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


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?

Asked By: Guy



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
Answered By: David Wolever

why don’t you just define a function?

def f(x):
    if x == 2:
        raise ValueError

there really is no justification to use lambda in this case.

Answered By: SilentGhost

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.

Answered By: Robert Rossney

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.

Answered By: Glenn Maynard

Probably the worst python line I’ve written so far:

f = lambda x: sys.stdout.write(["2n",][2*(x==2)-2])

If x == 2 you print,

if x != 2 you raise.

Answered By: jimifiki

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
Answered By: filotn

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
Answered By: Juan Pablo Lopez

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

Answered By: Victor Lucas

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)))
Answered By: Rahul

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


This snippet should help you:

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

Answered By: Adarsh Somasundar

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
Answered By: Paul Cysne

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")
  print("another logic")
Answered By: Anurag Bhakuni

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

>>> f = lambda x: print(x) if x == 2 else print("ERROR")
>>> f(23)
>>> f(2)
Answered By: Nagesh Jayaram

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
Answered By: abhi

it might be worth considering np.where

Answered By: ZakS
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