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
<         


        

Answer 1

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 2

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 3

Try it:

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

Out:

True
False

Answer 4

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 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

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
>>>