Define a lambda expression that raises an Exception

Question

How can I write a lambda expression that\'s equivalent to:

def x():
    raise Exception()

The following is not allowed:

y =         


        

Answer 1

There is more than one way to skin a Python:

y = lambda: (_ for _ in ()).throw(Exception('foobar'))

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Lambdas accept statements. Since raise ex is a statement, you could write a general purpose raiser:

def raise_(ex):
    raise ex

y = lambda: raise_(Exception('foobar'))

But if your goal is to avoid a def, this obviously doesn't cut it. It does, however allow you to conditionally raise exceptions, e.g.:

y = lambda x: 2*x if x < 10 else raise_(Exception('foobar'))

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Alternatively you can raise an exception without defining a named function. All you need is a strong stomach (and 2.x for the given code):

type(lambda:0)(type((lambda:0).func_code)(
  1,1,1,67,'|\0\0\202\1\0',(),(),('x',),'','',1,''),{}
)(Exception())

And a python3 strong stomach solution:

type(lambda: 0)(type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b''),{}
)(Exception())

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Thanks @WarrenSpencer for pointing out a very simple answer if you don't care which exception is raised: y = lambda: 1/0.