matplotlib-users Mailing List for matplotlib

On 8/8/2009 10:09 AM John Hunter apparently wrote:
> Here is the corrected example:
Seems perfect!
Time to study it.
Thanks!
Alan

On Thu, Aug 06, 2009 at 10:53:13AM -0400, Jae-Joon Lee wrote:
> The thread below might be helpful.
> http://thread.gmane.org/gmane.comp.python.matplotlib.general/16373
> This will work as far as you keep the aspect="auto".
> Also, if you're using matplotlib 0.99rc version, or matplotlib from
> svn, you may take a look at
> http://matplotlib.sourceforge.net/mpl_toolkits/axes_grid/users/overview.html#insetlocator
> This works even though aspect=1 (or other value).
Thanks a lot to both of you. I have fallen in a time gap, and thus have
had to switch to other work and haven't been able to use your advice, but
it will come in handy when I get back to what I was doing.
Gaël

On Sat, Aug 8, 2009 at 9:00 AM, John Hunter<jd...@gm...> wrote:
> """
> This example shows how to use a path patch to draw a bunch of
> rectangles for an animated histogram
> """
Oops, in my last post I have the timer logic wrong because the call is
non-blocking so the animation ran too fast -- a rare occurrence in the
world of mpl animation :-)
Here is the corrected example:
"""
This example shows how to use a path patch to draw a bunch of
rectangles for an animated histogram
"""
import time
import numpy as np
import matplotlib
matplotlib.use('TkAgg') # do this before importing pylab
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import matplotlib.path as path
fig = plt.figure()
ax = fig.add_subplot(111)
# histogram our data with numpy
data = np.random.randn(1000)
n, bins = np.histogram(data, 100)
# get the corners of the rectangles for the histogram
left = np.array(bins[:-1])
right = np.array(bins[1:])
bottom = np.zeros(len(left))
top = bottom + n
nrects = len(left)
# here comes the tricky part -- we have to set up the vertex and path
# codes arrays using moveto, lineto and closepoly
# for each rect: 1 for the MOVETO, 3 for the LINETO, 1 for the
# CLOSEPOLY; the vert for the closepoly is ignored but we still need
# it to keep the codes aligned with the vertices
nverts = nrects*(1+3+1)
verts = np.zeros((nverts, 2))
codes = np.ones(nverts, int) * path.Path.LINETO
codes[0::5] = path.Path.MOVETO
codes[4::5] = path.Path.CLOSEPOLY
verts[0::5,0] = left
verts[0::5,1] = bottom
verts[1::5,0] = left
verts[1::5,1] = top
verts[2::5,0] = right
verts[2::5,1] = top
verts[3::5,0] = right
verts[3::5,1] = bottom
barpath = path.Path(verts, codes)
patch = patches.PathPatch(barpath, facecolor='green',
edgecolor='yellow', alpha=0.5)
ax.add_patch(patch)
ax.set_xlim(left[0], right[-1])
ax.set_ylim(bottom.min(), top.max())
def animate():
 if animate.cnt>=100:
 return
 animate.cnt += 1
 # simulate new data coming in
 data = np.random.randn(1000)
 n, bins = np.histogram(data, 100)
 top = bottom + n
 verts[1::5,1] = top
 verts[2::5,1] = top
 fig.canvas.draw()
 fig.canvas.manager.window.after(100, animate)
animate.cnt = 0
fig.canvas.manager.window.after(100, animate)
plt.show()

On Sat, Aug 8, 2009 at 6:17 AM, Kaushik
Ghose<Kau...@hm...> wrote:
> (http://matplotlib.sourceforge.net/api/artist_api.html#matplotlib.patches.Rectangle)
> e.g. set_height() to change the rectangles?
>
> e.g. code
> ------------------------------
> import pylab
> x = pylab.rand(20)
> h = pylab.hist(x)
> h[2][0].set_height(1)
> pylab.draw()
> --------------------------------
This is the right idea, but it is likely to be slow for animation,
since each rectangle is a separate mpl object, each with its own
graphics contexs, rendering step, etc. bar and hist were very early
functions which I wrote before we had collections and compound paths.
If I were rewriting it from scratch, I would use a compound path.
It's a little more work upfront because you have to manually compute
the vertices and path codes, but it will be goo-gobs faster. Here is
the animated histogram for tk using compound paths
"""
This example shows how to use a path patch to draw a bunch of
rectangles for an animated histogram
"""
import numpy as np
import matplotlib
matplotlib.use('TkAgg') # do this before importing pylab
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import matplotlib.path as path
fig = plt.figure()
ax = fig.add_subplot(111)
# histogram our data with numpy
data = np.random.randn(1000)
n, bins = np.histogram(data, 100)
# get the corners of the rectangles for the histogram
left = np.array(bins[:-1])
right = np.array(bins[1:])
bottom = np.zeros(len(left))
top = bottom + n
nrects = len(left)
# here comes the tricky part -- we have to set up the vertex and path
# codes arrays using moveto, lineto and closepoly
# for each rect: 1 for the MOVETO, 3 for the LINETO, 1 for the
# CLOSEPOLY; the vert for the closepoly is ignored but we still need
# it to keep the codes aligned with the vertices
nverts = nrects*(1+3+1)
verts = np.zeros((nverts, 2))
codes = np.ones(nverts, int) * path.Path.LINETO
codes[0::5] = path.Path.MOVETO
codes[4::5] = path.Path.CLOSEPOLY
verts[0::5,0] = left
verts[0::5,1] = bottom
verts[1::5,0] = left
verts[1::5,1] = top
verts[2::5,0] = right
verts[2::5,1] = top
verts[3::5,0] = right
verts[3::5,1] = bottom
barpath = path.Path(verts, codes)
patch = patches.PathPatch(barpath, facecolor='green',
edgecolor='yellow', alpha=0.5)
ax.add_patch(patch)
ax.set_xlim(left[0], right[-1])
ax.set_ylim(bottom.min(), top.max())
def animate():
 # simulate new data coming in
 data = np.random.randn(1000)
 n, bins = np.histogram(data, 100)
 top = bottom + n
 verts[1::5,1] = top
 verts[2::5,1] = top
 fig.canvas.draw()
def run():
 for i in range(100):
 fig.canvas.manager.window.after(100, animate)
fig.canvas.manager.window.after(100, run)
plt.show()

The guide is based on mpl 0.99.
You may upgrade your mpl,
or take a look at the thread below. it is a workaround that will work
with older version.
-JJ
On Fri, Aug 7, 2009 at 8:29 PM, Eliezer, David<del...@kn...> wrote:
> Hi,
>     I am graphing several time series together on the same graph,
> and so it is important to have a legend. I am also stacking two graphs
> on top of one another, because some of the data is of a different type.
>
> I wrote the graphing code as follows (pardon the newbie code, this is my
> first matplotlib day...)
>
> fig = plt.figure()
> axU = fig.add_subplot(211)
> axU.set_ylabel('price,$', color='r')
> axU.grid(True)
> axL = fig.add_subplot(212)
> axL.set_ylabel('size, shs', color='b')
> axL.grid(True)
> stepPlotsU = []
> stepPlotsL = []
> for i in range( len( plotvalues ) ) :
>   if axisSide[i] == 'U' :
>     stepPlotsU.append(axU.step( T,plotvalues[i], where='post' ))
>   else :
>     stepPlotsL.append(axL.step( T,plotvalues[i], where='post' ))
>
> lU = axU.legend(tuple(stepPlotsU),tuple(legendArrayU),loc='upper right')
> lL = axL.legend( tuple(stepPlotsL),tuple(legendArrayL),loc='upper right')
> axU.axis([float(TMin),float(TMax),1.1 * minValueU -0.1 *
> maxValueU,1.1*maxValueU - 0.1*minValueU])
> axL.axis([float(TMin),float(TMax),0.8*minValueL,1.2*maxValueL])
> axU.set_xlabel("Time, secs")
> fig.savefig('test.png')
>
> My code works, and produces an almost perfect graph, BUT...
> My problem is that the legend is quite large, and covers up a lot of the
> graph.
> The solution seems to be in the user guide, here
>
> <http://matplotlib.sourceforge.net/users/plotting/legend.html#plotting-guide-legend>
>
> where it shows how to call legend with the kwarg bbox_to_anchor=(1.05,1),
> and in the
> example, the box is nicely moved.
>
> In my case, though, I am calling
>
> axU.legend(tuple(stepPlotsU),tuple(legendArrayU),loc='upper right',
> bbox_to_anchor=(1.05,1))
>
> but it says this is an "unexpected keyword)", I assume it is saying that
> axis.legend() doesn’t use this
> keyword, even if legend does.
>
>   lL = axL.legend( tuple(stepPlotsL),tuple(legendArrayL),loc='upper
> right',bbox_to_anchor=(1.05,1))
>  File
> "/etg/source/Linux/pkg/Python-2.6.2/lib/python2.6/site-packages/matplotlib/axes.py",
> line 3823, in legend
>   self.legend_ = mlegend.Legend(self, handles, labels, **kwargs)
> TypeError: __init__() got an unexpected keyword argument 'bbox_to_anchor'
>
> Does anyone out there know what I can do to get the legend off my graph?
> And is there a way
> to shrink the legend font size down? It is way too large, and I can’t find
> anything in the docs about this,
> that axis.legend() will accept...
>
>                         Dave
>
> PS: I am using Python 2.6.2, matplotlib 0.98.5.3, numpy 1.3.0, scipy 0.7.1
>
>
>
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>

Hi Alan,
Alan G Isaac wrote:
> This is a second plea for help.
> http://www.mail-archive.com/mat...@li.../msg12632.html
> 
> I have a figure.Figure embedded in a FigureCanvasTkAgg.
> Each iteration, new data are received,
> and I want an updated histogram.
> 
> Now I can at least see a way to do this with pyplot:
> I just clear my axes each iteration, call ax.hist,
> and then call plot.draw(). I think (?) my problem
> is finding an equivalent to plt.draw() in the object
> oriented interface.
> 
> What I'd really like to do each iteration is change
> only the heights of the rectangles.
> 
I don't do fancy stuff like embedding figures in GUIs, but assuming things work 
the same way regardless: have you tried grabbing the object handles from hist 
(which are patches) and using the rectangle patch functions 
(http://matplotlib.sourceforge.net/api/artist_api.html#matplotlib.patches.Rectangle) 
e.g. set_height() to change the rectangles?
e.g. code
------------------------------
import pylab
x = pylab.rand(20)
h = pylab.hist(x)
h[2][0].set_height(1)
pylab.draw()
--------------------------------
h is a tuple with the last element as a list of patch objects
In [82]: h
Out[82]:
(array([3, 3, 2, 1, 3, 0, 0, 1, 3, 4]),
 array([ 0.02262869, 0.11966418, 0.21669968, 0.31373517, 0.41077066,
 0.50780615, 0.60484164, 0.70187713, 0.79891263, 0.89594812,
 0.99298361]),
 <a list of 10 Patch objects>)
Best
-Kaushik

Zane Selvans wrote:
> Hey there,
> 
> I'm trying to plot a bunch-o-vectors, colored red or blue, depending on 
> whether their magnitude is positive or negative (they represent 
> stresses), and I'm doing something like this:
> 
> mag1 = evals[:,0]
> ex1 = evecs[:,0,1]
> ey1 = evecs[:,0,0]
> C1 = np.where(mag1 >= 0, 'red', 'blue')
> 
> mag2 = evals[:,1]
> ex2 = evecs[:,1,1]
> ey2 = evecs[:,1,0]
> C2 = np.where(mag2 >= 0, 'red', 'blue')
> 
The C argument to quiver is for an array to be used with a colormap; you 
could use that with a listed colormap, or, to use something very close 
to what you already have, try using the "color" kwarg for your C1 and 
C2. It takes a single value or a sequence of color specs.
Eric
> basemap_ax.quiver(np.degrees(calc_phis), 
> np.degrees(np.pi/2.0-calc_thetas), mag1*ex1, mag1*ey1, C1, lw=0, 
> width=0.002, scale=1e8)
> basemap_ax.quiver(np.degrees(calc_phis), 
> np.degrees(np.pi/2.0-calc_thetas), mag2*ex2, mag2*ey2, C2, lw=0, 
> width=0.002, scale=1e8)
> basemap_ax.quiver(np.degrees(calc_phis), 
> np.degrees(np.pi/2.0-calc_thetas), -mag1*ex1, -mag1*ey1, C1, lw=0, 
> width=0.002, scale=1e8)
> basemap_ax.quiver(np.degrees(calc_phis), 
> np.degrees(np.pi/2.0-calc_thetas), -mag2*ex2, -mag2*ey2, C2, lw=0, 
> width=0.002, scale=1e8)
> 
> And it gives me the stack trace below... I also tried just using 
> plain-old quiver (not via basemap) and got the same error. I tried 
> using rgba tuples and grayscale strings as colors as well, and still got 
> the same error. Anybody have any idea what the deal is? Or am I 
> misunderstanding what *C* is supposed to be (i.e. not just an array of 
> colors, of the same length as the number of vectors being plotted...)
> 
> /Library/Python/2.5/site-packages/mpl_toolkits/basemap/__init__.pyc in 
> quiver(self, x, y, u, v, *args, **kwargs)
> 2877 ax.hold(h)
> 2878 try:
> -> 2879 ret = ax.quiver(x,y,u,v,*args,**kwargs)
> 2880 try:
> 2881 plt.draw_if_interactive()
> 
> /Library/Python/2.5/site-packages/matplotlib/axes.pyc in quiver(self, 
> *args, **kw)
> 5850 def quiver(self, *args, **kw):
> 5851 if not self._hold: self.cla()
> -> 5852 q = mquiver.Quiver(self, *args, **kw)
> 5853 self.add_collection(q, False)
> 5854 self.update_datalim(q.XY)
> 
> /Library/Python/2.5/site-packages/matplotlib/quiver.pyc in 
> __init__(self, ax, *args, **kw)
> 367 **kw)
> 368 self.polykw = kw
> --> 369 self.set_UVC(U, V, C)
> 370 self._initialized = False
> 371
> 
> /Library/Python/2.5/site-packages/matplotlib/quiver.pyc in set_UVC(self, 
> U, V, C)
> 439 mask = ma.mask_or(U.mask, V.mask, copy=False, shrink=True)
> 440 if C is not None:
> --> 441 C = ma.masked_invalid(C, copy=False).ravel()
> 442 mask = ma.mask_or(mask, C.mask, copy=False, shrink=True)
> 443 if mask is ma.nomask:
> 
> /Library/Python/2.5/site-packages/numpy/ma/core.pyc in masked_invalid(a, 
> copy)
> 1996 """
> 1997 a = np.array(a, copy=copy, subok=True)
> -> 1998 condition = ~(np.isfinite(a))
> 1999 if hasattr(a, '_mask'):
> 2000 condition = mask_or(condition, a._mask)
> 
> TypeError: bad operand type for unary ~: 'NotImplementedType'
> 
> 
> -- 
> Zane A. Selvans
> Amateur Earthling
> http://zaneselvans.org
> +1 303 815 6866
> 
> 
> ------------------------------------------------------------------------
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Hi,
 I am graphing several time series together on the same graph,
and so it is important to have a legend. I am also stacking two graphs
on top of one another, because some of the data is of a different type.
I wrote the graphing code as follows (pardon the newbie code, this is my first matplotlib day...)
fig = plt.figure()
axU = fig.add_subplot(211)
axU.set_ylabel('price,$', color='r')
axU.grid(True)
axL = fig.add_subplot(212)
axL.set_ylabel('size, shs', color='b')
axL.grid(True)
stepPlotsU = []
stepPlotsL = []
for i in range( len( plotvalues ) ) :
 if axisSide[i] == 'U' :
 stepPlotsU.append(axU.step( T,plotvalues[i], where='post' ))
 else :
 stepPlotsL.append(axL.step( T,plotvalues[i], where='post' ))
lU = axU.legend(tuple(stepPlotsU),tuple(legendArrayU),loc='upper right')
lL = axL.legend( tuple(stepPlotsL),tuple(legendArrayL),loc='upper right')
axU.axis([float(TMin),float(TMax),1.1 * minValueU -0.1 * maxValueU,1.1*maxValueU - 0.1*minValueU])
axL.axis([float(TMin),float(TMax),0.8*minValueL,1.2*maxValueL])
axU.set_xlabel("Time, secs")
fig.savefig('test.png')
My code works, and produces an almost perfect graph, BUT...
My problem is that the legend is quite large, and covers up a lot of the graph.
The solution seems to be in the user guide, here
<http://matplotlib.sourceforge.net/users/plotting/legend.html#plotting-guide-legend>
where it shows how to call legend with the kwarg bbox_to_anchor=(1.05,1), and in the
example, the box is nicely moved.
In my case, though, I am calling
axU.legend(tuple(stepPlotsU),tuple(legendArrayU),loc='upper right', bbox_to_anchor=(1.05,1))
but it says this is an "unexpected keyword)", I assume it is saying that axis.legend() doesn't use this
keyword, even if legend does.
 lL = axL.legend( tuple(stepPlotsL),tuple(legendArrayL),loc='upper right',bbox_to_anchor=(1.05,1))
 File "/etg/source/Linux/pkg/Python-2.6.2/lib/python2.6/site-packages/matplotlib/axes.py", line 3823, in legend
 self.legend_ = mlegend.Legend(self, handles, labels, **kwargs)
TypeError: __init__() got an unexpected keyword argument 'bbox_to_anchor'
Does anyone out there know what I can do to get the legend off my graph? And is there a way
to shrink the legend font size down? It is way too large, and I can't find anything in the docs about this,
that axis.legend() will accept...
 Dave
PS: I am using Python 2.6.2, matplotlib 0.98.5.3, numpy 1.3.0, scipy 0.7.1
DISCLAIMER:
This e-mail, and any attachments thereto, is intended only for use by the addressee(s) named herein and may contain legally privileged and/or confidential information. If you are not the intended recipient of this e-mail, you are hereby notified that any dissemination, distribution or copying of this e-mail, and any attachments thereto, is strictly prohibited. If you have received this in error, please immediately notify me and permanently delete the original and any copy of any e-mail and any printout thereof. E-mail transmission cannot be guaranteed to be secure or error-free. The sender therefore does not accept liability for any errors or omissions in the contents of this message which arise as a result of e-mail transmission.
NOTICE REGARDING PRIVACY AND CONFIDENTIALITY Knight Capital Group may, at its discretion, monitor and review the content of all e-mail communications. http://www.knight.com