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

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

thanks

Answer 1

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.

Answer 2

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.

Answer 3

You can easily raise an exception in a lambda, if that's what you really want to do.

def Raise(exception):
    raise exception
x = lambda y: 1 if y < 2 else Raise(ValueError("invalid value"))

Is this a good idea? My instinct in general is to leave the error reporting out of lambdas; let it have a value of None and raise the error in the caller. I don't think this is inherently evil, though--I consider the "y if x else z" syntax itself worse--just make sure you're not trying to stuff too much into a lambda body.

Answer 4

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.

Answer 5

Lambdas in Python are fairly restrictive with regard to what you're allowed to use. Specifically, you can't have any keywords (except for operators like and, not, or, etc) in their body.

So, there's no way you could use a lambda for your example (because you can't use raise), but if you're willing to concede on that… You could use:

f = lambda x: x == 2 and x or None

Answer 6

note you can use several else...if statements in your lambda definition:

f = lambda x: 1 if x>0 else 0 if x ==0 else -1

Answer 7

If you still want to print you can import future module

from __future__ import print_function

f = lambda x: print(x) if x%2 == 0 else False

Answer 8

what you need exactly is

def fun():
    raise Exception()
f = lambda x:print x if x==2 else fun()

now call the function the way you need

f(2)
f(3)

Answer 9

This snippet should help you:

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

print(x(40))

Answer 10

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 11

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)))

Answer 12

Try it:

is_even = lambda x: True if x % 2 == 0 else False
print(is_even(10))
print(is_even(11))

Out:

True
False