Mailman 3 November 2020 - Matplotlib-users

tk and matplotlib
by Vincent Douce Mathoscope Nov. 30, 2020

Nov. 30, 2020

hi i am creating a matplotlib draw (musical sales on a guitar) in function of some parameters (tonic note, type of gamme or chord...) i am now creating tk radio buttons so that the user can choose the parameters i can get these parameters and create the matplotlib picture with them but the matplotlib will open a different window would it be possible to have a same window, divided as a table, with at the lift the radio buttons, and at the right the matplotlib result ? ---- https://lemathoscope.com/ cours de maths postbac par visio

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Set text.latex.preamble in style sheet
by Thibaut Appel Nov. 23, 2020

Nov. 23, 2020

Dear matplotlib developers, Suppose I define a style sheet 'foo.mplstyle' as text.usetex : True font.family : serif and setting the LaTeX preamble in the following MWE: import numpy as np import matplotlib.pyplot as plt plt.rcParams["text.latex.preamble"] = r"\usepackage{newpxtext} \usepackage{eulervm}" plt.style.use('foo') x = np.linspace(0.0,1.0,50) y = x plt.plot(x,y) plt.show() This works fine with matplotlib 3.3.0. But the MWE fails when I try to set the LaTeX preamble directly in the style sheet. What's the correct syntax for 'text.latex.preamble'? I tried text.latex.preamble : r"\usepackage{newpxtext} \usepackage{eulervm}" text.latex.preamble : "\usepackage{newpxtext}, \usepackage{eulervm}" text.latex.preamble : \usepackage{newpxtext},\usepackage{eulervm} without success, despite seeing the API change notice <https://matplotlib.org/3.3.3/api/prev_api_changes/api_changes_3.3.0.html> about the parameter. System info: Ubuntu 20.10 Linux 5.8.0-29-generic x86_64 matplotlib 3.3.0 installed with apt ghostscript 9.52 texlive 2020 Thanks for the support, Thibaut

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Finalizing Code of Conduct
by Hannah Nov. 15, 2020

Nov. 15, 2020

Hello community, We have been revising our code of conduct and settled on using the latest form of the contributor covenant with some minor customization for reporting: https://github.com/matplotlib/matplotlib/pull/18802 We chose this code of conduct based on these criteria: 1) lists the behaviors we think are appropriate and inappropriate 2) includes reporting and enforcement mechanisms 3) clear and concise We are seeking holistic feedback on whether this new CoC is an accurate statement of the behavior we expect in Matplotlib, our expectations are in line with yours, and if there are significant issues with the text, Please refrain from wordsmithing; we think that most problems are likely an issue with the philosophy of the CoC and not solvable via changing some words (plus or minus the reporting and enforcement sections). We will be taking feedback for the next week (deadline Tuesday Nov 24th, 3:00PM EST) on the mailing list, discourse, pull request, and in private as appropriate. Thank you, Hannah

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Title overwriting axis label
by Neal Becker Nov. 13, 2020

Nov. 13, 2020

The plot is produced by the following code: import matplotlib.pyplot as plt ax = plt.gca() ax.ticklabel_format(style='sci', scilimits=(-3,3), axis='y') ax.ticklabel_format(style='sci', scilimits=(-3,3), axis='x') plt.plot (np.arange (start=0, stop=len(freq)*opt.segment, step=opt.segment), freq) plt.xlabel ('time(symbols)') plt.title (r'$E_s/N_0=%s, \alpha=%s, \omega_n=%s, \Delta_f=%s$ mod=%s'%(opt.esnodB, opt.alpha, latex_float(opt.omega_n), opt.freq, opt.mod)) plt.ylabel ('freq(cycle/sym)') plt.tight_layout() Any suggestions how to prevent this overlap of the title and the axis label? This is mpl-3.3.3

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prohibited to create memoryview without Py_buffer
by Neal Becker Nov. 12, 2020

Nov. 12, 2020

fedora-33 mplcairo-0.3 rpm -qa python3-matplotlib-* python3-matplotlib-tk-3.3.2-1.fc33.x86_64 python3-matplotlib-qt4-3.3.2-1.fc33.x86_64 python3-matplotlib-data-3.3.2-1.fc33.noarch python3-matplotlib-data-fonts-3.3.2-1.fc33.noarch python3-matplotlib-qt5-3.3.2-1.fc33.x86_64 This happened when I tried so save a plot to pdf using the gui. I got a dialog error box with the message: prohibited to create memoryview without Py_buffer -- *Those who don't understand recursion are doomed to repeat it*

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Re: [Matplotlib-users] Arc drawing
by Paul Hobson Nov. 2, 2020

Nov. 2, 2020

Hey Luca, You're referencing some variables in your snippet that you haven't defined, so I can't really help you troubleshoot that. Here's a basic example to get you started: from matplotlib import pyplot from matplotlib import patches fig, ax = pyplot.subplots() arc = patches.Arc( xy=(2, 3), width=5, height=5, angle=0, theta1=45, theta2=180, ) ax.add_patch(arc) ax.set_aspect('equal') ax.autoscale() On Mon, Nov 2, 2020 at 8:36 AM Luca Bertolotti <luca72.bertolotti(a)gmail.com> wrote: > Hy Paul > I heve read it but i'm not able to draw it: > The data that i have are: > {'handle': '39D', 'owner': '1F', 'layer': 'AM_0', 'center': Vector(1000.0, > -1256.595067376733, 0.0), 'radius': 2100.0, 'start_angle': > 67.75727639121727, 'end_angle': 112.2427236087827} > > h= float(2* entity.dxf.radius) > > I have try this: > b = Arc((float(xr), float(yr)), (h, h),0, float(entity.dxf.end_angle), > float(entity.dxf.end_angle) , color='red', lw=1) > ax.add_patch(b) > > But i get: > Exception "unhandled TypeError" > can't multiply sequence by non-int of type 'float' > File: > c:\users\luca\appdata\local\programs\python\python38\lib\site-packages\matplotlib\patches.py, > Line: 1427 > > Can you help me > > Bets Regards > > Il giorno lun 2 nov 2020 alle ore 17:07 Paul Hobson <pmhobson(a)gmail.com> > ha scritto: > >> Hey Luca, >> >> Matplotllib has an Arc artist: >> https://matplotlib.org/3.3.2/api/_as_gen/matplotlib.patches.Arc.html >> >> -Paul >> >> On Mon, Nov 2, 2020 at 5:02 AM Luca Bertolotti < >> luca72.bertolotti(a)gmail.com> wrote: >> >>> Hello i have to draw an arc with matplotlib, i have the following data: >>> >>> {'handle': '39D', 'owner': '1F', 'layer': 'AM_0', 'center': >>> Vector(1000.0, -1256.595067376733, 0.0), 'radius': 2100.0, 'start_angle': >>> 67.75727639121727, 'end_angle': 112.2427236087827} >>> basically i have : >>> Center of the arc >>> Starting angle >>> end angle. >>> Do you have any suggestion >>> >>> Regards >>> _______________________________________________ >>> Matplotlib-users mailing list >>> Matplotlib-users(a)python.org >>> https://mail.python.org/mailman/listinfo/matplotlib-users >>> >>

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Arc drawing
by Luca Bertolotti Nov. 2, 2020

Nov. 2, 2020

Hello i have to draw an arc with matplotlib, i have the following data: {'handle': '39D', 'owner': '1F', 'layer': 'AM_0', 'center': Vector(1000.0, -1256.595067376733, 0.0), 'radius': 2100.0, 'start_angle': 67.75727639121727, 'end_angle': 112.2427236087827} basically i have : Center of the arc Starting angle end angle. Do you have any suggestion Regards

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