Buggy python behavior related to for loop and higher-order functions [duplicate]

This question already has answers here: Local variables in nested functions (4 answers) Lambda in a loop [duplicate] (4 answers) Closed 1 hour ago.

I'd like to ask about the behavior of python.

I am using python 3.10.8. I wrote the following code:

from typing import Callable

def buggy_swap_function_args(fs: list[Callable[[str, str], str]]):
    swaped_fs: list[Callable[[str, str], str]] = []
    for f in fs:
      开发者_如何学Python  swaped_fs.append(lambda x, y: f(y, x))
    return swaped_fs

fs = [lambda x,y: "1" + x + y, lambda x,y: "2" + x + y, lambda x,y: "3" + x + y]
swapped_fs = buggy_swap_function_args(fs)

for swapped_f in swapped_fs:
    print(swapped_f("a", "b"))

In my understanding, this program should print as follows:

1ba
2ba
3ba

However, the above program printed as follows:

3ba
3ba
3ba

I re-wrote as follows and obtained the desired print:

def swap_function_args(fs: list[Callable[[str, str], str]]):
    swaped_fs: list[Callable[[str, str], str]] = []

    def g(f: Callable[[str, str], str]) -> Callable[[str, str], str]:
        def h(x: str, y: str) -> str:
            return f(y, x)

        return h

    for f in fs:
        swaped_fs.append(g(f))
    return swaped_fs

fs = [lambda x,y: "1" + x + y, lambda x,y: "2" + x + y, lambda x,y: "3" + x + y]
swapped_fs = swap_function_args(fs)

for swapped_f in swapped_fs:
    print(swapped_f("a", "b"))

I think the above two functions are completely the same. What's the difference between them? Is this a bug in python?