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

Question

In python 2.6, I want to do:

f = lambda x: if x==2 print x else raise Exception()
f(2) #should print \"2\"
f(3) #should throw an exception
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Answer 1

Following sample code works for me. Not sure if it directly relates to this question, but hope it helps in some other cases.

a = ''.join(map(lambda x: str(x*2) if x%2==0 else "", range(10)))