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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

Try it   is even   lambda x  True if x   2    0 else False print is even 10   print is even 11     Out   True False

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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  lt  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

User · Answer

Hope this will help a little you can resolve this problem in the following way f   lambda x   x  2     if f 3     print  quot do logic quot   else    print  quot another logic quot

User · Answer

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

User · Answer

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

User · Answer

the solution for the given scenerio is  f   lambda x   x if x    2 else print  quot number is not 2 quot   f 30     number is not 2 f 2     2

User · Answer

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

User · Answer

note you can use several else   if statements in your lambda definition   f   lambda x  1 if x gt 0 else 0 if x   0 else -1

User · Answer

This snippet should help you   x   lambda age   Older  if age  gt  30 else  Younger   print x 40

User · Answer

I think this is what you were looking for  gt  gt  gt  f   lambda x   print x  if x  2 else print  quot ERROR quot    gt  gt  gt  f 23  ERROR  gt  gt  gt  f 2  2  gt  gt  gt

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

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