I made a modification so it can take 1, 3, 4, ... number of operands
nicely.
def symround(x, ndigits=0):
if hasattr(x,'operator') and x.operator():
o = map( lambda y : symround(y,ndigits=ndigits) , x.operands() )
r = o[0]
for i in xrange(1,x.nops()):
r = x.operator(
Hi Burcin,
On 25 Okt., 15:26, Burcin Erocal wrote:
> This initializes a list with a single element for objects which return
> None for operator() now. IMHO, this approach is inefficient. In this
> case, you should act on the object directly.
I didnt claim that it was efficient - my code was inte
Hi Simon,
On Mon, 25 Oct 2010 06:09:16 -0700 (PDT)
Simon King wrote:
> On 25 Okt., 14:39, Burcin Erocal wrote:
> > If we return an identity operator for these cases, how do you plan
> > to test for it in your code:
>
> Something like this:
>
> L = x.operands()
> if len(L)>1:
> return x.op
PS:
I know that testing "is None" is faster than "len(L)>1" and wouldn't
insist that there *has* to be an identity operator. One has to
consider two different cases anyway.
However, if there *is* an operator, s==s.operator()(*s.operands())
should hold.
Cheers,
Simon
--
To post to this group, s
Hi Burcin!
On 25 Okt., 14:39, Burcin Erocal wrote:
> If we return an identity operator for these cases, how do you plan to
> test for it in your code:
Something like this:
L = x.operands()
if len(L)>1:
return x.operator()(*map(lambda ..., L))
else:
try:
return x.operator()(round
Hi Simon,
On Mon, 25 Oct 2010 05:09:06 -0700 (PDT)
Simon King wrote:
> I only opened one ticket, namely #10169, aiming at making
> s==s.operator()(*s.operands()) work uniformely.
I don't think this makes sense for variables, numeric objects, or
constants, in other words, objects for which .ope
Hi Burcin,
On 25 Okt., 11:04, Burcin Erocal wrote:
> I suggest we raise a ValueError when there is no operator or operands.
> This is already done for iterators of symbolic expressions in #7537:
>
> http://trac.sagemath.org/sage_trac/ticket/7537
>
> Can you open a ticket to do the same for operan
Hi Simon,
On Mon, 25 Oct 2010 00:41:07 -0700 (PDT)
Simon King wrote:
> @symbolic experts (Burcin et al):
> Is it really necessary that x.operator() returns None and x.operands()
> returns []? What about an identity operator?
The operator and operands don't have a meaning if you have a single
va
Hi Cristóvão!
On 25 Okt., 03:30, Cristóvão Sousa wrote:
> ...
> It just has a minor bug when the operator has more than two operands,
> like x+y+z, but I'll try to fix it as I got the picture now.
Yes, to my surprise, the "add" operator only accepts two arguments,
but the list of operands of a s
On Oct 24, 12:34 pm, Simon King wrote:
> I guess it would be needed to define a recursive function for that
> purpose, using operator and operands of a symbolic expression. Such
> as:
>
> sage: def symround(x, ndigits=0):
> : if hasattr(x,'operator') and x.operator():
> : retur
Hi Cristóvão!
On 24 Okt., 00:50, Cristóvão Sousa wrote:
> But, is there any way of round reals even if they are inside a symbolic
> expression?
I guess it would be needed to define a recursive function for that
purpose, using operator and operands of a symbolic expression. Such
as:
sage: def sy
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