ConditionSet object not iterable

[amine-aboufirass]

I would like to generate a simple set based on a condition to mimic set builder notation, and then enumerate its contents. I tried the following based on an example in the docs:

from sympy import Symbol, S, ConditionSet
from sympy.abc import x

new_set = ConditionSet(x, x<7, S.Naturals)
iterable = iter(new_set)
print("done")

When I run this I get an error:

Traceback (most recent call last):
  File "test.py", line 5, in <module>
    iterable = iter(new_set)
TypeError: 'ConditionSet' object is not iterable

Why is it that I cannot enumerate the ConditionSet object? It has finite contents so I would assume this should be possible?

Thanks

[akshanshbhatt]

ConditionSet can represent your set in set builder form. It is useful only when you have to check whether a particular element is present in your set. These sets are not stored in their roster form. Although this is one specific case in which the set is countably finite. I would suggest you use iterable sets such as FiniteSet for such purposes.

[amine-aboufirass]

But you cannot construct FiniteSet objects based on conditionals... Ultimately I would like to construct a set like A = {1, 4, 7, 10, 13...} and check whether it is closed under operations like addition and multiplication, or perhaps user-defined operations like x+y+x*y. Perhaps I am going about this the wrong way?

[oscarbenjamin]

It would be possible to add an __iter__ method to ConditionSet but in a case like this the iteration would never terminate. You can do it yourself like:

def iter_cs(cs):
    for x in cs.base_set:
        if cs.contains(x):
            yield x

Ideally there would be a way to simplify this ConditionSet into a FiniteSet.

[amine-aboufirass]

I see. That is a slick workaround but I could see it totally break down for cases where the call to .contains is expensive. It also appears to not know when to raise a StopIteration...

I tried simplify but still get another ConditionSet on the other end... Any idea why it is not simplifying to FiniteSet in the current version of sympy?

[oscarbenjamin]

That's not a slick workaround. That's what __iter__ for a ConditionSet would have to be if it was defined. The fact that contains is expensive is only part of the problem. It is sometimes impossible to know whether an element of the base set does or does not satisfy the condition e.g.:

In [11]: ConditionSet(x, x<y, {1, 2, 3})
Out[11]: {x │ x ∊ {1, 2, 3} ∧ (x < y)}

There are not many functions for simplifying sets yet.

[smichr]

I posted this comment at SO:

ConditionSet does not try hard to resolve its expression with an infinite base set. For linear or quadratic univariate inequalities, it wouldn't be hard to do so at instantiation, however. But for now you have to do so on your own:

>>> c = ConditionSet(x, x < 7, S.Naturals)
>>> solveset(c.args[1], c.args[0], domain=S.Reals).intersection(c.args[-1])
Range(1, 7, 1)
>>> next(iter(_))
1

In theory you should be able to set domain=c.args[-1] but the solver does not handle such sets well, yet.

[smichr]

does not handle such sets well, yet.

This is fixed in the most up to date master.