🌐 Mailman 3 December 2009 - Python-list - python.org

Inspect module - getargspec raise TypeError for built-in functions
by Nadav Chernin Dec. 1, 2009

Dec. 1, 2009

Hi, all When I use getargspec(func) for user-defined function, all is working OK, but using it for built-in functions raise TypeError: >>> import sys >>> getargspec(sys.getsizeof) Traceback (most recent call last): File "<pyshell#15>", line 1, in <module> getargspec(sys.getsizeof) File "C:\Python26\lib\inspect.py", line 803, in getargspec raise TypeError('arg is not a Python function') TypeError: arg is not a Python function

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Questions about list-creation
by Manuel Graune Dec. 1, 2009

Dec. 1, 2009

Hello, in (most) python documentation the syntax "list()" and "[]" is treated as being more or less the same thing. For example "help([])" and "help(list())" point to the same documentation. Since there are at least two cases where this similarity is not the case, (see below) can someone explain the reasoning behind this and point to further / relevant documentation? (To clarify: I am not complaining about this, just asking.) 1.) when using local variables in list comprehensions, say a=[i for i in xrange(10)] the local variable is not destroyed afterwards: print "a",a print "i",i using the similar code b=list(j for j in xrange(10)) the local variable is destroyed after use: print "b",b print "j",j and 2) a=list([]) vs. b=[[]] Regards, Manuel Graune -- A hundred men did the rational thing. The sum of those rational choices was called panic. Neal Stephenson -- System of the world http://www.graune.org/GnuPG_pubkey.asc Key fingerprint = 1E44 9CBD DEE4 9E07 5E0A 5828 5476 7E92 2DB4 3C99

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Re: Feature request: String-inferred names
by Gregory Ewing Dec. 1, 2009

Dec. 1, 2009

> On Wed, 25 Nov 2009 20:44:36 -0600, The Music Guy > <musicguy(a)alphaios.net> declaimed the following in > gmane.comp.python.general: > >>I just posted to my blog about a feature that I'd like to see added to >>Python. >> >>http://alphaios.blogspot.com/2009/11/python-string-inferred-names-working.h… I don't think getattr and setattr are used anywhere near frequently enough to justify special syntax. To warrant having its own synax, a feature needs to be used *all the time*. If you find yourself using getattr and setattr that much, it's probably a symptom that you're abusing namespaces for things that would be better done using dictionaries. (A frequent question asked by newcomers from certain other kinds of languages is something like "How do I assign to a variable whose name is in another variable?" The answer is almost always "Don't do that, use a dictionary.") -- Greg

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PIL build error on Snow Leopard
by Xiao Dec. 1, 2009

Dec. 1, 2009

Hello, I haven't fully understood the nuances in the difference between Apple's system Python and MacPython. But I have just installed Python 2.6.4 from python.org. Now I'm trying to install a fresh downloaded PIL 1.1.6 but couldn't. python setup.py install gives: lipo: can't open input file: /var/tmp//ccfwpQd6.out (No such file or directory) Everything worked fine on Apple's Python 2.6.1

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getattr problem
by Victor Subervi Dec. 1, 2009

Dec. 1, 2009

Hi; I have the following code that execute without a problem: import sys,os sys.path.append(os.getcwd()) import options storesTables = [] junkStores = string.join(addStore(), ', ') for table in optionsTables(): if table not in ('particulars', junkStores): storesTables.append(table) for table in storesTables: try: fn = getattr(options, table) print fn() except: pass I need to change the obvious line to this or something similar (that actually works): fn = getattr(options, '%s("names")' % table) That is, I need to pass the variable "names" to each table as it is called. How do I do this? TIA, Victor

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Re: Can't print Chinese to HTTP
by Gnarlodious Dec. 1, 2009

Dec. 1, 2009

> you probably need to change the encoding of sys.stdout >>> sys.stdout.encoding 'UTF-8' >> #!/usr/bin/python > do you know what python version, exactly, that gets called by this hashbang? Verified in HTTP: >>> print(sys.version) 3.1.1 Is is possible modules are getting loaded from my old Python? I symlinked to the new Python, and no I do not want to roll it back because it is work (meaning I would have to type "sudo"). ls /usr/bin/python lrwxr-xr-x 1 root wheel 63 Nov 20 21:24 /usr/bin/python -> /Library/ Frameworks/Python.framework/Versions/3.1/bin/python3.1 Ugh, I have not been able to program in 11 days. Now I remember doing it that way because I could not figure out how to get Apache to find the new Python. ls /usr/local/bin/python3.1 lrwxr-xr-x 1 root wheel 71 Nov 20 08:19 /usr/local/bin/python3.1 - > ../../../Library/Frameworks/Python.framework/Versions/3.1/bin/ python3.1 So they are both pointing to the same Python. And yes, I would prefer easier http scripting, but don't know one. -- Gnarlie

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The Strong Relationship between MatLab and MatPlotLib? What One Needs to Know?
by W. eWatson Dec. 1, 2009

Dec. 1, 2009

Although MatPlotLib has plenty of examples, they do not seem to cover the fundamentals like figure. It seems as though in someway this is dependent upon a user's knowledge of MatLab. Is this true, or oes MatPlotLib provide some description of how forming a figure works?

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Python PIL and Vista/Windows 7 .. show() not working ...
by Esmail Dec. 1, 2009

Dec. 1, 2009

Hello all. I am using the PIL 1.1.6 and Python 2.6.x under XP without any problems. However, I can't display any images under Vista or Windows 7. I could understand Windows 7 as it's relatively new, but Vista has been around for a bit. Sample code: import Image im = Image.open('c://mypic.jpg') im.show() this will work fine under XP, but under Windows 7 and Vista the default image viewer will come up with some error message that the image can't be found. I tried with an external image view program and tried to supply it via the command parameter to show - but that too didn't work. Definition: im.show(self, title=None, command=None) Any suggestions/help/workarounds? If you can get this to work with Vista or Windows 7 I'd love to hear from you. Thanks! Esmail

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Re: How to prevent re.split() from removing part of string
by Jeremy Dec. 1, 2009

Dec. 1, 2009

On Nov 30, 5:24 pm, MRAB <pyt...(a)mrabarnett.plus.com> wrote: > Jeremy wrote: > > I am using re.split to... well, split a string into sections. I want > > to split when, following a new line, there are 4 or fewer spaces. The > > pattern I use is: > > > sections = re.split('\n\s{,4}[^\s]', lineoftext) > > > This splits appropriately but I lose the character matched by [^s]. I > > know I can put parentheses around [^s] and keep the matched character, > > but the character is placed in it's own element of the list instead of > > with the rest of the lineoftext. > > > Does anyone know how I can accomplish this without losing the matched > > character? > > First of all, \s matches any character that's _whitespace_, such as > space, "\t", "\n", "\r", "\f". There's also \S, which matches any > character that's not whitespace. Thanks for the reminder. I knew \S existed, but must have forgotten about it. > > But in answer to your question, use a look-ahead: > > sections = re.split('\n {,4}(?=\S)', lineoftext) Yep, that does the trick. Thanks for the help! Jeremy

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os.remove() permission problem
by Victor Subervi Dec. 1, 2009

Dec. 1, 2009

Hi; I get the following error when I try os.remove(file) *OSError*: [Errno 13] Permission denied: 'particulars.py' args = (13, 'Permission denied') errno = 13 filename = 'particulars.py' strerror = 'Permission denied' Here are the permissions: -rwxr-xr-x 1 root root 455 Nov 28 05:58 particulars.py When I go into the python interpreter and execute that statement, it succeeds. What have I missed? TIA, Victor

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