I should know this ...! Anyway, I have a list of 36 tuples, each with
x, y, z values I want to create a surface plot ...
Need help putting data into right format for matplot3D ...
This is a gmail account used by Keith D. Anthony
On Sat, Mar 16, 2019 at 12:03 PM wrote:
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> Today's Topics:
>
>1. Re: Question regarding the local function object (Terry Reedy)
>2. subprocess svn checkout password issue (Martin De Kauwe)
>3. RE: asyncio Question (Joseph L. Casale)
>4. Re: Implement C's Switch in Python 3 (jf...@ms4.hinet.net)
>5. Re: subprocess svn checkout password issue (dieter)
>6. Re: subprocess svn checkout password issue (Martin De Kauwe)
>7. Re: Question regarding the local function object (Gregory Ewing)
>8. Re: how to embed non-tkinter VLC player into grid of tkinter
> with python? (akashsahu...@gmail.com)
>9. Re: subprocess svn checkout password issue (Dan Sommers)
>
>
>
> -- Forwarded message --
> From: Terry Reedy
> To: python-list@python.org
> Cc:
> Bcc:
> Date: Fri, 15 Mar 2019 13:00:50 -0400
> Subject: Re: Question regarding the local function object
> On 3/15/2019 8:47 AM, Arup Rakshit wrote:
> > Hi,
> >
> > I am reading a book where it says that:
> >
> > Just like module-level function definitions, the definition of a local
> function happens at run time when the def keyword is executed.
> Interestingly, this means that each call to sort_by_last_letter results in
> a new definition of the function last_letter. That is, just like any other
> name bound in a function body, last_letter is bound separately to a new
> function each time sort_by_last_letter is called.
> >
> > If that above is true, why the below program shows the same object
> reference for last_letter every time I call function sort_by_last_letter.
> >
> > # file name is sample.py
> >
> > def sort_by_last_letter(strings):
> > def last_letter(s):
> > return s[-1]
> > print(last_letter)
> > return sorted(strings, key=last_letter)
> >
> > python3 -i sample.py
> sort_by_last_letter(['ghi', 'def', 'abc'])
> > .last_letter at 0x1051e0730>
> > ['abc', 'def', 'ghi']
> sort_by_last_letter(['ghi', 'def', 'abc'])
> > .last_letter at 0x1051e0730>
> > ['abc', 'def', 'ghi']
> sort_by_last_letter(['ghi', 'def', 'abckl'])
> > .last_letter at 0x1051e0730>
> > ['def', 'ghi', 'abckl']
>
> To build on Calvin's explanation ...
> intersperse other function definitions between the repeated calls
>
> sort_by_last_letter(['ghi', 'def', 'abc'])
> def a(): return 'skjsjlskjlsjljs'
> print(a)
> sort_by_last_letter(['ghi', 'def', 'abc'])
> def b(): return 546465465454
> print(b)
> sort_by_last_letter(['ghi', 'def', 'abc'])
>
> and memory gets reused a different way.
>
> .last_letter at 0x03A51D40>
> # <== is same memory as .
> .last_letter at 0x043C2710>
> # ditto
> .last_letter at 0x043C2768>
>
> Creating a new list or string did not have the same effect. I believe
> that CPython function objects must currently all have the same size or
> at least the same max size and conclude that CPython currently allocates
> them from a block of memory that is some multiple of that size. These
> are, of course, current internal implementation details, subject to
> change and even variation across hardware and OSes.
>
> --
> Terry Jan Reedy
>
>
>
>
>
> -- Forwarded message --
> From: Martin De Kauwe
> To: python-list@python.org
> Cc:
> Bcc:
> Date: Fri, 15 Mar 2019 15:17:22 -0700 (PDT)
> Subject: subprocess svn checkout password issue
> Hi,
>
> I'm trying to write a script that will make a checkout from a svn repo and
> build the result for the user. However, when I attempt to interface with
> the shell it asks the user for their filename and I don't know how to
> capture this with my implementation.
>
> user = "XXX578"
> root="https://trac.nci.org.au/svn/cable;
> repo_name = "CMIP6-MOSRS"
>
> cmd = "svn checkout %s/branches/Users/%s/%s" % (root, user, repo_name)
> p = subprocess.Popen(cmd, shell=True, stdin=subprocess.PIPE,
> stdout=subprocess.PIPE,
> stderr=subprocess.PIPE)
> error = subprocess.call(cmd, shell=True)
> if error is 1:
> raise("Error downloading repo"
>
> I tried adding .wait(timeout=60) to the subprocess.Popen command but that
> didn't work.
>
> Any advice on whether there is an augmentation to the above, or a better
> approach, would be much appreciated. I need to solve
