Is there a way to perform "if" in python's lambda

The syntax you're looking for:

lambda x: True if x % 2 == 0 else False

But you can't use print or raise in a lambda.

why don't you just define a function?

def f(x):
    if x == 2:
        print(x)
    else:
        raise ValueError

there really is no justification to use lambda in this case.

Probably the worst python line I've written so far:

f = lambda x: sys.stdout.write(["2\n",][2*(x==2)-2])

If x == 2 you print,

if x != 2 you raise.

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