On Fri, Oct 19, 2012 at 11:08 PM, elmar werling el...@net4werling.de wrote:
vmin=min(z), vmax=max(z)
A suggestion, when dealing with arrays, it is generally faster to use
the numpy function to compute the max and min, either np.max(z) or
z.max(), than the standard Python one.
I also think that would be useful. It would, for example, allow to
generate preview plots from other languages, without interfacing
them to Python. It must be said that MPL is actually quite nice
looking in the default settings for basic plotting, and this is an
nice feature that can exploit.
I
On Sun, Oct 14, 2012 at 3:42 AM, Paul Tremblay paulhtremb...@gmail.com wrote:
You don't want to install for python 2.6. Python 2.6 is out of date at
this point.
I disagree. Most of the tools and libraries for 2.6 are available for
2.7 and viceversa (they are quite similar), but not all of them.
in the terminal type:
sudo port install py26-python
sudo port install py26-matplotlib
and there you are!
On Sun, Oct 14, 2012 at 5:37 AM, lulu lauracur...@me.com wrote:
Okay -- that's good to know.
I've just installed macports, but am not sure how to go about using
matplotlib in my python
On Windows it works just fine. Just a wild guess, can you try to make
it a raw string?
plt.xlabel(r'Percent [%]')
On Tue, Oct 9, 2012 at 10:45 PM, Francesco Montesano
franz.berges...@gmail.com wrote:
Hi
2012/10/9 Nikolaus Rath nikol...@rath.org
Hello,
For some reason, my matplotlib isn't
On Sunday, September 9, 2012, Eric Firing wrote:
Regarding the need to pre-allocate: yes, matlab is slicker in this
regard, and every now and then there is discussion about implementing
equivalent behavior in numpy, or in an add-on module.
Technically, you don´t have to preallocate the
In the example you provide, bins is returned by the hist command,
whereas in your code, bins is a number that you defined as 20. So,
change:
bins = 20
plt.hist(C, bins, ...
by:
nbins = 20
n, bins, patches = plt.hist(C, nbins, ...
As a side comment, your data loading is too complex, and fail
On Fri, Jul 27, 2012 at 9:57 PM, surfcast23 surfcas...@gmail.com wrote:
y = mlab.normpdf( nbins, avg, sigma)
l = plt.plot(nbins, y, 'r--', linewidth=1)
plt.show()
You should not change bins there, as you are evaluating the gaussian
function at different values.
Also, sigma is a vector, but it
, surfcast23 surfcas...@gmail.com wrote:
Thanks for catching that sigma was still a vector! I am no longer getting the
errors, but the best fit line is not showing up.Is there something else I am
missing ?
BTW thanks for the heads up on the np.mean and np.standard functions.
Khary
Daπid
On Sat, Jul 28, 2012 at 12:22 AM, surfcast23 surfcas...@gmail.com wrote:
Am I reading (bins[1]-bins[0]) correctly as taking the difference between
what is in the second and first bin?
Yes. I am multipliying the width of the bins by their total height.
Surely there are cleaner and more general
On Thu, Jul 26, 2012 at 5:05 PM, Andreas Hilboll li...@hilboll.de wrote:
That's really easy :) I could live with this solution, applying some
external tool like pdfcrop to the result.
If you can use other output, you can generate a png image, which would
be easier to cut (even inside
On Mon, Jul 16, 2012 at 11:09 AM, Benjamin Jonen bjo...@gmail.com wrote:
2) The coloring and the way the lines curve around looks very nice to
me. I remember that the Excel charts did not have this nice look
before Excel 2007. Can I achieve similar effects with matplotlib? I'm
not really sure
I don't know if there is any reason for not having it, but as a
workaround, you could use np.hist to get the data (syntax is the same
as mpl.hist and returns the same numbers, but without drawing) and
then renormalise and plot with mpl.bars.
On Thu, Jul 12, 2012 at 4:42 PM, Alan G Isaac
Hello,
I am trying to type Erdős in the title of a figure. I am using the
LaTeX command Erd\H{o}s, as it works in normal latex documents both in
text and math mode, but it malfunctioning putting the two lines on the
d. To obtain the proper typeset, I have to write:
plt.title(r$Erdo\H s$)
This
If all your values are positive (and you are sure of it), you could
use the SymmetricalLogScale It uses log scale for large values (both
positive and negative), and linear for small ones.
http://matplotlib.sourceforge.net/devel/add_new_projection.html
I believe there is a way to do it without
First, this is another topic, so please, change the subject of the
message so it doesn't get messed up with others (and possible help
lost in the process).
Now, you are indeed plotting one dot at the time and generating a
label for it. If you don't want that, you have to plot the whole list
at
http://matplotlib.sourceforge.net/examples/api/demo_affine_image_00.html
This link of the gallery example is broken (error 404). It corresponds
to the third example of the fourth row.
--
Keep Your Developer Skills
I can run the script you provided up to 600 without noticing any
memory growth (TkAgg backend by default). Using the WxAgg, the memory
leak appears.
If you don't explicitly need WxAgg, I would recommend using TkAgg, or
better Agg if you are not showing the figures. If it is not the case,
more
You are importing pylab again before each plot. You have loaded it before.
for i,t in enumerate(times):
import pylab as pl # --ERASE THIS
On Wed, Jun 15, 2011 at 12:10 PM, Alain Pascal Frances
frances17...@itc.nl wrote:
Hi,
I have a script that creates and saves figures in a loop.
I am using MPL 1.0.1 with Python 2.6 over Windows XP and it works.
I would like to add an advice: range(n) creates a list of size n, and
stores it in memory. But in your code you are only using one number at
a time. Python has a better instruction: xrange. It works exactly like
range, but doesn't
I have checked with all the interpolation modes and the only one that
behaves badly is 'nearest'. There are them:
http://dl.dropbox.com/u/1351211/Interpolation_modes.zip
On Mon, Apr 18, 2011 at 12:46 PM, Emanuele Passera
emanuele.pass...@treuropa.com wrote:
Hello everybody,
I am experiencing a
Hello.
I want to plot the result of a calculation, show it, modify and
repeat. I have tried with pyplot interactive mode, see code:
http://pastebin.com/jsdsLN4z
The first plot shows fine, but from now on, all the new plots appear
in a new window and all of them are blank --including the first
Hello.
It is stated that show() should be the last function in a script, as
long as it will stop the execution of the program. Nevertheless, I
have seen that the current behavior is just a pause in the flow, and
it will be restored when the window is closed. The documentation says:
Many users
On Tue, Nov 2, 2010 at 12:49 PM, David Kremer david.kremer...@gmail.com wrote:
Personally I used show() yesterday, and it blocks the execution until
the figure's window is closed. It is of perfect convenience I think,
as a default behavior.
Yes, this is the optimal behavior, but it is not
If you are curious, here is what Mathematica can tell about the
integral (assuming everything constant but z):
http://pastebin.com/Gir3XZBe
On Wed, Oct 13, 2010 at 4:28 PM, Waléria Antunes David
waleriantu...@gmail.com wrote:
Hi all,
I know here is a group for matplotlib, but can anyone
On Mon, Oct 4, 2010 at 11:36 AM, nickj nickja...@gmail.com wrote:
To better illustrate my problem, here is a screenshot with added red lines
pointing to the odd slicing I get around the
contours: http://www.bigoceans.com/slices.png
A low-tech option is to plot first all the oceans in light
On Sat, Oct 2, 2010 at 7:39 PM, Jouni K. Seppänen j...@iki.fi wrote:
And yet, we still allow for saving to jpegs.
Wow, I didn't know. Last time I tried that I got a traceback, and
assumed that it was not supported exactly because jpeg is a nonsensical
format for most graphs.
Don't forget MPL
today. ;-)
On Thu, Sep 16, 2010 at 5:12 PM, Benjamin Root ben.r...@ou.edu wrote:
On Thu, Sep 16, 2010 at 12:03 AM, Daπid davidmen...@gmail.com wrote:
Does MPL support in any way the Z channel? If not, is there any
possibility to use it? For example, to create a parallel matrix of the
same
I think you can make it with pyplot.contourf() and the argument V
http://matplotlib.sourceforge.net/api/pyplot_api.html#matplotlib.pyplot.contour
contour(Z,V)
contour(X,Y,Z,V)
draw contour lines at the values specified in sequence V
On Wed, Sep 15, 2010 at 9:02 PM, Luke hazelnu...@gmail.com
Does MPL support in any way the Z channel? If not, is there any
possibility to use it? For example, to create a parallel matrix of the
same dimensions of the image with the values of Z in each pixel.
Thank you very much.
David.
If it is only one image, an easy low-tech workaround is to save it as
PNG and then put it into a PDF. This way, you only have to load one
element.
On Mon, Aug 30, 2010 at 11:36 PM, Jeremy Conlin jlcon...@gmail.com wrote:
I have a matplotlib plot that I saved to a pdf image. The plot
consists
Image Magick and Inkscape seem to work for this. Probably the first
one is easier to automatize in batch processing.
On Thu, Aug 26, 2010 at 4:21 PM, Michael Droettboom md...@stsci.edu wrote:
matplotlib does not have any built-in support for any color spaces other
than RGB. You would need to
I cannot see what is wrong, but after saving each figure you should
add plt.clf() in order to delete the image and preventing memory
leaks, because MPL stores one image in top of the other.
For the filecode I suggest you to use something like:
savefig(head+str(filecode).zfill(digits)+format,
What I use to create animations is plainly:
savefig(head+str(filecode).zfill(digits)+format, dpi=205)
plt.clf()
filecode+=1
where filecode is the name, digits an int and format usually .png.
clf() is important in order to prevent memory leaks, because otherwise
mpl stores all the figures one in
On Sat, Aug 14, 2010 at 10:19 PM, Eric Firing efir...@hawaii.edu wrote:
The problem is that the alpha kwarg can never be other than a scalar
in mpl at present, as far as I know. The error message from to_rgba was
intended to be informative, but in this case it is misleading.
Thanks for your
Hello.
I have had an issue trying to plot an histogram with Matplotlib. The line is:
plt.hist([SNIa.angles, SNIbc.angles, SNII.angles], 11, range=[-pi, pi],
normed=True,histtype='stepfilled',color=['g', 'r',
'b'],alpha=[1, 0.6, 1])
But the error is raised when I try to save the image.
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