Fix #764: wrong SymbolicOperator.isclose() result #1114
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Changes: 1. `isclose()` now accepts relative and absolute tolerances. This allows for more precise control over the comparison, just like in NumPy. 2. The comparison logic was corrected. The old implementation used a single tolerance value that it scaled in a way that could produce incorrect results. The new logic uses a common formula for checking if 1 numbers are close: ``` abs(a - b) <= abs_tol + rel_tol * max(abs(a), abs(b)) ``` 3. Updated `__eq__` in `SymbolicOperator` to use the tolerance values. 4. Added a new test `symbolic_operator_test.py`.
mpharrigan
reviewed
Aug 25, 2025
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| rel_tol(float): Relative tolerance. | ||
| abs_tol(float): Absolute tolerance. |
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consider rtol and atol for consistency with numpy/scipy
| return self.__class__(term=term, coefficient=coefficient) | ||
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| def isclose(self, other, tol=EQ_TOLERANCE): | ||
| def isclose(self, other, rel_tol=EQ_TOLERANCE, abs_tol=EQ_TOLERANCE): |
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this is technically a breaking API change. are we ok with that? weigh confusion vs. code compatibility to keeping one of them named tol
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Yeah, maybe we could keep 'tol' as a deprecated parameter and set atol=tol if it exists.
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gentle ping |
Per comments in PR review, this renames `rel_tol` and `abs_tol` to be `rtol` and `atol`, respectively, as well ask keeps the previous `tol` parameter but marks it as deprecated.
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Changes:
isclose()now accepts relative and absolute tolerances. This allows for more precise control over the comparison, just like in NumPy.__eq__inSymbolicOperatorto use the tolerance values.symbolic_operator_test.py.