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

Question

This question already has answers here:

Does Python have a ternary conditional operator? (32 answers)

Closed 2 years ago.

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?

• Lennart Regebro: You can't print or raise in a lambda. Lambdas are just functions, you can alwaya use a function instead.
• Guy: 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..
• S.Lott: 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.
• Aaron McMillin: @LennartRegebro lambdas are not functions in python, they are only expressions, that is why there are many things you can not do with them.
• Lennart Regebro: @AaronMcMillin Lambdas are functions. They are restricted to expressions for syntax reasons, but they ARE functions.

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.

• recursive: in python 3, you can use print
• Glenn Maynard: It's a horrible syntax--easily the worst Python language construct, approaching Perl levels of absurdity in its out-of-order evaluation--but it's what was asked for. You're seriously voting down answers for being correct?
• S.Lott: @Glenn Maynard: it's not the syntax that horrifying. It's any expression of the form True if expression else False. The if construct is totally redundant and therefore deeply and horrifyingly confusing. It's as bad as the statement form: if expression: return True.
• Robert Rossney: It's the correct answer to the question "How do I write a lambda function that tells me if a number is even?" It is not, however, a correct answer to the question that the OP originally asked. However much you don't like the example I contrived, my post does, in fact, clearly answer the OP's question.
• Glenn Maynard: It's painfully obvious that anyone suggesting "x%2==0"--or voting up a comment recommending it, which makes at least seven people--didn't even read the original question.

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.

• Lukáš Lalinský: print is not a function in 2.6 yet. :)
• newacct: @Lukáš Lalinský: it still works in 2.x. it will be treated as a pair of redundant parentheses
• Glenn Maynard: You don't know his actual use case, so there's no way you can say there's no reason to use a lambda.
• S.Lott: @Glenn Maynard: There's almost no reason to use a lambda, period. Assigning a lambda to a variable -- as a stand-in for def -- is generally a Very Bad Idea (tm). Just use a def so mere mortal programmers can read, interpret, understand and maintain it.
• Glenn Maynard: There are plenty of legitimate uses of lambdas. If you can't think of any, then that's not lambda's fault. (I'm not a fan of the syntax itself, of course--it's a clumsy workaround for the fact that Python's poorly-conceived indentation syntax can't handle inline functions like normal languages.)

Answer 3

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 4

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.

• Glenn Maynard: Possibly, but it depends heavily on the particular case. Of course, you can also decorate the lambda after creating it. x = RaiseValueErrorOnNone(x), again, depending on the case.

Answer 5

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 6

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

• Daniel Werner: lambda's specific restriction is that you are not allowed to use statements, only expressions.
• Coddy: Aren't lambdas meant to be short and concise else won't you write a separate function anyway?

Answer 7

This snippet should help you:

x = lambda age: 'Older' if age > 30 else 'Younger'
print(x(40))

Answer 8

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 9

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

• 0xc0de: It doesn't go with python's philosophy in terms of clarity. Even though logically equivalent, the if syntax is always preferred over this. The obvious way of checking conditions.
• Victor Lucas: Thank you! I used this in a work of functional language in college because of restrictions that the professor has imposed who says that I couldn't use if statement, so I found this not obvious way.

Answer 10

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

• wovano: The question was specifically about Python 2.6, not 3.x.

Answer 11

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)

• wovano: This gives SyntaxError: invalid syntax. I've tested this in Python 2.7 since I don't have Python 2.6 installed, but I really can't imagine this is valid in Python 2.6, since you can't use print in a lambda expression.

Answer 12

the solution for the given scenerio is:

f = lambda x : x if x == 2 else print("number is not 2")
f(30) # number is not 2
f(2) #2

• wovano: This is valid in Python 3.x, but the question was about Python 2.6. In Python 2.x this answer is not correct.

Answer 13

Here's the solution if you use Python 3.x!

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

• wovano: This is valid in Python 3.x, but the question was about Python 2.6. In Python 2.x this answer does not work, since you can't use print in a lambda expression.

Answer 14

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 15

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")

Answer 16

it might be worth considering np.where