How to compare two Python ASTs, ignoring arguments?

Question

I want to elegantly compare two Python expressions ignoring any differences in the arguments. For example comparing plt.show() and plt.show(*some-args) should return True.

I've tried parsing the expressions using ast as follows:

import ast

node1 = ast.parse("plt.show()").body[0]
node2 = ast.parse("plt.show(*some-args)").body[0]
print(node1 == node2)

Returns: False

Answer 1

Some slight modifications to @Seanny123's answer:

from itertools import zip_longest
from typing import Union


def compare_ast(node1: Union[ast.expr, list[ast.expr]], node2: Union[ast.expr, list[ast.expr]], ignore_args=False) -> bool:
    """Compare two AST nodes for equality."""
    if type(node1) is not type(node2):
        return False

    if isinstance(node1, ast.AST):
        for k, v in vars(node1).items():
            if k in {"lineno", "end_lineno", "col_offset", "end_col_offset", "ctx"}:
                continue
            if ignore_args and k == "args":
                continue
            if not compare_ast(v, getattr(node2, k), ignore_args):
                return False
        return True

    elif isinstance(node1, list) and isinstance(node2, list):
        return all(compare_ast(n1, n2, ignore_args) for n1, n2 in zip_longest(node1, node2))
    else:
        return node1 == node2

Example:

import ast

node1 = ast.parse("plt.show()").body[0]
node2 = ast.parse("plt.show(*some-args)").body[0]
print(compare_ast(node1, node2, ignore_args=True))

Returns: True