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From: Eric F. <ef...@ha...> - 2007-11-20 21:15:35
|
Jeff Whitaker wrote:
> If I draw two images with imshow, then set_zorder for one of them to be
> higher than the other, should that one be the one that displays?
Jeff,
It is a wart. Images are handled separately from other artists, and
always drawn first, either as a composite or in the order in which they
were added to the list. Here is the relevant code in the Axes.draw()
method:
if len(self.images)<=1 or renderer.option_image_nocomposite():
for im in self.images:
im.draw(renderer)
else:
# make a composite image blending alpha
# list of (mimage.Image, ox, oy)
mag = renderer.get_image_magnification()
ims = [(im.make_image(mag),0,0)
for im in self.images if im.get_visible()]
im = mimage.from_images(self.bbox.height()*mag,
self.bbox.width()*mag,
ims)
im.is_grayscale = False
l, b, w, h = self.bbox.get_bounds()
# composite images need special args so they will not
# respect z-order for now
renderer.draw_image(l, b, im, self.bbox)
Maybe the set_zorder method for images should raise a warning, since it
can't do what one might reasonably expect.
Mike may be able to comment on the prospects for removing this wart at
some point in the transforms branch; he is looking at related questions
right now.
Eric
>
> for example, with
>
> from pylab import *
> delta = 0.025
> x = y = arange(-3.0, 3.0, delta)
> X, Y = meshgrid(x, y)
> Z1 = bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
> Z2 = bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
> Z = Z2-Z1 # difference of Gaussians
> im2 = imshow(Z2, interpolation='bilinear', cmap=cm.gray,
> origin='lower', extent=[-3,3,-3,3])
> im2.set_zorder(2)
> im1 = imshow(Z1, interpolation='bilinear', cmap=cm.gray,
> origin='lower', extent=[-3,3,-3,3])
> im1.set_zorder(1)
> show()
>
> I expected Z2 to be plotted, but I actually see Z1. In fact, I always
> see the last one that was plotted, regardless of what I set the zorder to.
>
> I'm using the latest SVN, with GTKAgg.
>
> -Jeff
>
|
|
From: Jeff W. <js...@fa...> - 2007-11-20 20:58:27
|
If I draw two images with imshow, then set_zorder for one of them to be
higher than the other, should that one be the one that displays?
for example, with
from pylab import *
delta = 0.025
x = y = arange(-3.0, 3.0, delta)
X, Y = meshgrid(x, y)
Z1 = bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
Z = Z2-Z1 # difference of Gaussians
im2 = imshow(Z2, interpolation='bilinear', cmap=cm.gray,
origin='lower', extent=[-3,3,-3,3])
im2.set_zorder(2)
im1 = imshow(Z1, interpolation='bilinear', cmap=cm.gray,
origin='lower', extent=[-3,3,-3,3])
im1.set_zorder(1)
show()
I expected Z2 to be plotted, but I actually see Z1. In fact, I always
see the last one that was plotted, regardless of what I set the zorder to.
I'm using the latest SVN, with GTKAgg.
-Jeff
--
Jeffrey S. Whitaker Phone : (303)497-6313
Meteorologist FAX : (303)497-6449
NOAA/OAR/PSD R/PSD1 Email : Jef...@no...
325 Broadway Office : Skaggs Research Cntr 1D-124
Boulder, CO, USA 80303-3328 Web : http://tinyurl.com/5telg
|
|
From: Eric F. <ef...@ha...> - 2007-11-20 17:15:52
|
John Hunter wrote: > On Nov 19, 2007 10:08 PM, Jeff Whitaker <js...@fa...> wrote: >> Bryan Fodness wrote: >>> I would like to use pcolor with the colors going from white (zero >>> values) to black (largest value). > >> pcolor(a, shading = 'flat', cmap=cm.binary) > > It may not be a binary map you want (that has just two colors) but a > reversed grayscale map which is continuous from white to black, in > which case you would use cmap=cm.gray_r Agreed, logically it should be cm.gray_r. But, surprisingly, cm.binary looks identical. Maybe cm.binary is not doing what you intended it to do? (I have always found LinearSegmentedColormap hard to understand, in spite of what seems to be abundant explanation in the docstrings.) Eric |
|
From: John H. <jd...@gm...> - 2007-11-20 16:52:26
|
On Nov 20, 2007 10:43 AM, Jordan Atlas <jc...@co...> wrote:
> Can someone recomend a way to save the data in such a way that the
> columns (or rows) are labeled? In otherwords, it would be nice to be
> able to open the saved data and know what each row is without having to
> refer to the script that created it. (referring to the creating script
> feels error prone when you have many rows of data being saved). I'm
> currently using the 'pylab.save' function to save the data.
I suggest using numpy record arrays for this -- the columns have names
and the data can be of different types. You can save and load them
using pickle (numpy.load and numpy.save will use pickle under the
hood).
If you want to stick with ASCII flat file representation (eg for use
with other programs), in matplotlib svn there are two nice functions
to help here: rec2csv and csv2rec. They support saving numpy record
arrays to CSV files with column names, and loading these back up later
doign type introspection to figure out the types (datetime, str,
float, int).
In [1]: import numpy as n
In [2]: import matplotlib.mlab as mlab
In [3]: x = n.random.rand(20,4)
In [4]: r = n.rec.fromrecords(x, names='age,weight,height,cash')
In [5]: r.dtype
Out[5]: dtype([('age', '<f8'), ('weight', '<f8'), ('height', '<f8'),
('cash', '<f8')])
In [7]: mlab.rec2csv(r, 'mydata.csv')
In [8]: !head mydata.csv
age,weight,height,cash
0.0449935,0.252057,0.316116,0.0635711
0.777189,0.155186,0.0537382,0.233598
0.731376,0.654577,0.977792,0.0171022
0.685975,0.373741,0.714592,0.620079
0.634548,0.956708,0.360962,0.885379
0.431011,0.359094,0.21484,0.961865
0.115155,0.78767,0.352753,0.769402
0.984747,0.720163,0.887608,0.316844
0.0478857,0.813668,0.882535,0.8837
In [9]: newr = mlab.csv2rec('mydata.csv')
In [10]: newr.dtype
Out[10]: dtype([('age', '<f8'), ('weight', '<f8'), ('height', '<f8'),
('cash', '<f8')])
In [11]: newr
Out[11]:
recarray([ (0.044993499999999999, 0.25205699999999998,
0.31611600000000001, 0.063571100000000005),
(0.77718900000000002, 0.15518599999999999, 0.0537382, 0.233598),
(0.73137600000000003, 0.65457699999999996, 0.97779199999999999,
0.017102200000000001),
(0.685975, 0.37374099999999999, 0.714592, 0.62007900000000005),
(0.634548, 0.956708, 0.36096200000000001, 0.88537900000000003),
(0.43101099999999998, 0.35909400000000002, 0.21484, 0.96186499999999997),
(0.11515499999999999, 0.78766999999999998, 0.35275299999999998,
0.76940200000000003),
(0.98474700000000004, 0.720163, 0.88760799999999995,
0.31684400000000001),
(0.047885700000000003, 0.81366799999999995,
0.88253499999999996, 0.88370000000000004),
(0.044475599999999997, 0.89918900000000002,
0.076484499999999997, 0.114994),
(0.75139299999999998, 0.70954300000000003, 0.458505,
0.33839900000000001),
(0.14619299999999999, 0.907717, 0.24915200000000001,
0.67030400000000001),
(0.89663199999999998, 0.61957300000000004,
0.0060039200000000003, 0.048883500000000003),
(0.20794000000000001, 0.56046499999999999,
0.078303899999999996, 0.216032),
(0.28726000000000002, 0.14282500000000001, 0.51740200000000003,
0.553037),
(0.96326999999999996, 0.21327299999999999, 0.72040999999999999,
0.181446),
(0.31984000000000001, 0.39338299999999998, 0.45787899999999998,
0.33919199999999999),
(0.42086200000000001, 0.98801499999999998, 0.53429000000000004,
0.074105699999999997),
(0.104211, 0.15845100000000001, 0.13339200000000001,
0.99228300000000003),
(0.73563299999999998, 0.948407, 0.44708900000000001,
0.79521399999999998)],
dtype=[('age', '<f8'), ('weight', '<f8'), ('height', '<f8'),
('cash', '<f8')])
You can also work with non floating point data
In [14]: url = 'http://ichart.finance.yahoo.com/table.csv?s=GE&d=10&e=20&f=2007&g=d&a=0&b=2&c=1962&ignore=.csv'
In [15]: import urllib
In [16]: urllib.urlretrieve(url, 'ge.csv')
Out[16]: ('ge.csv', <httplib.HTTPMessage instance at 0x8fa7b2c>)
In [17]: r = mlab.csv2rec('ge.csv')
In [18]: !head ge.csv
Date,Open,High,Low,Close,Volume,Adj Close
2007-11-19,38.48,38.51,38.00,38.16,35415000,38.16
2007-11-16,38.50,38.67,37.87,38.65,50181100,38.65
2007-11-15,38.93,38.93,38.13,38.31,41590000,38.31
2007-11-14,39.90,39.95,38.82,39.01,39650800,39.01
2007-11-13,38.50,39.25,38.25,39.21,42053400,39.21
2007-11-12,38.24,39.04,38.17,38.25,36968000,38.25
2007-11-09,38.52,38.75,38.11,38.38,42662200,38.38
2007-11-08,39.20,39.32,37.50,39.02,52970300,39.02
2007-11-07,39.90,39.93,38.99,39.08,46720100,39.08
In [20]: r[:10]
Out[20]:
recarray([ (datetime.datetime(2007, 11, 19, 0, 0), 38.479999999999997,
38.509999999999998, 38.0, 38.159999999999997, 35415000,
38.159999999999997),
(datetime.datetime(2007, 11, 16, 0, 0), 38.5,
38.670000000000002, 37.869999999999997, 38.649999999999999, 50181100,
38.649999999999999),
(datetime.datetime(2007, 11, 15, 0, 0), 38.93, 38.93,
38.130000000000003, 38.310000000000002, 41590000, 38.310000000000002),
(datetime.datetime(2007, 11, 14, 0, 0), 39.899999999999999,
39.950000000000003, 38.82, 39.009999999999998, 39650800,
39.009999999999998),
(datetime.datetime(2007, 11, 13, 0, 0), 38.5, 39.25, 38.25,
39.210000000000001, 42053400, 39.210000000000001),
(datetime.datetime(2007, 11, 12, 0, 0), 38.240000000000002,
39.039999999999999, 38.170000000000002, 38.25, 36968000, 38.25),
(datetime.datetime(2007, 11, 9, 0, 0), 38.520000000000003,
38.75, 38.109999999999999, 38.380000000000003, 42662200,
38.380000000000003),
(datetime.datetime(2007, 11, 8, 0, 0), 39.200000000000003,
39.32, 37.5, 39.020000000000003, 52970300, 39.020000000000003),
(datetime.datetime(2007, 11, 7, 0, 0), 39.899999999999999,
39.93, 38.990000000000002, 39.079999999999998, 46720100,
39.079999999999998),
(datetime.datetime(2007, 11, 6, 0, 0), 40.200000000000003,
40.490000000000002, 39.969999999999999, 40.18, 42131000, 40.18)],
dtype=[('date', '|O4'), ('open', '<f8'), ('high', '<f8'),
('low', '<f8'), ('close', '<f8'), ('volume', '<i4'), ('adj_close',
'<f8')])
|
|
From: Jordan A. <jc...@co...> - 2007-11-20 16:43:23
|
Can someone recomend a way to save the data in such a way that the columns (or rows) are labeled? In otherwords, it would be nice to be able to open the saved data and know what each row is without having to refer to the script that created it. (referring to the creating script feels error prone when you have many rows of data being saved). I'm currently using the 'pylab.save' function to save the data. Thanks, --Jordan Eric Firing wrote: > Jordan Atlas wrote: > >> Is it possible to 'save' a matplotlib figure object using >> something like the python pickle module? Basically, I'd like to save >> the 'figure' as a file so that I can open it and manipulate it if >> something is wrong. Alternative suggestions are welcome. >> > No, this has been requested before but it is not an easy modification > of mpl to make. The alternative suggestion is to always encapsulate > the making of a figure in a script, and then save the script and the > data. If you want to save the combination in a single file, then > arrange for the script and data to reside in a single subdirectory, > and zip or tar that subdirectory. |
|
From: John H. <jd...@gm...> - 2007-11-20 14:21:12
|
On Nov 19, 2007 10:08 PM, Jeff Whitaker <js...@fa...> wrote: > Bryan Fodness wrote: > > I would like to use pcolor with the colors going from white (zero > > values) to black (largest value). > pcolor(a, shading = 'flat', cmap=cm.binary) It may not be a binary map you want (that has just two colors) but a reversed grayscale map which is continuous from white to black, in which case you would use cmap=cm.gray_r JDH |
|
From: Michael D. <md...@st...> - 2007-11-20 13:08:10
|
Which version(s) of matplotlib are you using (on both platforms)...
Cheers,
Mike
Jun She wrote:
> <mailto:mat...@li...>Dear matplotlib users,
>
> The operating system of my PC is Ubuntu dapper.
> I installed matplotlib with Python 2.4.
>
> When I use /imshow/ with /axis('equal') /, the axis of the output figure
> is not equal.
> However, when I run same script in my laptop with Windows XP, the figure
> seems OK.
>
> Why?
> Many thanks!
>
> BR/Jun
>
>
> ------------------------------------------------------------------------
>
> -------------------------------------------------------------------------
> This SF.net email is sponsored by: Microsoft
> Defy all challenges. Microsoft(R) Visual Studio 2005.
> http://clk.atdmt.com/MRT/go/vse0120000070mrt/direct/01/
>
>
> ------------------------------------------------------------------------
>
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
--
Michael Droettboom
Science Software Branch
Operations and Engineering Division
Space Telescope Science Institute
Operated by AURA for NASA
|
|
From: Jun S. <sh...@gm...> - 2007-11-20 07:53:18
|
<mat...@li...>Dear matplotlib users,
The operating system of my PC is Ubuntu dapper.
I installed matplotlib with Python 2.4.
When I use *imshow* with *axis('equal')*, the axis of the output figure is
not equal.
However, when I run same script in my laptop with Windows XP, the figure
seems OK.
Why?
Many thanks!
BR/Jun
|
|
From: Jeff W. <js...@fa...> - 2007-11-20 04:08:13
|
Bryan Fodness wrote: > I would like to use pcolor with the colors going from white (zero > values) to black (largest value). > > I am using, > > pcolor(a, shading = 'flat') > colorbar() > > I do not see how to do this. > > > pcolor(a, shading = 'flat', cmap=cm.binary) -Jeff -- Jeffrey S. Whitaker Phone : (303)497-6313 NOAA/OAR/CDC R/PSD1 FAX : (303)497-6449 325 Broadway Boulder, CO, USA 80305-3328 |
|
From: Bryan F. <bry...@gm...> - 2007-11-20 03:37:22
|
I would like to use pcolor with the colors going from white (zero values) to black (largest value). I am using, pcolor(a, shading =3D 'flat') colorbar() I do not see how to do this. --=20 "The game of science can accurately be described as a never-ending insult to human intelligence."=09- Jo=E3o Magueijo |
|
From: Chris D. <di...@al...> - 2007-11-20 02:26:02
|
Hello All, When I launch ipython (by typing "ipython -pylab") and issue the first matplotlib command, such as simply "figure()", I get the following error messages: TkMacOSXDoHLEvent failed : cppe 1 ,kHighLevelEvent 44464458 0000 XDFD,-1708 TkMacOSXDoHLEvent failed : cppe 1 ,kHighLevelEvent 44464458 0000 XDFD,-1708 Any subsequent calls yield no problems. While this is not a show stopper, it is certainly annoying. Any thoughts on what might be causing the problem? Thanks in advance for your assistance! Chris |