>
> it looks like in the first example that the expression is returned
> directly while in the second case it is not(?)
Ideally, for routines with one expr, `routine` and `routine[0]` will be
identical. Not sure if this is possible, and I may decide that's too much
magic. The main purpose behind all this is to make a more composable way of
combining underlying compiled code.
After playing around for a bit, it looks like this is going to be
complicated. Even combinatorial operations (add, mul, etc...) call methods
that refer to the underlying expression being computed. I'm not sure of the
best way to do it, but right now I'm thinking I'll make a list of
operations I want to be aliased to the underlying expression. Then I'll
redefine `__getattribute__`, and redirect requests to the expression if
they're in that list. That still may end up being too complicated, and I
may have to rethink my approach.
On Fri, Sep 19, 2014 at 9:00 AM, Chris Smith <smi...@gmail.com> wrote:
> The assumptions system will try to call `_eval_is_foo` if it exists for
> the object. So if you add definitions for ``_eval_is_integer` to your
> Routine object you can return the value of `is_integer` for the return
> value. I suspect there will be issues with how you define Routine, however,
> since it looks like in the first example that the expression is returned
> directly while in the second case it is not(?).
>
>
> On Thursday, September 18, 2014 10:32:01 PM UTC-5, James Crist wrote:
>>
>> I have a new SymPy type that serves to represent a unit of computation
>> (contains an expression/expressions that is/are being computed). I'd like
>> to be able to alias queries on the assumptions of the element to
>> assumptions of the underlying expression it represents. Example:
>>
>> >>> a, b, c = symbols('a, b, c', integer=True)
>> >>> expr = a + b + c
>>
>> # Create the tree element
>> >>> r = Routine((a, b, c), (expr))
>>
>> # r now represents a computational routine.
>> # We can use this as a function in other expressions.
>> >>> new_expr = 1 + 2 + r(a, b, c)
>> >>> new_expr
>> 1 + 2 + r(a, b, c)
>>
>> # The following should work
>> >>> new_expr.is_integer
>> True
>> >>> r(1, 2, 3).is_integer
>> True
>>
>> For `Routine` objects with multiple returns, the results are indexed to
>> select the output element:
>>
>> >>> exprs = (a + b + c, a*b*c + 4.1)
>> >>> r = Routine((a, b, c), exprs)
>>
>> # Use it in a new expression
>> >>> (1 + r(1, 2, 3)[0]).is_integer
>> True
>> >>> (1 + r(1, 2, 3)[1]).is_integer
>> False
>>
>> Any idea how to go about doing this? I have little to no understanding of
>> the assumption system, so before I start digging through the code I thought
>> I'd ask if anyone had thoughts on how to tackle this. The `Routine` type,
>> the `AppliedRoutine` type, and the results/arguments are all done. I just
>> need to figure out (if possible) how I can make this play well with the
>> assumption system.
>>
>> -Jim
>>
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