matplotlib-users Mailing List for matplotlib (Page 8)

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
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Laboratoire des Sciences du Climat et de l'Environnement (LSCE-IPSL)
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Hi all,
it seems that whenever I plot something, a window opens.
from matplotlib import pylab
import numpy
pylab.plot(numpy.random.normal(size=100))
Now, I have code that is supposed to produce diagnostic plots as PDFs. Only
when I pass a command line option the script should actually open a
window and also display
the results on the screen.
I am pretty sure that in earlier times the window would only open when
I call pylab.show().
It may well be that I messed too much with my matplotlibrc, if you
could just point me in the right direction ...
I use matplotlib 1.0.1 on a Mac with the MacOSX backend.
Maximilian

Mark Gurling:
>
> I have two bar graph scripts (good.py and bad.py). 
...
> while in good.py the y-axis ends precisely at -30.0, in bad.py the 
> y-axis ends below -30.0 despite the yticks setting specified on line 
> 20. Is there an explanation for this behavior? How might I remedy this?
What do you really want?
Tony Yu tries to explain the behaviour.
I would suggest using an explicit axis, e.g.
plt.axis([0,0.10,-25,80])
(or between -30 and 100, or 'tight', etc.)
Perhaps you wanted axis('tight') rather than your tight_layout?
Jerzy Karczmarczuk

On Thu, Jun 7, 2012 at 3:44 PM, Mark Gurling <mag...@gm...> wrote:
> I am on Ubuntu 11.10
> matplotlib version 1.1.0
> numpy version 1.5.1
>
> I have two bar graph scripts (good.py and bad.py). Each generates a graph
> that contains two bars: one bar that extends along the positive y-axis and
> another bar that extends along the negative y-axis. The only difference
> between the two scripts is that in good.py the positive bar extends to
> 69.0, but in bad.py it extends to 70.0; however, while in good.py the
> y-axis ends precisely at -30.0, in bad.py the y-axis ends below -30.0
> despite the yticks setting specified on line 20. Is there an explanation
> for this behavior? How might I remedy this?
>
>
`yticks()` just sets the ticks, which shows up correctly in both plots. It
sounds like what you want to specify is the axis limit. You can add the
following (e.g. after the call to `yticks`):
 plt.ylim(ymin=-30)
As for the reason, it has to do with creating axes sizes that fit all the
elements within the plot area and also allow ticks that are "nicely"
spaced. You just happen to be near the threshold of two different spacings,
I think.
-Tony

Is it possible to clip an image like the matplotlib clip example, but by using a polycollection. I'd like to avoid looping through and creating paths/patches for each element and then combining them all at the end, is that the only way?
Cheers,
Alex
Alexander Crosby
RPS ASA
55 Village Square Drive
South Kingstown, RI 02879-8248
USA
Tel: +1 (401) 789-6224
Fax: +1 (401) 789-1932
Email: cr...@rp...<mailto:cr...@rp...>
www: asascience.com<http://www.asascience.com/> | rpsgroup.com<http://www.rpsgroup.com/>
A member of the RPS Group plc

Thank you very much. It works at last.
Kamel
2012年6月7日 Tony Yu <ts...@gm...>
>
> On Thu, Jun 7, 2012 at 12:14 PM, kamel maths <kam...@gm...>wrote:
>
>> Thanks for your answers.
>>
>> It is not very clear for me yet. This a script I tested.
>> ----------------------------------------------
>> from pylab import *
>>
>> fig = figure()
>> ax = fig.add_subplot(111)
>> ax.axis('equal')
>>
>> x = linspace(-2, 3, 50)
>> ax.plot(x, sin(x))
>>
>> ylim = ax.get_ylim()
>> print(ylim)
>>
>> show()
>>
>> ylim = ax.get_ylim()
>> print(ylim)
>> ------------------------------------
>> With the first print, I obtain (-1,1) and with the second one, after
>> having closed the figure showed, I obtain (-1.94,1.94). But I would like to
>> obtain (-1.94,1.94) with the first one in order to use this data in the
>> figure.
>> Is it possible ?
>>
>>
> Hi Kamel,
>
> (I'm recopying to the list; unfortunately, the mailing list doesn't do
> this automatically)
>
> Well, this is strange: I swear I tested this yesterday, and I got (-1.94,
> 1.94) when calling `get_ylim` before `show`. But when I test this today, I
> get (-1, 1) with the call before `show`.
>
> I guess the initial call to `get_ylim` returns that data limits (which are
> (-1, 1) in this case), but the call to `axis('equal')` results in a
> rescaling at draw time.
>
> If you're just trying to readjust the plot before the plot blocks
> execution, you can run `draw()` right before calling `get_ylim`. There may
> be a better approach, though.
>
> -Tony
>
>
>

On Thu, Jun 7, 2012 at 12:14 PM, kamel maths <kam...@gm...> wrote:
> Thanks for your answers.
>
> It is not very clear for me yet. This a script I tested.
> ----------------------------------------------
> from pylab import *
>
> fig = figure()
> ax = fig.add_subplot(111)
> ax.axis('equal')
>
> x = linspace(-2, 3, 50)
> ax.plot(x, sin(x))
>
> ylim = ax.get_ylim()
> print(ylim)
>
> show()
>
> ylim = ax.get_ylim()
> print(ylim)
> ------------------------------------
> With the first print, I obtain (-1,1) and with the second one, after
> having closed the figure showed, I obtain (-1.94,1.94). But I would like to
> obtain (-1.94,1.94) with the first one in order to use this data in the
> figure.
> Is it possible ?
>
>
Hi Kamel,
(I'm recopying to the list; unfortunately, the mailing list doesn't do this
automatically)
Well, this is strange: I swear I tested this yesterday, and I got (-1.94,
1.94) when calling `get_ylim` before `show`. But when I test this today, I
get (-1, 1) with the call before `show`.
I guess the initial call to `get_ylim` returns that data limits (which are
(-1, 1) in this case), but the call to `axis('equal')` results in a
rescaling at draw time.
If you're just trying to readjust the plot before the plot blocks
execution, you can run `draw()` right before calling `get_ylim`. There may
be a better approach, though.
-Tony

Le jeudi 07 juin 2012 à 14:07 +0100, David Craig a écrit :
> Hi,
> I trying to define an area in a pcolor plot (several plots) using the 
> ginput(). However since it is an irregular shape and will be different 
> in each plot so I cant define how many points there will be before hand, 
> I've tried the following but it requires a double click at each point, 
> which I would like to avoid as it duplicates points
> 
> |x = randn(10,10)
> imshow(x)
> 
> button = False
> points = []
> while button == False:
> points.append(ginput(1))
> button = waitforbuttonpress()
> |
> 
> Anyone know a better way to go about this??
> thanks
> Dave
ginput has a way to do it. From docs:
 If n is zero or negative, accumulate clicks until a middle click
 (or potentially both mouse buttons at once) terminates the
 input.
http://matplotlib.sourceforge.net/api/pyplot_api.html#matplotlib.pyplot.ginput
Pressing [enter] also terminates the points acquisition.
-- 
Fabrice Silva <si...@lm...>
LMA UPR CNRS 7051

On Thu, Jun 7, 2012 at 9:07 AM, David Craig <dcd...@gm...> wrote:
> Hi,
> I trying to define an area in a pcolor plot (several plots) using the
> ginput(). However since it is an irregular shape and will be different in
> each plot so I cant define how many points there will be before hand, I've
> tried the following but it requires a double click at each point, which I
> would like to avoid as it duplicates points
>
> x = randn(10,10)
> imshow(x)
>
> button = False
> points = []
> while button == False:
> points.append(ginput(1))
> button = waitforbuttonpress()
>
> Anyone know a better way to go about this??
> thanks
> Dave
>
>
I would utilize the event handling mechanism within mpl to do this. You
can create a function that would append the xy coordinates to points at
each call. Here is a mockup.
def on_press(event) :
 on_press.points.append((event.xdata, event.ydata))
on_press.points = []
Then, you would need some way to indicate that you want to start and
finish, such as some sort of key press:
def key_press(event) :
 if event.key == 'a' :
 if key_press.active :
 key_press.mouse_cid =
key_press.fig.canvas.mpl_connect('button_press_event', on_press)
 else :
 key_press.fig.canvas.mpl_disconnect(key_press.mouse_cid)
 key_press.active = not key_press.active
key_press.active = False
key_press.fig = plt.figure()
key_press.fig.canvas.mpl_connect('key_press_event', key_press)
There are some more elegant ways of doing this, particularly by creating
some sort of object to hold these variables, but this is a quick and dirty
way of doing it.
I hope that helps!
Ben Root

Hi,
I trying to define an area in a pcolor plot (several plots) using the 
ginput(). However since it is an irregular shape and will be different 
in each plot so I cant define how many points there will be before hand, 
I've tried the following but it requires a double click at each point, 
which I would like to avoid as it duplicates points
|x = randn(10,10)
imshow(x)
button = False
points = []
while button == False:
 points.append(ginput(1))
 button = waitforbuttonpress()
|
Anyone know a better way to go about this??
thanks
Dave

On Wed, Jun 6, 2012 at 6:12 PM, Tony Yu <ts...@gm...> wrote:
>
>
> On Wed, Jun 6, 2012 at 3:32 PM, kamel maths <kam...@gm...> wrote:
>
>> Hi,
>>
>> for this script:
>> ------------------------------------
>> from pylab import *
>>
>> fig = figure()
>> ax = fig.add_subplot(111)
>> ax.axis('equal')
>>
>> x = linspace(-2, 3, 50)
>> ax.plot(x, sin(x))
>>
>> show()
>> ---------------------------------
>> If I try to get ymax with ax.get_ylim(), i obtain 1.0 whereas I observe
>> it is 2.0.
>> How can I obtain 2.0 for ymax ?
>>
>> Thanks.
>>
>> Kamel
>>
>
> Hi Kamel,
>
> I'm not seeing the same result: I actually get back (-1.94, 1.94) from
> `get_ylim`. When do you call `get_ylim`? Do you call it *after* calling
> `plot`?
>
> -Tony
>
>
Or, more likely, are you calling it *after* you close the figure? If so,
then the axes has already been cleared and you are merely finding the
limits for a newly created (but unshown) figure.
Make sure you get the limits after the plotting, but before the show().
Ben Root

On Wed, Jun 6, 2012 at 3:32 PM, kamel maths <kam...@gm...> wrote:
> Hi,
>
> for this script:
> ------------------------------------
> from pylab import *
>
> fig = figure()
> ax = fig.add_subplot(111)
> ax.axis('equal')
>
> x = linspace(-2, 3, 50)
> ax.plot(x, sin(x))
>
> show()
> ---------------------------------
> If I try to get ymax with ax.get_ylim(), i obtain 1.0 whereas I observe it
> is 2.0.
> How can I obtain 2.0 for ymax ?
>
> Thanks.
>
> Kamel
>
Hi Kamel,
I'm not seeing the same result: I actually get back (-1.94, 1.94) from
`get_ylim`. When do you call `get_ylim`? Do you call it *after* calling
`plot`?
-Tony

Hi,
for this script:
------------------------------------
from pylab import *
fig = figure()
ax = fig.add_subplot(111)
ax.axis('equal')
x = linspace(-2, 3, 50)
ax.plot(x, sin(x))
show()
---------------------------------
If I try to get ymax with ax.get_ylim(), i obtain 1.0 whereas I observe it
is 2.0.
How can I obtain 2.0 for ymax ?
Thanks.
Kamel

> From: Eric Firing [mailto:ef...@ha...] 
> Sent: Wednesday, June 06, 2012 13:41
> To: mat...@li...
> Subject: Re: [Matplotlib-users] scatter plot with constant x
> 
> On 06/06/2012 06:42 AM, Ethan Gutmann wrote:
> >> ...
> >> No, but you can do this:
> >>
> >> plt.plot([3] * 4, [60, 80, 120, 180], ...)
> >
> > This started from a simple enough question, but it got me 
> thinking about what the fastest way to do this is (in case 
> you have HUGE arrays, or many loops over them).
[...]
> Since we end up needing float64 anyway:
> 
> In [3]: %timeit l=np.empty(10000,dtype=np.float64); l.fill(3)
> 100000 loops, best of 3: 14.1 us per loop
> 
> In [4]: %timeit l=np.zeros(10000,dtype=np.float64);l[:]=3
> 10000 loops, best of 3: 26.6 us per loop
> 
> Eric
Numpy's as_strided came to mind; it can make a large array that's really a
view of a one-element array:
 In [1]: as_strided = np.lib.stride_tricks.as_strided
 In [2]: s = as_strided(np.array([3], dtype=np.float64), shape=(10000,),
 ...: strides=(0,))
 In [3]: s[0] = 4
 In [4]: s[9999] # all elements share data
 Out[4]: 4.0
It's somewhat slower to create the base array and the view than to create and
fill a 10000-element array:
 In [5]: %timeit l = np.empty(10000, dtype=np.float64); l.fill(3)
 100000 loops, best of 3: 10.1 us per loop
 In [6]: %timeit s = as_strided(np.array([3], dtype=np.float64), 
 shape=(10000,), strides=(0,)) # line broken for email
 10000 loops, best of 3: 21.6 us per loop
However, once created, its contents may be changed much more quickly:
 In [7]: l = np.empty(10000, dtype=np.float64)
 In [8]: %timeit l.fill(3)
 100000 loops, best of 3: 7.71 us per loop
 In [9]: %timeit s[0] = 3
 10000000 loops, best of 3: 116 ns per loop
Numpy's broadcast_arrays uses as_strided under the hood. Code could look
like:
 x, y = np.broadcast_arrays(3, [60, 80, 120, 180])
 plt.plot(x, y, '+')
 x[0] = 21 # new x for all samples
 plt.plot(x, y, 'x')

On Jun 6, 2012, at 11:41 AM, Eric Firing wrote:
> Since we end up needing float64 anyway:
> 
> In [3]: %timeit l=np.empty(10000,dtype=np.float64); l.fill(3)
> 100000 loops, best of 3: 14.1 us per loop
nice, fill and empty seem to be responsible for about half the speed up each, good tools to know about. 

On 06/06/2012 06:42 AM, Ethan Gutmann wrote:
>> ...
>> No, but you can do this:
>>
>> plt.plot([3] * 4, [60, 80, 120, 180], ...)
>
> This started from a simple enough question, but it got me thinking about what the fastest way to do this is (in case you have HUGE arrays, or many loops over them). This may be old news to some of you, but I thought it was interesting:
>
> In ipython --pylab
>
> In [1]: %timeit l=[3]*10000
> 10000 loops, best of 3: 53.3 us per loop
>
> In [2]: %timeit l=np.zeros(10000)+3
> 10000 loops, best of 3: 26.9 us per loop
>
> In [3]: %timeit l=np.ones(10000)*3
> 10000 loops, best of 3: 32.9 us per loop
>
> In [4]: %timeit l=(np.zeros(1)+3).repeat(10000)
> 10000 loops, best of 3: 87.4 us per loop
>
> In [5]: %timeit l=np.zeros(10000);l[:]=3
> 10000 loops, best of 3: 21.6 us per loop
>
> In [6]: %timeit l=np.zeros(10000,dtype=np.uint8);l[:]=3
> 100000 loops, best of 3: 13.9 us per loop
>
> Using int16, int32, float32 get progressively slower to the default float64 case listed on line [5], changing the datatype in other methods doesn't result in nearly as large a speed up as it does in the last case.
>
> Ben's method is probably the most elegant for small arrays, but does any one else have a faster way to do this? (I'm assuming no use of blitz, inline C, f2py, but if you think you can do it faster in one of those, show me the way).
>
Since we end up needing float64 anyway:
In [3]: %timeit l=np.empty(10000,dtype=np.float64); l.fill(3)
100000 loops, best of 3: 14.1 us per loop
In [4]: %timeit l=np.zeros(10000,dtype=np.float64);l[:]=3
10000 loops, best of 3: 26.6 us per loop
Eric
> Sorry, maybe this is more appropriate on the numpy list.
>
> Ethan
>
>
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On 06/06/2012 12:54 PM, Ethan Gutmann wrote:
> On Jun 6, 2012, at 10:49 AM, Michael Droettboom wrote:
>> Interesting result. Note, however, that matplotlib will eventually turn
>> all data arrays into float64 at rendering time, so any speed advantage
>> to using integers will be lost by the subsequent conversion, I suspect.
> I don't think it does if you pass uint8 to imshow, but otherwise you might be right.
Sure. I was referring to scatter here.
With imshow, of course, everything is ultimately turned into 
8-bits-per-plane rgba.
Mike

On Jun 6, 2012, at 10:49 AM, Michael Droettboom wrote:
> Interesting result. Note, however, that matplotlib will eventually turn 
> all data arrays into float64 at rendering time, so any speed advantage 
> to using integers will be lost by the subsequent conversion, I suspect.
I don't think it does if you pass uint8 to imshow, but otherwise you might be right. 
ethan

On 06/06/2012 12:42 PM, Ethan Gutmann wrote:
>> ...
>> No, but you can do this:
>>
>> plt.plot([3] * 4, [60, 80, 120, 180], ...)
> Using int16, int32, float32 get progressively slower to the default float64 case listed on line [5], changing the datatype in other methods doesn't result in nearly as large a speed up as it does in the last case.
>
Interesting result. Note, however, that matplotlib will eventually turn 
all data arrays into float64 at rendering time, so any speed advantage 
to using integers will be lost by the subsequent conversion, I suspect.
Mike

> ...
> No, but you can do this:
> 
> plt.plot([3] * 4, [60, 80, 120, 180], ...)
This started from a simple enough question, but it got me thinking about what the fastest way to do this is (in case you have HUGE arrays, or many loops over them). This may be old news to some of you, but I thought it was interesting: 
In ipython --pylab
In [1]: %timeit l=[3]*10000
10000 loops, best of 3: 53.3 us per loop
In [2]: %timeit l=np.zeros(10000)+3
10000 loops, best of 3: 26.9 us per loop
In [3]: %timeit l=np.ones(10000)*3
10000 loops, best of 3: 32.9 us per loop
In [4]: %timeit l=(np.zeros(1)+3).repeat(10000)
10000 loops, best of 3: 87.4 us per loop
In [5]: %timeit l=np.zeros(10000);l[:]=3
10000 loops, best of 3: 21.6 us per loop
In [6]: %timeit l=np.zeros(10000,dtype=np.uint8);l[:]=3
100000 loops, best of 3: 13.9 us per loop
Using int16, int32, float32 get progressively slower to the default float64 case listed on line [5], changing the datatype in other methods doesn't result in nearly as large a speed up as it does in the last case. 
Ben's method is probably the most elegant for small arrays, but does any one else have a faster way to do this? (I'm assuming no use of blitz, inline C, f2py, but if you think you can do it faster in one of those, show me the way). 
Sorry, maybe this is more appropriate on the numpy list. 
Ethan

On Tue, Jun 5, 2012 at 11:53 AM, Ulrich vor dem Esche <
ulr...@go...> wrote:
> Hey! :o)
> This should be simple, but i cant manage: I need to plot many dots with
> the same x, like
>
> plt.plot([3,3,3,3],[60,80,120,180],'+',markersize=8,mec='k')
>
> The array for x values is silly, especially since the number of y values
> may be rather large. Is there a way to enter a constant there?
>
> Cheers to you all!
> Ulli
>
>
No, but you can do this:
plt.plot([3] * 4, [60, 80, 120, 180], ...)
Does that help?
Ben Root

On Wed, Jun 6, 2012 at 8:01 AM, Guillaume Gay
<gui...@mi...> wrote:
> Le 05/06/2012 16:25, Tom Dimiduk a écrit :
>> Is any of this stuff I should be looking to upstream or split off into
>> the start of a scientific imaging library for python?
> Have you had a look at skimage https://github.com/scikits-image ?
>
>
> BTW I uses matplotlib (and the whole pylab suite) in my projects for all
> the visualisation.
> A (peer reviewed published) example here:
> https://github.com/Kinetochore-segregation
>
> Best
>
> Guillaume
The Spyder (http://code.google.com/p/spyderlib/) python-based matlab
clone uses matplotlib for plotting.
Python(X,Y) (http://code.google.com/p/pythonxy/) is an integrated
windows python release that includes a ton of science, engineering,
and mathematics-oriented python packages, including matplotlib.
Numpy uses small bits of matplotlib when building the documentation,
but I don't know if that counts (I think it may even use it for
building matplotlib-related parts of the documentation, in which case
it really doesn't count).
I know someone is working on a pure python backend for the Cantor
advanced mathematics software (http://edu.kde.org/cantor/). The
project only started recently, however (see
http://blog.filipesaraiva.info/?p=779 ). There is also already a sage
backend for Cantor, which of course uses matplotlib for plotting
because that is what sage uses.
-Todd

Le 05/06/2012 16:25, Tom Dimiduk a écrit :
> Is any of this stuff I should be looking to upstream or split off into
> the start of a scientific imaging library for python?
Have you had a look at skimage https://github.com/scikits-image ?
BTW I uses matplotlib (and the whole pylab suite) in my projects for all 
the visualisation.
A (peer reviewed published) example here: 
https://github.com/Kinetochore-segregation
Best
Guillaume

On Tue, Jun 5, 2012 at 10:18 AM, Eric Firing <ef...@ha...> wrote:
> In oceanography: it is used in the shipboard ADCP data acquisition and
> processing systems, presently installed on 20 ships.
Suggestion: let's have for mpl something like what we created long ago
for IPython, an official page listing projects that use it (and btw,
if your project uses IPython as a component/library and you're not
already listed here, please do so!):
http://wiki.ipython.org/Projects_using_IPython
While I'm not a huge fan of wikis for everything, for this it's
actually a good solution, as it's very low overhead for others to
update. And it comes in handy as an official list whenever we do
presentations about IPython, to show that it's actually useful for
something.
I don't think we have a MPL wiki, but if it's just for a page or two
we could just use the one at github.
Cheers,
f

I got pretty good results with the code below. Note that I am reading the
FLIP_COLORS from a gui checkbox.
 FLIP_COLORS = self.dark_background_flag.get()
 if FLIP_COLORS:
 matplotlib.rcParams['figure.facecolor'] = '0.0'
 matplotlib.rcParams['axes.edgecolor'] = 'grey'
 matplotlib.rcParams['text.color'] = 'white'
 matplotlib.rcParams['ytick.color'] = '#00ff00'
 matplotlib.rcParams['xtick.color'] = '#0ED5D5'
 matplotlib.rcParams['axes.labelcolor'] = '#0ED5D5'
 matplotlib.rcParams['axes.facecolor'] = 'black'
 matplotlib.rcParams['grid.color'] = '0.3'
 matplotlib.rcParams['grid.linestyle'] = '-'
 matplotlib.rcParams['lines.markeredgewidth'] = 0.0
 else:
 matplotlib.rcdefaults()
 ## I seems like setting matplotlib.rcParams['figure.facecolor']
isn't
 ## enough. I think this is a bug and set_facecolor() is a
work-around.
 fig.set_facecolor(matplotlib.rcParams['figure.facecolor'])
This will flip and also flip back. I found a few colors didn't follow
the crowd. For example, axes.ylabel.color doesn't seem to have
an entry in rcParams. For this, I have to modify the plot generation
statments something like:
 ax.set_ylabel('Voltage (volts)',
color=matplotlib.rcParams['ytick.color'])
That sets the ylabel text to be the same as the tick marks. I also have to
do things like that to change line colors and such when
going flipping colors.
DavidS