On Mon, Nov 10, 2008 at 10:15 PM, Brian Granger <[EMAIL PROTECTED]> wrote:
>
> OK, sounds great. Now a follow on....or two...
>
> * How can I get the non-commuting part of the Mul? I see that
> Mul.flatten knows about it, but I don't see how to get it back.
You need to do something like:
[p for p in a.args if not p.is_commutative]
where "a" is the (Mul) expression containing commutative and
non-commutative parts. If you want to get it as a Mul, do:
Mul(*[p for p in a.args if not p.is_commutative])
>
> * I then need to take an expression, find all the instances of Mul in
> the tree and replace those by something else, one by one. Is there a
> simple way of doing this? the most common usage case that I will have
> is a sum of terms;
>
> 3*A*B*|state1> + 4*C*D*|state>
The best thing is to loop through the expression, so something like:
from sympy.interactive import *
e = x**2+(y*x+2)**3*z + z*x
def change_mul(a):
return a**2
def walk(node):
args_new = []
for arg in node.args:
if arg.is_Mul:
args_new.append(change_mul(arg))
else:
args_new.append(walk(arg))
if len(args_new) > 0:
return node.__class__(*args_new)
else:
return node
print e
print walk(e)
This prints:
x*z + x**2 + z*(2 + x*y)**3
x**2 + x**2*z**2 + z**2*(2 + x*y)**6
i.e. each Mul is powered to 2. I agree that we should maybe even
simplify this --- if you (or anyone!) has any ideas how to make this
even simpler, let's implement it. Maybe even something like:
def walk(node):
args_new = []
for arg in node.args:
if arg.is_Mul:
args_new.append(change_mul(arg))
else:
args_new.append(walk(arg))
return node.__class__(*args_new)
This would require Symbols to have the actual name of the symbol in
.args(). But so far our policy was to only have Basic objects as args,
which imho is a good policy. Any thoughts are welcome.
Ondrej
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