In python, why does 0xbin() return False?

If you disassemble the code, you'll see that Yself's answer, which mentions that 0xbin() is interpreted as 0xb in (), is confirmed:

>>> import dis
>>> dis.dis('0xbin()')
  1           0 LOAD_CONST               0 (11)
              2 BUILD_TUPLE              0
              4 COMPARE_OP               6 (in)
              6 RETURN_VALUE

You can use Python's own tokenizer to check!

import tokenize
import io
line = b'0xbin()'
print(' '.join(token.string for token in tokenize.tokenize(io.BytesIO(line).readline) if token.type!=59))

This prints the tokens in your string, separated by spaces. In this case, the result will be:

0xb in ( ) 

In other words, it returns False because the number 11 (0xb) is not in the empty tuple (()).

(Thanks to Roman Odaisky for suggesting the use of tokenize in the comments!)

EDIT: To explain the code a bit more thoroughly: the tokenize function expects input in a bit of a weird format, so io.BytesIO(line).readline is a function that turns a sequence of bytes into something tokenize can read. tokenize then tokenizes it and returns a series of namedtuples; we take the string representing each one and join them together with spaces. The type != 59 part is used to ignore the encoding specifier that would otherwise show up at the beginning.

Python seems to interpret 0xbin() as 0xb in (), meaning is eleven in an empty tuple. The answer is no, therefore False.