How do you choose math.isclose rel_tol and abs_tol when comparing drifted totals?

## The exact rule `math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)` returns True when ```python abs(a - b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol) ``` Three details are worth reading off that line. It is **symmetric**, because it scales by the larger of the two magnitudes rather than by "the expected value" - swapping the arguments cannot change the verdict. The two tolerances are combined with `max`, so they are alternatives, not a sum: a pair passes if *either* the relative or the absolute test passes. And both are **keyword-only**, so `math.isclose(a, b, 0.01)` is a `TypeError`, not a silently different tolerance. ## What each tolerance is for `rel_tol` is a fraction of the numbers being compared. At the default 1e-09 you are allowing about a millionth of a percent of slack, which is generous next to the roughly 1.1e-16 relative error of a single float operation and tight enough to catch a real disagreement. Relative tolerance is what you want for values produced by arithmetic, because the absolute size of the error scales with the size of the numbers: an error of 0.001 is noise on a total of a million and catastrophic on a total of 0.002. `abs_tol` is a flat floor in the units of the quantity. Its default of 0.0 is the single most common trap in this API. Comparing a small value against exactly zero, the relative branch computes `rel_tol * max(abs(x), 0.0)`, which is a fraction of `x` and therefore always smaller than `x` itself. So `math.isclose(1e-300, 0.0)` is False, and every "is this effectively zero" check needs an explicit `abs_tol`. ## Choosing the numbers on real data Take an inventory sync where quantities move between two systems through a 17-service dependency graph, and each hop applies its own unit conversion in float. The two sides then report totals that differ in the last few digits - a pure floating-point rounding drift, not a lost record. Two questions set the tolerances. *How much drift can arithmetic alone explain?* Each rounding contributes at most about 1.1e-16 relative error, and a chain of a few dozen conversions and a summation over a few million rows keeps you many orders of magnitude below 1e-9. So the default `rel_tol` is already a loose bound: if the two totals differ by more than that, something other than rounding is wrong, which is exactly the alarm you want. Deliberately widening `rel_tol` to 1e-6 to "make the check pass" converts a data-integrity alarm into silence. *What difference would a human act on?* That sets `abs_tol`. For counts of physical units, a difference below half a unit cannot represent a real discrepancy, so `abs_tol=0.5` is defensible and also rescues the near-zero case where one side legitimately reports 0.0. Write the reasoning next to the call, because a bare `1e-6` in code is unreviewable. ## What isclose is not It is **not transitive**. `isclose(a, b)` and `isclose(b, c)` do not imply `isclose(a, c)`, so tolerance comparison cannot be used to group, dedupe or key values - there is no consistent equivalence class, and building a dict or a set on "close enough" produces order-dependent results. It is **not an exactness fix**. If the two totals are money or counts that must reconcile exactly, the answer is to change the representation - integer smallest units, or an exact base-ten decimal type - not to widen a tolerance until the mismatch stops being reported. It says nothing about **ordering**. `isclose` answers one question; if you need "a is meaningfully greater than b", write that comparison with the same tolerance explicitly. ## Special values The IEEE special values follow the standard rather than intuition. `math.isclose(float('nan'), float('nan'))` is False - nan is close to nothing, including itself - so a tolerance check on a pipeline that can produce nan silently reports "not close" instead of raising. Each infinity is close only to itself with the same sign, and `math.isclose(math.inf, -math.inf)` is False. Guard with `math.isnan()` or `math.isfinite()` before comparing if either side can be a special value. ## The interview point The interviewer is checking whether tolerances are chosen or copied. A strong answer states the formula, explains why `abs_tol` exists and why its 0.0 default breaks comparisons against zero, ties each number to a property of the data rather than to a habit, and knows when to stop tuning and change the representation instead.