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

## The Question :

391 people think this question is useful

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

• You can’t print or raise in a lambda. Lambdas are just functions, you can alwaya use a function instead.
• I disagree with you. I need 4 different, very short functions like the one above that need to be put in a list/dictionary so I can iterate over them and select which ones to use in each iteration. Instead of many lines of code of just inits, before the iteration, itself I can bring it down to only 4 lines of init code. The less the merrier..
• 4 lines of code is not a laudable solution when other people have to read, interpret, understand and maintain the code. Further, the “print/raise” problem in the example shows this which cannot and should not be done in lambdas.
• @LennartRegebro lambdas are not functions in python, they are only expressions, that is why there are many things you can not do with them.
• @AaronMcMillin Lambdas are functions. They are restricted to expressions for syntax reasons, but they ARE functions.

712 people think this answer is useful

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.

44 people think this answer is useful

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.

30 people think this answer is useful

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.

23 people think this answer is useful

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.

15 people think this answer is useful

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



15 people think this answer is useful

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



2 people think this answer is useful

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



2 people think this answer is useful

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

2 people think this answer is useful

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)



2 people think this answer is useful

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

print(x(40))



1 people think this answer is useful

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



0 people think this answer is useful

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)))



0 people think this answer is useful

Try it:

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



Out:

True
False



0 people think this answer is useful

Hope this will help a little

you can resolve this problem in the following way

f = lambda x:  x==2

if f(3):
print("do logic")
else:
print("another logic")



0 people think this answer is useful

I think this is what you were looking for

>>> f = lambda x : print(x) if x==2 else print("ERROR")
>>> f(23)
ERROR
>>> f(2)
2
>>>



f = lambda x : x if x == 2 else print("number is not 2")