我怎样才能在 lambda ZC1C425268E68385D1AB5074F14C1

Question

Sorry if this is very lame, but I'm pretty new to Python.

As in Python everything is an object, I assume in every object the object itself can be get somehow. In object methods the self variable contains it. From the object reference the class object can be get (like type(self) ). But how this could be got inside a lambda?

I could figure out for a normal function:

import inspect

def named_func():
    func_name = inspect.stack()[0].function
    func_obj = inspect.stack()[1].frame.f_locals[func_name]
    print(func_name, func_obj, func_obj.xxx)

named_func.xxx = 15
named_func()

The output looks like this:

named_func <function named_func at 0x7f56e368c2f0> 15

Unfortunately in a lambda the inspect.stack()[0].function gives <lambda> inside the lambda.

Is there a way to get the function object inside a lambda? Is there a way to get function object directly (not using the name of the function)? I imagined __self__ , but it does not work.

UPDATE

I tried something like this in lambda:

lambda_func = lambda : inspect.stack()[0]
lambda_func.xxx = 2
print(lambda_func())

This prints:

FrameInfo(frame=<frame at 0x7f3eee8a6378, file './test_obj.py', line 74, code <lambda>>, filename='./test_obj.py', lineno=74, function='<lambda>', code_context=['lambda_func = lambda : inspect.stack()[0]\n'], index=0)

But for example is there a way to get the lambda object field xxx in this case? For this the lambda object should be got somehow.

Answer 1

So I think I've figured out how to do this in a very hacky way. I believe this will only work completely correctly if you follow the advice from the answer I found here

Basically, given the stack you can find the code object, and then using the gc module you can find the reference to the lambda.

Example with @Tomalak 's factorial lambda!

import gc
import inspect


def current_lambda():
    lambda_code = inspect.stack()[1].frame.f_code
    lambda_obj = gc.get_referrers(lambda_code)[0]
    return lambda_obj


print((lambda n: 1 if n < 2 else n * current_lambda()(n - 1))(5))

Outputs

120

Revisiting your example:

lambda_func = lambda: current_lambda().xxx
lambda_func.xxx = 10
print(lambda_func())

Outputs:

10

Answer 2

We can now use a new python syntax to make it shorter and easier to read, without the need to define a new function for this purpose.

You can find two examples below:

Fibonacci:

(f:=lambda x: 1 if x <= 1 else f(x - 1) + f(x - 2))(5)

Factorial:

(f:=lambda x: 1 if x == 0 else x*f(x - 1))(5)

We use := to name the lambda: use the name directly in the lambda itself and call it right away as an anonymous function.

So in your particular use-case it would give something like that:

print((f:=lambda: f.__hash__())())  # prints the hash for example

You can do whatever you want with that f variable now (inside the lambda).

But in fact, if you don't mind multi-lines for your code, you could also just use the name directly and do something like that:

f = lambda : f.xxx
f.xxx = 2
print(f())

(see https://www.python.org/dev/peps/pep-0572 for more information about this := operator)