On 08/12/2011 08:14 PM, Chris Smith wrote:
My solution is as follows -
def reduce_sqr(expr):
args = expr.args
if len(args) == 0:
return(sqrt(expr))
else:
if str(type(expr)) == "<class 'sympy.core.power.Pow'>":
args = expr.args
if args[1] == 2:
e_mag = args[0]
else:
e_mag = sqrt(expr)
else:
e_mag = 1
for arg in args:
if str(type(arg)) == "<class
'sympy.core.power.Pow'>":
x = arg.args
if len(x) == 2 and x[1] == 2:
e_mag *= x[0]
else:
return(sqrt(expr))
return(e_mag)
There are a few extra properties/function that you can use to make life easier.
Mul.make_args will treat an expression like a Mul so you can then
process args in a single loop
If you are dealing with a Pow (.is_Pow) then you can access the base
with .base, the exp with .exp and get both with .as_base_exp()
def red_sqrt(s):
... out = []
... inn = []
... for m in Mul.make_args(s.base):
... if m.is_Pow:
... b, e = m.as_base_exp()
if e == 2:
... out.append(b)
... else:
... inn.append(m)
... return Mul(*out)*sqrt(Mul(*inn))
epath can be used to select all Pow at the top level of an expression, too:
eq
x + (x**(1/2)*y**4*cos(x)*cos(x*y)**2)**(1/2)
epath('/Pow', eq)
[(x**(1/2)*y**4*cos(x)*cos(x*y)**2)**(1/2)]
Notice that epath didn't pull out powers from inside the main sqrt and
it didn't return the x from eq
/c
Thank you.
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