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?
thanks
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.