latter.
Thank you for your help gentlemen!
Larry Wigton
On Thursday, September 27, 2012 8:54:19 AM UTC-7, Larry Wigton wrote:
>
> ***
> Python code:
>
> a= Symbol('a')
> b= Symbol('b')
> c= S
t from code using sympy 7.1 and also bleeding-edge sympy:
x= a*b + a*c
factor(x)= a*(b + c)
simplify(x)= a*(b + c)
x= 1.2*b + 1.2*c
factor(x)= 1.2*(b + c)
simplify(x)= 1.2*b + 1.2*c
simplify(factor(x))= 1.2*b + 1.2*c
*
why does simp
*pg01/pg00, pg01*(1.0 - pg01/pg00)]
[(x0, 1/pg00), (x1, pg01*x0)]
[x1**2, -2*x1 + 1.0, pg01*(-x1 + 1.0)]
I think sympy 7.1 result is better.
Larry Wigton
On Thursday, September 6, 2012 12:55:44 PM UTC-7, Larry Wigton wrote:
>
> Python code using cse from sympy:
>
> from sympy import
want to replace the tt
guys.
So for example if I replaced t1 by something I would clobber t11, t12 etc.
That is why I made my own iterator.
Larry Wigton
On Wednesday, September 26, 2012 10:42:21 AM UTC-7, smichr wrote:
>
> > class Myname:
> > def __init__(self):
> &
x27;str'
*
So bleeding edge code fails is this case when I try using my own
names for intermediate variables. I should point out this only happens
sometimes. Most of the time the bleeding edge code behaves properly.
I a
On Thursday, September 6, 2012 6:26:38 PM UTC-7, smichr wrote:
>
> On Fri, Sep 7, 2012 at 1:40 AM, Larry Wigton >
> wrote:
> > Python code using cse from sympy:
> >
> > from sympy import *
> > x=Symbol('x')
> > y=Symbol('y'
Python code using cse from sympy:
from sympy import *
x=Symbol('x')
y=Symbol('y')
eq1 = 5*x**3*y**2 + y**3
eq2 = 4*x**2*y**3 + y**2
eq = [eq1,eq2]
print eq
(red,rep) = cse(eq)
print red
print rep
eq = [eq2,eq1]
print eq
(red,rep) = cse(eq)
print red
print rep
*
