How fast is this? Does it scale well for large expressions with many
variables? Can it be modified to return the mapping?
By the way, Dummy is a subclass of Symbol, so doing .atoms(Symbol,
Dummy) is redundant.
Aaron Meurer
On Sun, Mar 10, 2013 at 10:44 PM, smichr <smi...@gmail.com> wrote:
> If two expression trees are structurally the same they should unify under
> some mapping of symbols. usympy makes this very easy to test for:
>
>>>> def structurally_equal(a, b):
> ... from sympy.unify import usympy
> ... if isinstance(a, (list, tuple)):
> ... if type(a) != type(b): return False
> ... a = Tuple(*a)
> ... b = Tuple(*b)
> ... # disambiguate symbols that are in common
> ... bsym = b.atoms(Symbol, Dummy)
> ... common = a.atoms(Symbol, Dummy) & bsym
> ... if common:
> ... b = b.xreplace(dict(zip(common, [Dummy() for c in common])))
> ... try:
> ... next(usympy.unify(a, b, {}, variables = list(bsym)))
> ... return True
> ... except:
> ... return False
> ...
>>>> structurally_equal([1], [1])
> True
>>>> structurally_equal([1], (1,))
> False
>>>> structurally_equal(Sum(x, (x, 1, 2)), Sum(y, (y, 1, 2)))
> True
>>>> structurally_equal(Sum(x, (x, 1, 2)), Sum(y, (y, 1, 21)))
> False
>
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