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pythonlambdaconditional-operator

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

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

This clearly isn't the syntax. Is it possible to perform an if in lambda and if so how to do it?

Solution

  • 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.