) in *title* works
 fine.

 Here is some example code:

 ## BEGIN #
 import matplotlib
 from matplotlib import rc, rcParams, rcdefaults
 from matplotlib.pyplot import plot, show, figure, title

 print matplotlib.__version__
 print matplotlib.get_configdir()
 print
 print rcParams['font.sans-serif']:, rcParams['font.sans-serif']
 print 'family is:', rcParams['font.family']
 print
 figure(4)
 title('This should be in a sans-serif font')
 show()

 rcParams['font.sans-serif'] = rcParams['font.monospace']
 print rcParams['font.sans-serif']:, rcParams['font.sans-serif']
 print 'family is:', rcParams['font.family']
 print
 figure(1)
 title('This should be in a monospace font')
 show()

 rcParams['font.sans-serif'] = rcParams['font.serif']
 print rcParams['font.sans-serif']:, rcParams['font.sans-serif']
 print 'family is:', rcParams['font.family']
 print
 figure(2)
 title('This should be in a serif font')
 show()

 rcdefaults()
 print rcParams['font.sans-serif']:, rcParams['font.sans-serif']
 print 'family is:', rcParams['font.family']
 print
 figure(3)
 title('This should be back to a sans-serif font')
 show()
 # END 

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Re: [Matplotlib-users] different Locator for opposite axes

2009-02-08 Thread Jae-Joon Lee
Since you call twinx then twiny, you're creating two additional axes, not one.
And I guess this is why labels are drawn twice. You may do

def twin(ax):
ax2 = ax.figure.add_axes(ax.get_position(True),
   frameon=False)
ax2.yaxis.tick_right()
ax2.yaxis.set_label_position('right')
ax.yaxis.tick_left()

ax2.xaxis.tick_top()
ax2.xaxis.set_label_position('top')
ax.xaxis.tick_bottom()

return ax2

ax = gca()
ax2 = twin(ax)
ax.scatter([0.4],[0.6])
ax2.scatter([10.],[10.])

draw()


Note that you need to manually adjust the view limits of each axes. If
you use sharex or sharey parameters for the axes, you can share their
view limits (this is how axes is created when twinx and twiny is
called). But then you cannot have different tick locators.

In case you need an axes with a same viewlimit as the original one but
just want to place ticks at different position, you may check my
related post.

http://sourceforge.net/mailarchive/forum.php?thread_name=4985DED6.90108%40head.cfa.harvard.eduforum_name=matplotlib-users

-JJ


On Sun, Feb 8, 2009 at 5:07 PM, Thomas Robitaille
thomas.robitai...@gmail.com wrote:
 Hi everyone,

 I am plotting a figure where I need two independent x axes and two
 independent y axes. I've tried to use both twinx and twiny at the
 same time, and this works to some extent, but it looks like it is
 plotting the labels for the bottom x axis and the right-hand y axis
 twice, which makes me think that I must be doing something wrong (the
 numbers appear more 'bold'). The code is below. Is there a better way
 to do this?

 In reality, I don't need a different scale for the opposite axes, but
 I want to specify different Locator functions, but I assume that
 creating a new axes instance as done below is the only way to do this?

 Thanks for any advice,

 Thomas

 ###
 fig = figure()
 ax  = fig.add_subplot(111)
 ax2 = ax.twinx().twiny()
 for tick in ax.yaxis.get_major_ticks():
tick.label1On = True
tick.label2On = False
tick.tick1On = True
tick.tick2On = False
 for tick in ax.xaxis.get_major_ticks():
tick.label1On = True
tick.label2On = False
tick.tick1On = True
tick.tick2On = False
 for tick in ax2.yaxis.get_major_ticks():
tick.label1On = False
tick.label2On = True
tick.tick1On = False
tick.tick2On = True
 for tick in ax2.xaxis.get_major_ticks():
tick.label1On = False
tick.label2On = True
tick.tick1On = False
tick.tick2On = True
 ax.scatter([0.4],[0.6])
 ax2.scatter([10.],[10.])
 draw()
 ###


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[Matplotlib-users] title pad?

2009-02-08 Thread C M
Is there anything like a title pad, similar to the xaxis.LABELPAD?
I'd like to see if the plot would look better with the title a bit higher
off the plot.

Thanks,
Che

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Re: [Matplotlib-users] title pad?

2009-02-08 Thread Jae-Joon Lee
As far as I know, there is no user settable attribute. But something
like below will work.

ax = gca()
title(My title)

ax.titleOffsetTrans._t = (0., 10.0/72.) # x, y offset in points/72.
default is (0., 5/72.)
ax.titleOffsetTrans.invalidate()

draw()

Alternatively, you may use

ax.title.set_y(1.1) # y position of the title in the normalized axes
coordinate. default 1.0

-JJ




On Mon, Feb 9, 2009 at 2:08 AM, C M cmpyt...@gmail.com wrote:
 Is there anything like a title pad, similar to the xaxis.LABELPAD?
 I'd like to see if the plot would look better with the title a bit higher
 off the plot.

 Thanks,
 Che

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