What's the meaning of "(1,) == 1," in Python?

Question

This is just operator precedence. Your first

(1,) == 1,

groups like so:

((1,) == 1),

so builds a tuple with a single element from the result of comparing the one-element tuple 1, to the integer 1 for equality They're not equal, so you get the 1-tuple False, for a result.

Other answers have already shown you that the behaviour is due to operator precedence, as documented here.

I'm going to show you how to find the answer yourself next time you have a question similar to this. You can deconstruct how the expression parses using the ast module:

>>> import ast
>>> source_code = '(1,) == 1,'
>>> print(ast.dump(ast.parse(source_code), annotate_fields=False))
Module([Expr(Tuple([Compare(Tuple([Num(1)], Load()), [Eq()], [Num(1)])], Load()))])

From this we can see that the code gets parsed as Tim Peters explained:

Module([Expr(
    Tuple([
        Compare(
            Tuple([Num(1)], Load()), 
            [Eq()], 
            [Num(1)]
        )
    ], Load())
)])

When you do

>>> (1,) == 1,

it builds a tuple with the result from comparing the tuple (1,) with an integer and thus returning False.

Instead when you assign to variables, the two equal tuples are compared with each other.

You can try:

>>> x = 1,
>>> x
(1,)