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

This snippet should help you:

x = lambda age: 'Older' if age > 30 else 'Younger'

print(x(40))

Answer 2

You can also use Logical Operators to have something like a Conditional

func = lambda element: (expression and DoSomething) or DoSomethingIfExpressionIsFalse

You can see more about Logical Operators here

Answer 3

An easy way to perform an if in lambda is by using list comprehension.

You can't raise an exception in lambda, but this is a way in Python 3.x to do something close to your example:

f = lambda x: print(x) if x==2 else print("exception")

Another example:

return 1 if M otherwise 0

f = lambda x: 1 if x=="M" else 0

Answer 4

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.