To confirm, if you mean that it is free from any Symbol (free or bound) then `not expr.has(Symbol)` will be best. But if you consider `Integral(x, (x, 1, 2))` as a number then you should use `is_number` or `free_symbols`, with `expr.is_number` failing sooner than `not expr.free_symbols` if the expression has a free symbol. (So if you suspect the expression has free symbols then use `is_number`, else `free_symbols`).
`f.is_number != (not bool(f.free_symbols))` should be an invariant for Expr, but SymPy also deals with Booleans, so `S.true.is_number` is False and `S.true.free_symbols` is empty. /c On Sunday, September 12, 2021 at 11:56:23 PM UTC-5 [email protected] wrote: > Are there any cases when f.is_number != (not bool(f.free_symbols))? > > If I have an arbitrary expression, what is the correct way to check > whether it has variables? > > Thank you. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/9af46205-ce22-494c-a604-c27b6682fa96n%40googlegroups.com.
