This is a case where you know the magnitude of a is less than 1 so for all
the cases where you are working with constants in the integer range,
replaceing `a` with a symbolic `1/2` will do the work for you. (Even
stating that `a` is positive will work for several cases.)
```
>>> a = var('a',positive=True);eqs=Abs(a**2 + 2),Abs(2*a**2 - 1),Abs(a - 1)
>>> eqs
(a**2 + 2, Abs(2*a**2 - 1), Abs(a - 1))
>>> t = var('two',even=True,prime=True)
>>> [i.subs(a,1/two).subs(two,1/a) for i in eqs]
[a**2 + 2, 1 - 2*a**2, 1 - a]
>>> eqs = (Abs(a - 1),
... Abs(a - 2),
... Abs(2*a - 3),
... Abs(a**2 - 2),
... Abs(a**2 - 3),
... Abs(a**2 - 11),
... Abs(2*a**2 - 1),
... Abs(2*a**2 - 3),
... Abs(3*a**2 - 2),
... Abs(3*a**2 - 16),
... Abs(4*a**2 - 3),
... Abs(5*a**2 - 4),
... Abs(5*a**2 - 7),
... Abs(5*a**2 - 16))
>>> eqs
(Abs(a - 1), Abs(a - 2), Abs(2*a - 3), Abs(a**2 - 2), Abs(a**2 - 3),
Abs(a**2 -
11), Abs(2*a**2 - 1), Abs(2*a**2 - 3), Abs(3*a**2 - 2), Abs(3*a**2 - 16),
Abs(4*
a**2 - 3), Abs(5*a**2 - 4), Abs(5*a**2 - 7), Abs(5*a**2 - 16))
>>> [i.subs(a,1/two).subs(two,1/a) for i in eqs]
[1 - a, 2 - a, 3 - 2*a, 2 - a**2, 3 - a**2, 11 - a**2, 1 - 2*a**2, 3 -
2*a**2, 2
- 3*a**2, 16 - 3*a**2, 3 - 4*a**2, 4 - 5*a**2, 7 - 5*a**2, 16 - 5*a**2]
```
If you want to target `Abs` in an arbitrary expression, the using `replace`
would be a good way to go: `expr.replace(lambda x: isinstance(x, Abs),
lambda x: x.subs(a,1/two).subs(two,1/a)`
/c
On Tuesday, March 30, 2021 at 9:44:51 AM UTC-5 balle...@googlemail.com
wrote:
> Could anyone suggest a solution for this? I can make a list of
> substitutions by hand (as below) and pass that to the .subs() method, but
> surely there is a better way.
>
> s_list={
> Abs(a - 1) : S(1) - a,
> Abs(a - 2) : S(2) - a,
> Abs(2*a - 3) : S(3) - S(2)*a,
> Abs(a**2 - 2) : S(2) - a**2,
> Abs(a**2 - 3) : S(3) - a**2,
> Abs(a**2 - 11) : S(11) - a**2,
> Abs(2*a**2 - 1) : S(1) - S(2)*a**2,
> Abs(2*a**2 - 3) : S(3) - S(2)*a**2,
> Abs(3*a**2 - 2) : S(2) - S(3)*a**2,
> Abs(3*a**2 - 16) : S(16) - S(3)*a**2,
> Abs(4*a**2 - 3) : S(3) - S(4)*a**2,
> Abs(5*a**2 - 4) : S(4) - S(5)*a**2,
> Abs(5*a**2 - 7) : S(7) - S(5)*a**2,
> Abs(5*a**2 - 16) : S(16) - S(5)*a**2,
> ...
> }
>
> On Sunday, March 28, 2021 at 11:14:57 AM UTC+2 B A wrote:
>
>> I am a sympy beginner, and fairly new to python, so I suspect that my
>> question has a simple answer, but have not been able to figure it out
>> myself.
>>
>> I have sympy expressions containing the built-in Abs function. The
>> arguments of Abs() are polynomials in a=symbol('a',
>> real=True,positive=True) . Here are a three examples:
>>
>> Abs(a**2 + 2)
>> sqrt(2)*Abs(2*a**2 - 1)/3
>> sqrt(2)*(a + 1)*Abs(a - 1)/3
>>
>> Here's the point: I know that 'a' is approximately 0.6. So in cases
>> where the argument of Abs has an unambiguous sign (within floating-point
>> precision) I would like to simplify the absolute value. For example in the
>> three cases above I would like:
>>
>> Abs(a**2 + 2) -> a**2 + 2
>> sqrt(2)*Abs(2*a**2 - 1)/3 -> sqrt(2)*(1- 2*a**2)/3
>> sqrt(2)*(a + 1)*Abs(a - 1)/3 -> sqrt(2)*(a + 1)*(1-a)/3
>>
>> where the right arrow '->' indicates replacement or simplification. Note
>> the change of sign in the second and third examples, because (for example)
>> I know that (a-1) is negative.
>>
>> Is there a (simple) way to implement this?
>>
>> Thank you!
>> Bruce
>>
>
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