Re: [Tutor] design advice for function
Danny Yoo said unto the world upon 2005-12-18 15:19: >> > This is caused by the line: print adder(). Obviously >> > if adder() doesn't receive any arguments, it can't >> > build the lists resulting in an IndexError. >> >>Right. > > > Hello! > > Just wanted to clarify the situation: argsList ends up being the empty > list, which is a perfectly good value: Danny is, of course, perfectly correct. Apologies to the op for the imprecision. Once again, I vow never to post after 3:00. Oh, wait, it is 3:23! Brian vdB ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] RegEx query
Hi Kent,
I apologise for the not overly helpful initial post.
I had six possible uris to deal with -
/thread/28742/
/thread/28742/?s=1291819247219837219837129
/thread/28742/5/
/thread/28742/5/?s=1291819247219837219837129
/thread/28742/?goto=lastpost
/thread/28742/?s=1291819247219837219837129&goto=lastpost
The only one I wanted to match was the first two.
My initial pattern /thread/[0-9]*?/(\?s\=.*)?(?!lastpost)$
matched the first two and the last in redemo.py (which I've got
stashed as a py2exe bundle, should I ever find myself sans Python but
having to use regexes).
I managed to sort it by using
/thread
/[0-9]*?/
(\?s\=\w*)?$
The s avoids the fourth possibility, and the \w precludes the & in the last uri.
But, circumventing the problem irks me no end, as I haven't fixed what
I was doing wrong, which means I'll probably do it again, and avoiding
problems instead of resolving them feels too much like programming for
the Win32 api to me.
(Where removing a service from the service database doesn't actually
remove the service from the service database until you open and close
a handle to the service database a second time...)
So yes, any advice on how to use negative lookaheads would be great. I
get the feeling it was the .* before it.
As for my problem with BeautifulSoup, I'm not sure what was happening
there. It was happening in interactive console only, and I can't
replicate it today, which suggests to me that I've engaged email
before brain again.
I do like BeautifulSoup, however. Although people keep telling about
some XPath programme that's better, apparently, I like BeautifulSoup,
it works.
Regards,
Liam Clarke
On 12/18/05, Kent Johnson <[EMAIL PROTECTED]> wrote:
> Liam Clarke wrote:
> > Hi all,
> >
> > Using Beautiful Soup and regexes.. I've noticed that all the examples
> > used regexes like so - anchors = parseTree.fetch("a",
> > {"href":re.compile("pattern")} ) instead of precompiling the pattern.
> >
> > Myself, I have the following code -
> >
> z = []
> x = q.findNext("a", {"href":re.compile(".*?thread/[0-9]*?/.*",
> >
> > re.IGNORECASE)})
> >
> >
> while x:
> >
> > ... num = x.findNext("td", "tableColA")
> > ... h = (x.contents[0],x.attrMap["href"],num.contents[0])
> > ... z.append(h)
> > ... x = x.findNext("a",{"href":re.compile(".*?thread/[0-9]*?/.*",
> > re.IGNORECASE)})
> > ...
> >
> > This gives me a correct set of results. However, using the following -
> >
> >
> z = []
> pattern = re.compile(".*?thread/[0-9]*?/.*", re.IGNORECASE)
> x = q.findNext("a", {"href":pattern)})
> >
> >
> while x:
> >
> > ... num = x.findNext("td", "tableColA")
> > ... h = (x.contents[0],x.attrMap["href"],num.contents[0])
> > ... z.append(h)
> > ... x = x.findNext("a",{"href":pattern} )
> >
> > will only return the first found tag.
> >
> > Is the regex only evaluated once or similar?
>
> I don't know why there should be any difference unless BS modifies the
> compiled regex
> object and for some reason needs a fresh one each time. That would be odd and
> I don't see
> it in the source code.
>
> The code above has a syntax error (extra paren in the first findNext() call)
> - can you
> post the exact non-working code?
> >
> > (Also any pointers on how to get negative lookahead matching working
> > would be great.
> > the regex (/thread/[0-9]*)(?!\/) still matches "/thread/28606/" and
> > I'd assumed it wouldn't.
>
> Putting these expressions into Regex Demo is enlightening - the regex matches
> against
> "/thread/2860" - in other words the "not /" is matching against the 6.
>
> You don't give an example of what you do want to match so it's hard to know
> what a better
> solution is. Some possibilities
> - match anything except a digit or a slash - [^0-9/]
> - match the end of the string - $
> - both of the above - ([^0-9/]|$)
>
> Kent
>
> >
> > Regards,
> >
> > Liam Clarke
> > ___
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> >
> >
>
>
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Re: [Tutor] Accessing next and previous items during iteration
On 18/12/05, Kent Johnson <[EMAIL PROTECTED]> wrote: > Ed Singleton wrote: > > Is it possible to access the next and previous items during an iteration? > > This just came up on c.l.python. Bengt Richter has a nice generator-based > solution. > http://groups.google.com/group/comp.lang.python/browse_thread/thread/2e4533f108fbf172/90d87c91dac844d3?hl=en#90d87c91dac844d3 That's perfect! It's a lovely piece of code as well. Obvious once you've seen it, but you wouldn't have thought of it before hand. Thanks Ed ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] RegEx query
Liam Clarke wrote: > Hi Kent, > > I apologise for the not overly helpful initial post. > > I had six possible uris to deal with - > > /thread/28742/ > /thread/28742/?s=1291819247219837219837129 > /thread/28742/5/ > /thread/28742/5/?s=1291819247219837219837129 > /thread/28742/?goto=lastpost > /thread/28742/?s=1291819247219837219837129&goto=lastpost > > The only one I wanted to match was the first two. > > My initial pattern /thread/[0-9]*?/(\?s\=.*)?(?!lastpost)$ > > matched the first two and the last in redemo.py (which I've got > stashed as a py2exe bundle, should I ever find myself sans Python but > having to use regexes). > > I managed to sort it by using > > /thread > /[0-9]*?/ > (\?s\=\w*)?$ > > The s avoids the fourth possibility, and the \w precludes the & in the last > uri. This seems like a good solution to me. The patterns you want to accept and reject are pretty similar so your regex has to be very specific to discriminate them. > > But, circumventing the problem irks me no end, as I haven't fixed what > I was doing wrong, which means I'll probably do it again, and avoiding > problems instead of resolving them feels too much like programming for > the Win32 api to me. > (Where removing a service from the service database doesn't actually > remove the service from the service database until you open and close > a handle to the service database a second time...) > > So yes, any advice on how to use negative lookaheads would be great. I > get the feeling it was the .* before it. I think you may misunderstand how * works. It will match as much as possible but it will backtrack and match less if that makes the whole match work. For example the regex ab*d will match abbd with b* matching bb. If I change the regex to ab*(?!d) then it will still match abbd but the b* will just match one b and the d doesn't participate in the match. So b* doesn't mean "match all the b's no matter what" it means "match as many b's as you can and still have the rest of the match succeed". In the case of ab*(?!d) this means, match an 'a', then a sequence of 'b', then something that is not 'd'. By shortening the match for b*, the 'not d' can match against the last 'b'. > > As for my problem with BeautifulSoup, I'm not sure what was happening > there. It was happening in interactive console only, and I can't > replicate it today, which suggests to me that I've engaged email > before brain again. > > I do like BeautifulSoup, however. Although people keep telling about > some XPath programme that's better, apparently, I like BeautifulSoup, > it works. Which XPath program is that? I haven't found one I really like. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] List-question
I have one function that finds some values. Then I want that function to find new values based on the values it found first. However, by just looping, it starts on an eternal job. As illustrated in: >>> list = [1,2,3] >>> list2 = list >>> list2 [1, 2, 3] >>> for i in list: ... print i ... list2.append(4) ... 1 2 3 4 4 4 and it will forever continue with 4's. Why would list be expanded with the values of list2? How can I copy the result from one list, and do things with the list without getting it to expand? -- This email has been scanned for viruses & spam by Decna as - www.decna.no Denne e-posten er sjekket for virus & spam av Decna as - www.decna.no ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] List-question
On 19/12/05, Øyvind <[EMAIL PROTECTED]> wrote: > I have one function that finds some values. Then I want that function to > find new values based on the values it found first. However, by just > looping, it starts on an eternal job. > > As illustrated in: > >>> list = [1,2,3] > >>> list2 = list > >>> list2 > [1, 2, 3] > >>> for i in list: > ... print i > ... list2.append(4) > ... > 1 > 2 > 3 > 4 > 4 > 4 and it will forever continue with 4's. > > Why would list be expanded with the values of list2? How can I copy the > result from one list, and do things with the list without getting it to > expand? Because they point to the same thing. Type "list2 is list" after your other code and see. You want list2 to be a COPY of list not a pointer to it. Do this by using list2 = list.copy() Slices create a copy, so a shortcut is: list2 = list[:] Ed ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] List-question
On 19/12/05, Ed Singleton <[EMAIL PROTECTED]> wrote: > On 19/12/05, Øyvind <[EMAIL PROTECTED]> wrote: > > I have one function that finds some values. Then I want that function to > > find new values based on the values it found first. However, by just > > looping, it starts on an eternal job. > > > > As illustrated in: > > >>> list = [1,2,3] > > >>> list2 = list > > >>> list2 > > [1, 2, 3] > > >>> for i in list: > > ... print i > > ... list2.append(4) > > ... > > 1 > > 2 > > 3 > > 4 > > 4 > > 4 and it will forever continue with 4's. > > > > Why would list be expanded with the values of list2? How can I copy the > > result from one list, and do things with the list without getting it to > > expand? > > Because they point to the same thing. > > Type "list2 is list" after your other code and see. > > You want list2 to be a COPY of list not a pointer to it. Do this by using > > list2 = list.copy() > > Slices create a copy, so a shortcut is: > > list2 = list[:] Sorry, you need to: from copy import copy before you can use copy. Also, you shouldn't use "list" as a variable name. as it is a built-in function that converts things to lists. Use "list1". That way you can also use list2 = list(list1). Ed ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Name tables questions
Bernard Lebel wrote:
> Hello,
>
> I have a few theoric questions regarding name tables. I wish to better
> understand a few things about this aspect of Python, in particular how
> module names and the import statements fit into the picture of name
> tables.
>
> - First of all, I understand each scope has its "local" name table,
> containing all the names defined in this scope, no matter the origin
> of the name. So, if you import a module in a scope, its name is added
> to the local scope name table, correct?
Yes, with a couple of notes:
- The correct term is "namespace", not "name table".
- Instead of talking about defining a name, a better term is "binding". Binding
a name in
a namespace means associating the name with a value in the context of the
namespace. This
is usually implemented by making an entry in some dictionary that implements
the namespace.
- "All the names defined in this scope" could be interpreted to mean "all the
names
accessible in this scope." With this interpretation your statement is *not*
correct. The
local namespace includes all names that are actually bound in the local scope,
not all
names accessible in the local scope.
For example:
>>> import sys
>>> a = 1
>>> def f():
... import os
... b = 2
... print 'locals:', locals()
... print 'globals:', globals()
...
>>> f()
locals: {'b': 2, 'os': }
globals: {'a': 1, 'f': , '__builtins__': , '__file__': 'C:\\Documents and Settings\\ktjohns
on\\pythonstartup.py', 'sys': , '__name__':
'__main__',
'__doc__': None}
The local namespace of f() includes os and b, which are bound in f(). The
global namespace
includes sys, a and f, all names bound at global scope, and several other names
inserted
automatically by the Python runtime.
>
> - When you import a module, I understand the module name is appended
> to Python dictionary of imported modules. I can see the name of the
> module if I print out the sys.modules attribute. If the module name is
> also added to the local scope name table, it does mean that the module
> is added in two tables: the local scope name table and the imported
> modules table. Is that correct?
Yes. The module object itself is cached in sys.modules. sys.modules is a dict
whose keys
are module names and values are the actual module objects. Subsequent imports
will find
the cached module and bind a name to it.
>
> - I *think* I have read in Learning Python (sorry I don't have the
> book near me) that during execution, when Python doesn't find a name
> in the current scope name table, then it will look up every
> encompassing name table one after another, until it can find it, and
> if it can't, will raise an error.
Yes.
> If I remember correctly about my reading, this lookup into an
> encompassing name table has a performance cost. The author suggests
> that, without regard to the practical considerations, it would not be
> such a bad idea to pass down all possible names to functions and
> encapsulated functions so name lookup into other tables could be
> avoided.
Yes, there is a performance cost. No, it doesn't make sense to "pass down" all
possible
names. If a name is not used in a function, then you have incurred the cost of
looking it
up and binding it in the local namespace without any benefit. If the name is
used once in
the function, it is still probably cheaper to just look it up once globally and
skip the
local rebinding and lookup.
Where this makes sense is in a loop that is a known performance bottleneck that
you are
trying to optimize. Then there is a benefit to making local copies of any
global names
that are in the loop. It is also beneficial to make local copies of attributes
that are
accessed in the loop such as methods.
> Now, let say I would do something like that, and be zealous at that,
> to the point where I would write import statements for the same
> modules in every possible function to define the module name there.
> Since the module name is already in the module names dictionary but
> not in the global name table, would there be any advantage, or
> disadvantage in doing so?
Again the only advantage of this would be if the module is used repeatedly in a
loop. But
in that case you are better off making a local reference to the module
attribute you
actually use, rather than the module itself. For example something like this to
generate
the paths to all files in a directory:
import os
def allFiles(path):
join = os.path.join
isfile = os.path.isfile
for f in os.listdir(path):
p = join(path, f)
if isfile(p):
yield p
This is not a very good example because the performance of this fn is likely to
be
dominated by the I/O, not the name lookup, but it gives the idea.
You might be interested in this recipe which automatically performs this kind
of optimization:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/277940
But I should stress that this is an *optimization*, which you should use on
Re: [Tutor] List-question
On Mon, 19 Dec 2005, Ed Singleton wrote: > On 19/12/05, Ed Singleton <[EMAIL PROTECTED]> wrote: > > > > list2 = list.copy() > > > > Slices create a copy, so a shortcut is: > > > > list2 = list[:] > > Sorry, you need to: > > from copy import copy > > before you can use copy. It should also be, after the import: list2=copy.copy(list1) rather than list2=list1.copy()# or list2=list.copy() ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] bnf
>>> mystring = 'laLA' >>> mystring.upper() 'LALA' >>> mystring.lower() 'lala' ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] question !
How can I remove and add record ( dictionary type) to a file. This is the program that I'm working on: the program should create a text file, print the contents of the text file, read the file after it's been created, add a record and print the contents of the file, remove a record(s) from the specified file, write it again, read it again, print the contens of the new file. Here is what I have so far: import cPickle, shelve def write_file(): CIT101 = ["Academic Computer Skills"] CIT111 = ["Database Management"] CIT115 = ["Intro to Computer scince"] CIT127 = ["ACCESS"] CIT211 = !
["Systems
Analysis and Design"] CIT216 = ["Visual Basic"] CIT218 = ["Intermediate Visual Basic"] CIT234 = ["Decision Support Using Excel"] pickle_file = open("pickles1.dat","w") cPickle.dump(CIT101, pickle_file) cPickle.dump(CIT111, pickle_file) cPickle.dump(CIT115, pickle_file) cPickle.dump(CIT127, pickle_file) cPickle.dump(CIT211, pickle_file) cPickle.dump(CIT216, pickle_file) cPickle.dump(CIT218, pickle_file) cPickle.dump(CIT234, pickle_file) print "A file has been created and the required specifications have been added" pickle_file.closedef read_file(): pickle_file = open("pickles1.dat","r")
CIT101 = cPickle.load(pickle_file) CIT111 = cPickle.load(pickle_file) CIT115 = cPickle.load(pickle_file) CIT127 = cPickle.load(pickle_file) CIT211 = cPickle.load(pickle_file) CIT216 = cPickle.load(pickle_file) CIT218 = cPickle.load(pickle_file) CIT234 = cPickle.load(pickle_file) pickle_file.close() pickles = shelve.open("pickles2.dat") pickles["CIT101"] = ["Academic Computer Skills"] pickles["CIT111"] = ["Database Management"] pickles["CIT115"] = ["Intro to Computer scince"] pickles["CIT127"] = ["ACCESS"] pickles["CIT211"] = ["Systems Analysis and Design"] pickles["CIT216"] = ["Visual Basic!
"]
pickles["CIT218"] = ["Intermediate Visual Basic"] pickles["CIT234"] = ["Decision Support Using Excel"] pickles.sync() for key in pickles.keys(): print key, "-", pickles[key]def dele_file(): word_dele = raw_input("Which record do u want to delete?: ") if word_dele in picles.keys(): del word_dele else: print "There is no such record in file pickles2.dat" pickles.close() def add_record(): CIT236 = ["SQL Programming"] CIT240 = ["Database programming"] pickle_file = open("pickles1.dat","a")
cPickle.dump(CIT236, pickle_file) cPickle.dump(CIT240, pickle_file) print "New data was added to the file" pickle_file.close def display_instructions(): """Display the Main menue""" print \ """ Main Manue: 1. Exit 2. Create a new file and add specifications 3. (not working)Add more courses to the file 4. Read the file 5. (not working)Delete f!
ile
""" # exit the program >>> 1 <<< def over_program(): """Exit the program""" print "Good Bye!" def main(): choice = None display_instructions() while choice != 1: choice = raw_input("\nChoice: ") if choice == "1": over_program() break
elif choice == "2": write_file() elif choice == "3": add_to_file() elif choice == "4": read_file() elif choice == "5": delete_file() else: print "\nSorry, but", choice, "isn!
't a
valid choice."main() raw_input("Press Enter Key to Exit.") __Do You Yahoo!?Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com ___
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[Tutor] float question
what does 2 mean in %2.4f ? ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] synchronized enumerate
Kent wrote: | Take a look at this thread on c.l.py for some discussion and | possibilities. The part you | are interested in starts around message 14. | http://groups.google.com/group/comp.lang.python/browse_frm/thread/ab1658dca4023e2b?hl=en&; | After looking at that I found something on python-dev that showed how to create an object that is sliceable but returns a slice object: sslice[::-1] --> slice(None, None, -1). It would be interesting if a single object could be sliced or accessed as a function, but I'm not sure that is possible. If it were, you could do: sslice(3) --> slice(None, 3, None) sslice[:3]--> slice(None, 3, None) Here's the code from python-dev: class MetaSlice(object): def __getitem__(cls, item): return item def __init__(self, *args, **kw): return super(MetaSlice,self).__init__(self, *args, **kw) class sslice(slice): __metaclass__=MetaSlice Anway, after seeing that, I decided to re-write the enumerate so it takes a slice argument and then I get the start and step values from that rather than having those passed in as arguments. I also initiated the discussion on python-dev (and they said that comp.lang.python was a better place for it) and c.l.p. The updated version of the enuerate can be found on compl.lang.python under "synchronized enumerate" at http://tinyurl.com/e35mh Thanks for the pointers and advice, Kent. /c ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
