Q: atoms(S(1))

[Audrius-St]

Hello,

import sympy as sp

In the following example code, it is not clear to me why the first expression returns

expr.atoms(sp.S(1)) {1}

whereas the second expression returns

expr.atoms(sp.S(1)) set()

def main():

    x = sp.symbols('x')

    expr = x + 1

    print(f'expr.atoms(sp.S(1)) {expr.atoms(sp.S(1))}')

    expr = x - 1

    print(f'expr.atoms(sp.S(1)) {expr.atoms(sp.S(1))}')

if __name__ == "__main__":

    main()

[Aaron Meurer]

atoms() works at the expression tree level, meaning it will only
return something if it exactly appears in the expression. In the
example x - 1, the expression is Add(x, -1), which doesn't contain 1.
It instead contains -1. Note that SymPy doesn't have a subtraction
class. Subtraction is represented as addition of a negative, where the
negative is represented as multiplication by -1 (or in the case of
number, just the negative number).

Aaron Meurer

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[Audrius-St]

Understood. Thank you for your explanation.