Re: [Tutor] Python 2.7 on Ubuntu 11.10 - Do not unintall

2011-11-02 Thread Joel M.
On Wed, Nov 2, 2011 at 11:39 AM, Prasad, Ramit wrote:

> >I use OS X / Windows and I have not noticed any dependency issues. So I
> was not aware about the same with respect to Ubuntu. I am glad that I
> learnt something from this discussion.
>
> I might be wrong, but Ubuntu itself is not dependent on Python. You can
> probably run a minimalist headless Ubuntu server without Python. Especially
> since the OP seemed able to login to his machine without it. X/Desktop
> requires it.
>
>
Ubuntu is not dependent on Python, but the environments (Unity, GNOME,
etc) will brake if you don't have Python.



> I am not sure if OS X needs it, but I do know it comes with Python
> installed.
>
> Joel M: Pretty much the only things that require restarting (from my
> experience with Debian) are kernel changes and hardware changes. You just
> had to restart X / login manager (gdm).
>
>
Yea I know, but I restarted anyhow.


> Ramit
>
>
> Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology
> 712 Main Street | Houston, TX 77002
> work phone: 713 - 216 - 5423
>
> --
>
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Re: [Tutor] login window using Tk

2011-11-01 Thread Joel M.
I am also intrested in this topic.

Chris were you thinking of using the window.hide() method?

-Joel M
On Nov 1, 2011 1:21 PM, "Chris Hare"  wrote:

>
> I am working on a python Tk program which involves a login window and I am
> looking for some advice.
>
> Currently the code I have creates a window (Toplevel) where the login
> controls are and I am running that using a main loop for the window.  The
> root window is hidden.  The objective is that when the user ha successfully
> authenticated, the login window is closed or the main loop os exited and
> then the root window is shown and the main loop started for the actual
> application.
>
> Questions:
> 1.  Is this the best way of doing this or is there a better way?
> 2.  How do I exit the main loop when the user has authenticated?
>
> Thanks
>
> Chris Hare
> ch...@labr.net
> http://www.labr.net
>
>
>
> Chris Hare
> ch...@labr.net
> http://www.labr.net
>
>
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Re: [Tutor] Paper Rock Scissors game - User's choice not returned properly

2011-10-31 Thread Joel M.
On Mon, Oct 31, 2011 at 1:09 PM, Joel Montes de Oca
wrote:

>  On Mon 31 Oct 2011 12:14:40 PM EDT, Hugo Arts wrote:
>
>
> On Mon, Oct 31, 2011 at 4:41 PM, Joel Montes de Oca
>   wrote:
>
>
> Hello everyone,
>
> I am having a little trouble with a block of code that isn't behaving the
> way I would expect. Maybe you can give me a hand and point where it is
> going
> wrong.
>
> The function that is not working correctly belongs to a Paper Rock Scissor
> game I am making.
>
> This particular function is responsible to:
> a) Get the user's choice (Paper, Rock, or Scissors)
> b) Return the user's choice within the variable choice to the function
> that called it.
>
> The function works correctly as long as the user does not try to enter a
> string other than 'P', 'R', or 'S'.
>
> Logic:
> Take the user's string and put it in the variable choice.
> If choice is not 'P', 'R', or 'S' then pass a message to the user and
> call the function again.
> If the choice is 'P', 'R', or 'S' then return choice to where it was
> called from.
>
> The problem is this.
>
> When the user enters a string other than the valid ones, the if statements
> catches it and calls the same function again so the user can enter a valid
> string. But the variable choice does not get assigned the new string
> entered
> by the user, instead it is empty.
>
> I would expect if the function runs again, the variable choice would be
> updated to the last value given by the user.
>
> The function: ( http://dpaste.com/644857/)
>
> def UserChoice (): # The function that returns the choice from the user
> print 'Please select (P) for paper, (R) for Rock, or (S) for Scissors.'
> choice = raw_input('What is your selection?: ')
>
> if choice.lower() not in ('p', 'r','s'): # Converts the user's choice to
> lowercase and confirms the choice is valid
> print 'I am sorry, you entered \'' + choice.upper() + '\' which is an
> invalid response. Please try again.'
> raw_input('Press Enter to try again.')
> UserChoice () # If the choice is not valid, run the function over
> else:
> return choice
>
>
>
> Your problem is that you don't quite understand recursion.
>
> When a function calls itself, it's not actually any different from
> when a function A calls another function B. Once function B is done
> running and returns, control goes back to function A. It doesn't
> matter if function B is actually the same function as A. In fact, even
> if A and B are the same function, they don't "share" any variables and
> are totally separate as if they were two different functions that just
> happened to do the same thing.
>
> So, how does this apply to your function? Let's go through a run of
> it. We call UserChoice (this is function A) and we input a wrong
> letter. So, A calls UserChoice again (this is function B), and this
> time we input something valid. So, function B runs the line "return
> choice."
>
> Control now returns to function A, right at the point where we called
> function B. So what do we do here with the choice that was just
> returned from function B? Well, looking at the line where it's called,
> it's just "UserChoice()". So we do nothing. We just throw it away.
> Then, we continue on with function A, move out of the if statement,
> and "fall off the end" of the function. And when that happens, Python
> returns None from function A to show you that nothing was returned.
>
> If you understood all that, you should be able to fix your problem.
>
> Hugo
>
>
> Hey Hugo,
>
> I think I understand your explanation. Let me see if I get it.
>
> The code:
>
> def UserChoice (): *# I WILL CALL THIS FUNCTION A*
>
>
> print 'Please select (P) for paper, (R) for Rock, or (S) for Scissors.'
> choice = raw_input('What is your selection?: ')
>
> if choice.lower() not in ('p', 'r','s'):
>
> print 'I am sorry, you entered \'' + choice.upper() + '\' which is an 
> invalid response. Please try again.'
> raw_input('Press Enter to try again.')
>
> UserChoice () *# I WILL CALL THIS FUNCTION B*
> else:
> return choice *# I WILL CALL THE ORIGINAL CALLING FUNCTION, FUNCTION 
> MAIN*
>
>
> OK so when the user uses a valid letter, the variable choice gets
> returned to FUNCTION MAIN. Everything works fine.
>
> When the user e

Re: [Tutor] Method to create small and simple database

2011-10-30 Thread Joel M.
On Sat, Oct 29, 2011 at 6:58 PM, Alan Gauld wrote:

> On 29/10/11 19:28, Joel Montes de Oca wrote:
>
>  After looking at the Python module documentation for sqlite3
>> (http://docs.python.org/**library/sqlite3.html#module-**sqlite3<http://docs.python.org/library/sqlite3.html#module-sqlite3>),
>> it seems
>> to me it's the best way to make the small database that I am looking for.
>>
>
> SQLlite is a great way to build small scale SQL databases.
> However for this app I'd probably second the advice to use shelve.
> Shelve acts like a dictionary in a file so you can associate a list of
> items with a user very very easily.
>
>
>
>  Now I need to look for a basic tutorial on constructing a simple
>> database (tables, rows, keys) and how to connect all that stuff
>> together. If anyone happens to know of a good intro tutorial or
>> documentation to database concepts, please forward it to me.
>>
>
> You can try the database topic in my tutorial(see below).
> It's only available for Python v2 at present but the translation
> to Python v3 (if that's what you are using) is trivial.
>
> --
> Alan G
> Author of the Learn to Program web site
> http://www.alan-g.me.uk/
>
>
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I will take a look into Shelve and Alan's tutorial sometime this coming
week.

Keep the suggestions coming if there's more. :)

-Joel M.
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