Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 11:52 am, eric  wrote:
> On Dec 18, 8:37 pm, [email protected] wrote:
>
> > I am trying to write a simple application to factor polynomials. I
> > wrote (simple) raw_input lines to collect the a, b, and c values from
> > the user, but I dont know how to implement the quadratic equation
>
> > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > into python. Any ideas?
>
> with numpy:
> from numpy import *
>
> s=[1,-1]
> x = -b+s*sqrt( b**2-4*a*c )/(2*a)
>
> Erichttp://codeslash.blogspot.com

Ahh. Great.. thank you. I didnt know about the sqrt function.. saves
me from doing "^1/2". Thanks again.
-CD
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 1:09 pm, Mark Dickinson  wrote:
> On Dec 18, 8:47 pm, Scott David Daniels  wrote:
>
> >      else: # a single result (discriminant is zero)
> >          return (-b / (2 * a),)
>
> Maybe make that (-b / (2. * a)) to avoid getting funny results
> when a and b are integers.  (Or do a from __future__ import
> division, or use Python 3.0, or )
>
> And to make the function more bullet-proof, you might want to
> do something like (untested):
>
>     from math import copysign
>
>     [rest of example as in Scott's post]
>
>     if discriminant: # two results
>         root1 = (-b - copysign(discriminant, b))/(2*a)
>         root2 = c/(a*root1)
>         return (root1, root2)
>
> to avoid numerical problems when b*b is much larger
> than abs(a*c). Compare with the results of the usual
> formula when a = c = 1, b = 10**9, for example.  But
> that still doesn't help you when the computation
> of the discriminant underflows or overflows...
>
> Isn't floating-point a wonderful thing!  :)
>
> Mark

Thanks for all your help! Its good to know how to do it w/ without
numpy.
And yes, floating point is the best thing since sliced bread. ^^

-CD
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 4:41 pm, "James Mills" 
wrote:
> On Fri, Dec 19, 2008 at 10:11 AM, Gabriel Genellina
>
>  wrote:
> > En Thu, 18 Dec 2008 17:37:35 -0200,  escribió:
>
> >> I am trying to write a simple application to factor polynomials. I
> >> wrote (simple) raw_input lines to collect the a, b, and c values from
> >> the user, but I dont know how to implement the quadratic equation
>
> >> x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> Why is this so hard ? This is simple simple
> expression. Reading through the Python
> tutorial and reading up on how to define
> functions is all you need! :)
>
> Here goes:
>
> >>> def f(a, b, c):
>
> ...     x = (-1 * b) + ((b**2 - (4 * a * c)) / (2 * a))
> ...     return (-1 * x), x
> ...
>
> >>> f(1, 2, 3)
> (6, -6)
>
> cheers
> James

Hiya James!

Just one small problem with your equation above...
The quadratic formula is:
x = -b +or- (b**2 - (4 * a * c))^1/2 / 2a

You just forgot the square root which makes quadratic a bit more
complicated.
You would have to download and import sqrt() from numpy or **.5

Also.. I need to build in functionality so the user does not have to
directly call the function like:
f(a,b,c)

Instead.. they should be able to just raw_input their values.
Also.. as with some of the examples above its a good idea to analyze
the discriminant to make sure we have a real solution.
Of course.. thats all pretty simple to build in. Thanks a lot!
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 5:30 pm, "James Mills" 
wrote:
> UPDATE:
>
> jmi...@atomant:~/tmp$ cat polycalc.py
> #!/usr/bin/env python
>
> from math import sqrt
>
> def f(a, b, c):
>     if (b**2 - (4 * a * c)) < 0:
>         return None, None # Can't solve
>     x1 = -b - (sqrt(b**2 - (4 * a * c)) / (2 * a))
>     x2 = -b + (sqrt(b**2 - (4 * a * c)) / (2 * a))
>     return x1, x2
>
> print "Polynomial Solver..."
> print
>
> while True:
>     a = float(raw_input("a: "))
>     b = float(raw_input("b: "))
>     c = float(raw_input("c: "))
>
>     x = f(a, b, c)
>     if None in x:
>         print "Can't solve!"
>     else:
>         print "x = (%0.2f, %0.2f)" % x
> jmi...@atomant:~/tmp$ ./polycalc.py
> Polynomial Solver...
>
> a: 1
> b: 8
> c: 5
> x = (-11.32, -4.68)

Ahh. Great.. that answers a lot of questions.
Originally I was using just a = raw_input('a: ')
And was getting errors because you cant perform mathmatical operations
on strings. >.<
Thanks again!
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 5:10 pm, Steven D'Aprano  wrote:
> On Thu, 18 Dec 2008 11:37:35 -0800, collin.day.0 wrote:
> > I am trying to write a simple application to factor polynomials. I wrote
> > (simple) raw_input lines to collect the a, b, and c values from the
> > user, but I dont know how to implement the quadratic equation
>
> > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > into python. Any ideas?
>
> def quadratic_solution(a, b, c):
>     sol1 = (-b + (b**2 - 4*a*c)**0.5)/2*a
>     sol2 = (-b - (b**2 - 4*a*c)**0.5)/2*a
>     return (sol1, sol2)
>
> Because this looks like homework, I've deliberately left in two errors in
> the above. One of them is duplicated in the two lines above the return,
> and you must fix it or you'll get radically wrong answers.
>
> The second is more subtle, and quite frankly if this is homework you
> could probably leave it in and probably not even lose marks. You will
> need to do significant research into numerical methods to learn what it
> is, but you will then get significantly more accurate results.
>
> --
> Steven

The corrected function is:
def quadratic_solution(a,b,c)
sol1 = -1*b + ((b**2 - 4*a*c)**.5)/2*a
sol1 = -1*b - ((b**2 - 4*a*c)**.5)/2*a
return (sol1, sol2)

Squaring the -b would give you some strange solutions :D

-CD
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 11:37 am, [email protected] wrote:
> I am trying to write a simple application to factor polynomials. I
> wrote (simple) raw_input lines to collect the a, b, and c values from
> the user, but I dont know how to implement the quadratic equation
>
> x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> into python. Any ideas?

I completed the code:

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
if b**2 - 4*a*c < 0:
return 'No real solution.'
else:
sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
return (sol1, sol2)

# execute
print solver(a,b,c)

Thanks to everyone who helped...
This really expanded my knowledge on some of the mathematical
functions in Python.
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 6:12 pm, Collin D  wrote:
> On Dec 18, 11:37 am, [email protected] wrote:
>
> > I am trying to write a simple application to factor polynomials. I
> > wrote (simple) raw_input lines to collect the a, b, and c values from
> > the user, but I dont know how to implement the quadratic equation
>
> > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > into python. Any ideas?
>
> I completed the code:
>
> #import
> from math import sqrt
>
> # collect data
> a = float(raw_input('Type a value: '))
> b = float(raw_input('Type b value: '))
> c = float(raw_input('Type c value: '))
>
> # create solver
> def solver(a,b,c):
>     if b**2 - 4*a*c < 0:
>         return 'No real solution.'
>     else:
>         sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
>         sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
>         return (sol1, sol2)
>
> # execute
> print solver(a,b,c)
>
> Thanks to everyone who helped...
> This really expanded my knowledge on some of the mathematical
> functions in Python.

UPDATE:
'

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
if b**2 - 4*a*c < 0:
return 'No real solution.'
else:
sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
return (sol1, sol2)

# execute
print solver(a,b,c)

--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 6:23 pm, "Russ P."  wrote:
> On Dec 18, 6:17 pm, Collin D  wrote:
>
>
>
>
>
> > On Dec 18, 6:12 pm, Collin D  wrote:
>
> > > On Dec 18, 11:37 am, [email protected] wrote:
>
> > > > I am trying to write a simple application to factor polynomials. I
> > > > wrote (simple) raw_input lines to collect the a, b, and c values from
> > > > the user, but I dont know how to implement the quadratic equation
>
> > > > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > > > into python. Any ideas?
>
> > > I completed the code:
>
> > > #import
> > > from math import sqrt
>
> > > # collect data
> > > a = float(raw_input('Type a value: '))
> > > b = float(raw_input('Type b value: '))
> > > c = float(raw_input('Type c value: '))
>
> > > # create solver
> > > def solver(a,b,c):
> > >     if b**2 - 4*a*c < 0:
> > >         return 'No real solution.'
> > >     else:
> > >         sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
> > >         sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
> > >         return (sol1, sol2)
>
> > > # execute
> > > print solver(a,b,c)
>
> > > Thanks to everyone who helped...
> > > This really expanded my knowledge on some of the mathematical
> > > functions in Python.
>
> > UPDATE:
> > '
>
> > #import
> > from math import sqrt
>
> > # collect data
> > a = float(raw_input('Type a value: '))
> > b = float(raw_input('Type b value: '))
> > c = float(raw_input('Type c value: '))
>
> > # create solver
> > def solver(a,b,c):
> >     if b**2 - 4*a*c < 0:
> >         return 'No real solution.'
> >     else:
> >         sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
> >         sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
> >         return (sol1, sol2)
>
> > # execute
> > print solver(a,b,c)
>
> You need to put your denominator, 2*a, in parens. The way it stands,
> you are dividing by 2, then multiplying by a. That's not what you
> want.
>
> Also, for better style, I suggest you compute the discriminanat once
> and store it for reuse rather than repeating the expression three
> times.- Hide quoted text -
>
> - Show quoted text -

I see what you mean on the denominator and discriminant. Ill do that.
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 6:27 pm, Collin D  wrote:
> On Dec 18, 6:23 pm, "Russ P."  wrote:
>
>
>
>
>
> > On Dec 18, 6:17 pm, Collin D  wrote:
>
> > > On Dec 18, 6:12 pm, Collin D  wrote:
>
> > > > On Dec 18, 11:37 am, [email protected] wrote:
>
> > > > > I am trying to write a simple application to factor polynomials. I
> > > > > wrote (simple) raw_input lines to collect the a, b, and c values from
> > > > > the user, but I dont know how to implement the quadratic equation
>
> > > > > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > > > > into python. Any ideas?
>
> > > > I completed the code:
>
> > > > #import
> > > > from math import sqrt
>
> > > > # collect data
> > > > a = float(raw_input('Type a value: '))
> > > > b = float(raw_input('Type b value: '))
> > > > c = float(raw_input('Type c value: '))
>
> > > > # create solver
> > > > def solver(a,b,c):
> > > >     if b**2 - 4*a*c < 0:
> > > >         return 'No real solution.'
> > > >     else:
> > > >         sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
> > > >         sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
> > > >         return (sol1, sol2)
>
> > > > # execute
> > > > print solver(a,b,c)
>
> > > > Thanks to everyone who helped...
> > > > This really expanded my knowledge on some of the mathematical
> > > > functions in Python.
>
> > > UPDATE:
> > > '
>
> > > #import
> > > from math import sqrt
>
> > > # collect data
> > > a = float(raw_input('Type a value: '))
> > > b = float(raw_input('Type b value: '))
> > > c = float(raw_input('Type c value: '))
>
> > > # create solver
> > > def solver(a,b,c):
> > >     if b**2 - 4*a*c < 0:
> > >         return 'No real solution.'
> > >     else:
> > >         sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
> > >         sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
> > >         return (sol1, sol2)
>
> > > # execute
> > > print solver(a,b,c)
>
> > You need to put your denominator, 2*a, in parens. The way it stands,
> > you are dividing by 2, then multiplying by a. That's not what you
> > want.
>
> > Also, for better style, I suggest you compute the discriminanat once
> > and store it for reuse rather than repeating the expression three
> > times.- Hide quoted text -
>
> > - Show quoted text -
>
> I see what you mean on the denominator and discriminant. Ill do that.- Hide 
> quoted text -
>
> - Show quoted text -

UPDATE:

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# find discriminant
disc = b**2 - 4*a*c

# create solver
def solver(a,b,c):
if disc < 0:
return 'No real solution.'
else:
sol1 = (-1 * b + (sqrt(disc))) / (2*a)
sol2 = (-1 * b - (sqrt(disc))) / (2*a)
return (sol1, sol2)

# execute
print solver(a,b,c)
--
http://mail.python.org/mailman/listinfo/python-list

Re: Factoring Polynomials

2008-12-18 Thread Collin D 
On Dec 18, 6:41 pm, "Russ P."  wrote:
> On Dec 18, 6:31 pm, Collin D  wrote:
>
>
>
>
>
> > On Dec 18, 6:27 pm, Collin D  wrote:
>
> > > On Dec 18, 6:23 pm, "Russ P."  wrote:
>
> > > > On Dec 18, 6:17 pm, Collin D  wrote:
>
> > > > > On Dec 18, 6:12 pm, Collin D  wrote:
>
> > > > > > On Dec 18, 11:37 am, [email protected] wrote:
>
> > > > > > > I am trying to write a simple application to factor polynomials. I
> > > > > > > wrote (simple) raw_input lines to collect the a, b, and c values 
> > > > > > > from
> > > > > > > the user, but I dont know how to implement the quadratic equation
>
> > > > > > > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > > > > > > into python. Any ideas?
>
> > > > > > I completed the code:
>
> > > > > > #import
> > > > > > from math import sqrt
>
> > > > > > # collect data
> > > > > > a = float(raw_input('Type a value: '))
> > > > > > b = float(raw_input('Type b value: '))
> > > > > > c = float(raw_input('Type c value: '))
>
> > > > > > # create solver
> > > > > > def solver(a,b,c):
> > > > > >     if b**2 - 4*a*c < 0:
> > > > > >         return 'No real solution.'
> > > > > >     else:
> > > > > >         sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
> > > > > >         sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
> > > > > >         return (sol1, sol2)
>
> > > > > > # execute
> > > > > > print solver(a,b,c)
>
> > > > > > Thanks to everyone who helped...
> > > > > > This really expanded my knowledge on some of the mathematical
> > > > > > functions in Python.
>
> > > > > UPDATE:
> > > > > '
>
> > > > > #import
> > > > > from math import sqrt
>
> > > > > # collect data
> > > > > a = float(raw_input('Type a value: '))
> > > > > b = float(raw_input('Type b value: '))
> > > > > c = float(raw_input('Type c value: '))
>
> > > > > # create solver
> > > > > def solver(a,b,c):
> > > > >     if b**2 - 4*a*c < 0:
> > > > >         return 'No real solution.'
> > > > >     else:
> > > > >         sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
> > > > >         sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
> > > > >         return (sol1, sol2)
>
> > > > > # execute
> > > > > print solver(a,b,c)
>
> > > > You need to put your denominator, 2*a, in parens. The way it stands,
> > > > you are dividing by 2, then multiplying by a. That's not what you
> > > > want.
>
> > > > Also, for better style, I suggest you compute the discriminanat once
> > > > and store it for reuse rather than repeating the expression three
> > > > times.- Hide quoted text -
>
> > > > - Show quoted text -
>
> > > I see what you mean on the denominator and discriminant. Ill do that.- 
> > > Hide quoted text -
>
> > > - Show quoted text -
>
> > UPDATE:
>
> > #import
> > from math import sqrt
>
> > # collect data
> > a = float(raw_input('Type a value: '))
> > b = float(raw_input('Type b value: '))
> > c = float(raw_input('Type c value: '))
>
> > # find discriminant
> > disc = b**2 - 4*a*c
>
> > # create solver
> > def solver(a,b,c):
> >     if disc < 0:
> >         return 'No real solution.'
> >     else:
> >         sol1 = (-1 * b + (sqrt(disc))) / (2*a)
> >         sol2 = (-1 * b - (sqrt(disc))) / (2*a)
> >         return (sol1, sol2)
>
> > # execute
> > print solver(a,b,c)
>
> A couple of style points. I would use -b rather than -1 * b. Also, you
> don't need the else clause. You can simplify it to
>
> def solver(a, b, c):
>
>     disc = b**2 - 4 * a * c
>
>     if disc < 0: return "No real solution."
>
>     sol1 = (-b + sqrt(disc)) / (2*a)
>     sol2 = (-b - sqrt(disc)) / (2*a)
>
>     return sol1, sol2- Hide quoted text -
>
> - Show quoted text -

UPDATE:

#import
from math import sqrt

# collect data
a = float(raw_input('Type a value: '))
b = float(raw_input('Type b value: '))
c = float(raw_input('Type c value: '))

# create solver
def solver(a,b,c):
disc = b**2 - 4*a*c
if disc < 0:
return 'No real solution.'
else:
sol1 = (-b + (sqrt(disc))) / (2*a)
sol2 = (-b - (sqrt(disc))) / (2*a)
return (sol1, sol2)

# execute
print solver(a,b,c)
--
http://mail.python.org/mailman/listinfo/python-list

Re: Tkinter - problem closing window

2009-01-05 Thread Collin D 
On Jan 5, 6:25 am, "Djames Suhanko"  wrote:
> Hello!
> I'm sorry my terrible english (my native language is portuguese).
> I has a litle program that open another window. When I close de root
> window in quit button, I need clicking 2 times to close. is where the
> problem?
>
> The source:
>   1 #!/usr/bin/env python
>   2 from Tkinter import *
>   3 import sys
>   4 import random
>   5 class App:
>   6  def __init__(self, master):
>   7    frame = Frame(master)
>   8    frame.pack()
>   9    rotulo = Label(frame, text="Clique em 'Gerar' e boa
> sorte!",borderwidth=2,bg="gray",justify=C    ENTER,relief=SUNKEN)
>  10    rotulo.pack()
>  11
>  12    self.button = Button(frame, text="Sair", fg="red",
> command=frame.quit,borderwidth=1)
>  13    self.button.pack(side=LEFT)
>  14    self.hi_there = Button(frame, text="Gerar Numero",
> command=self.say_hi,borderwidth=1)
>  15    self.hi_there.pack(side=RIGHT,padx=2,pady=2)
>  16
>  17  def gera_seis(self):
>  18    a = {}
>  19    for i in range(6):
>  20       a[i] = "%02d" %  int (random.randint(0,60))
>  21    resultadoA = "%s-%s-%s-%s-%s-%s" %
> (str(a[0]),str(a[1]),str(a[2]),str(a[3]),str(a[4]),str(a[5]))
>  22    return resultadoA
>  23
>  24  def say_hi(self):
>  25    resultado = self.gera_seis()
>  26    raiz = Tk()
>  27    F = Frame(raiz)
>  28    F.pack()
>  29    hello = Label(F, text=resultado)
>  30    hello.pack()
>  31    F.mainloop()
>  32
>  33 root = Tk()
>  34 root.title("$$$ Loteria $$$")
>  35 app = App(root)
>  36 root.mainloop()
>
> --
> Djames Suhanko
> LinuxUser 158.760

Also for style, you might want to group the import lines so they look
like this:

from Tkinter import *
import sys, random

A bit more pythonic. :P
--
http://mail.python.org/mailman/listinfo/python-list

Re: Tkinter - problem closing window

2009-01-05 Thread Collin D 
On Jan 5, 9:21 am, Roger  wrote:
> On Jan 5, 11:52 am, Collin D  wrote:
>
>
>
> > On Jan 5, 6:25 am, "Djames Suhanko"  wrote:
>
> > > Hello!
> > > I'm sorry my terrible english (my native language is portuguese).
> > > I has a litle program that open another window. When I close de root
> > > window in quit button, I need clicking 2 times to close. is where the
> > > problem?
>
> > > The source:
> > >   1 #!/usr/bin/env python
> > >   2 from Tkinter import *
> > >   3 import sys
> > >   4 import random
> > >   5 class App:
> > >   6  def __init__(self, master):
> > >   7    frame = Frame(master)
> > >   8    frame.pack()
> > >   9    rotulo = Label(frame, text="Clique em 'Gerar' e boa
> > > sorte!",borderwidth=2,bg="gray",justify=C    ENTER,relief=SUNKEN)
> > >  10    rotulo.pack()
> > >  11
> > >  12    self.button = Button(frame, text="Sair", fg="red",
> > > command=frame.quit,borderwidth=1)
> > >  13    self.button.pack(side=LEFT)
> > >  14    self.hi_there = Button(frame, text="Gerar Numero",
> > > command=self.say_hi,borderwidth=1)
> > >  15    self.hi_there.pack(side=RIGHT,padx=2,pady=2)
> > >  16
> > >  17  def gera_seis(self):
> > >  18    a = {}
> > >  19    for i in range(6):
> > >  20       a[i] = "%02d" %  int (random.randint(0,60))
> > >  21    resultadoA = "%s-%s-%s-%s-%s-%s" %
> > > (str(a[0]),str(a[1]),str(a[2]),str(a[3]),str(a[4]),str(a[5]))
> > >  22    return resultadoA
> > >  23
> > >  24  def say_hi(self):
> > >  25    resultado = self.gera_seis()
> > >  26    raiz = Tk()
> > >  27    F = Frame(raiz)
> > >  28    F.pack()
> > >  29    hello = Label(F, text=resultado)
> > >  30    hello.pack()
> > >  31    F.mainloop()
> > >  32
> > >  33 root = Tk()
> > >  34 root.title("$$$ Loteria $$$")
> > >  35 app = App(root)
> > >  36 root.mainloop()
>
> > > --
> > > Djames Suhanko
> > > LinuxUser 158.760
>
> > Also for style, you might want to group the import lines so they look
> > like this:
>
> > from Tkinter import *
> > import sys, random
>
> > A bit more pythonic. :P
>
> In that case you probably want to take out the 'from' import and:
>
> import Tkinter, sys, random
>
> in order to avoid any namespace issues especially if you have a large
> project with lots of gui manipulations.  But that's just me being
> pedantic. ;)

I agree... you could have conflicting functions.. not fun. XD
--
http://mail.python.org/mailman/listinfo/python-list

Command parsing... best module to use?

2009-11-02 Thread Collin D 
Hey everyone.

I am writing a game in python, and it includes a text console somewhat
like the one in WoW and Runescape. I want to be able to include "/"
commands, like IRC, and was wondering what the best module would be to
parse these.

Thanks a lot,
Collin D
-- 
http://mail.python.org/mailman/listinfo/python-list

Re: Command parsing... best module to use?

2009-11-03 Thread Collin D 
Thanks for the replies. Pyparsing looks just like what I need.

-- 
http://mail.python.org/mailman/listinfo/python-list